- Find where the first derivative is zero. - What does the derivative do on either side of its only zero? - Compare the maximum y-value with \(0\).

1. Differentiate: \(f'(x)=2-e^{x+1}\). The only critical number satisfies \(e^{x+1}=2\), so \(x=\ln(2)-1\). 2. The second derivative is \(f''(x)=-e^{x+1}<0\) for every real \(x\), and \(f''(\ln(2)-1)=-2<0\). The second derivative test gives a local maximum. Because \(f'\) is strictly decreasing, it is positive before its only zero and negative after; the maximum is global. 3. The maximum value is \(f(\ln(2)-1)=2\ln(2)-4<0\). 4. Every function value is at most this negative number, so \(f\) has no real zeros.

Global maximum: \(\left(\ln(2)-1,2\ln(2)-4\right)\). Since \(2\ln(2)-4<0\), the function has no real zeros.

- Locate the graph's peaks and valleys. - Look for where the bending direction changes. - Count the x-axis crossings. - Use y-axis symmetry to check paired values.

1. The graph has three local extrema: local minima near \(x=-2.2\) and \(x=2.2\), and a local maximum at \(x=0\). 2. The graph changes concavity near \(x=-1.3\) and \(x=1.3\). Thus, the left inflection point lies between \(-2\) and \(-1\), and the right one lies between \(1\) and \(2\). 3. The graph crosses the x-axis four times on \([-4, 4]\), so four zeros are visible.

a) Three local extrema: two local minima and one local maximum b) Approximately \(x=\pm1.3\); between \(-2\) and \(-1\), and between \(1\) and \(2\) c) Four zeros

- Identify which graph approaches the x-axis for large positive \(x\). - Compare behavior as \(x\to-\infty\). - Use zeros and y-intercepts to distinguish the functions. - Differentiate when an extremum is a useful identifying feature.

1. For \(g_1(x)=4xe^{-x}\), the graph has a zero at \(x=0\), approaches \(0\) as \(x\to\infty\), and approaches \(-\infty\) as \(x\to-\infty\). Also, \(g_1'(x)=4(1-x)e^{-x}\), so it has a maximum at \(x=1\), with value \(4/e\approx1.47\). This matches Graph A. 2. For \(g_2(x)=e^x-x-1\), \(g_2(0)=0\) and \(g_2'(x)=e^x-1\), which changes from negative to positive at \(x=0\). Thus, \((0, 0)\) is a minimum. The function approaches \(\infty\) in both directions. This matches Graph B. 3. For \(g_3(x)=2-e^{0.5x}\), the graph approaches the horizontal asymptote \(y=2\) as \(x\to-\infty\), has y-intercept \(1\), and approaches \(-\infty\) as \(x\to\infty\). This matches Graph C.

Graph A: \(g_1(x)=4xe^{-x}\); Graph B: \(g_2(x)=e^x-x-1\); Graph C: \(g_3(x)=2-e^{0.5x}\)

- How could zeros and changes in direction help locate the features of the sketch? - What information does each derivative contribute to the graph description? - Which parts of the domain need special attention when naming graph features?

1. Factor \(f(x)=(x-3)(x+1)\), giving x-intercepts at \(x=-1\) and \(x=3\). 2. Differentiate: \(f'(x)=2x-2\), which vanishes at \(x=1\). Its sign is negative before \(1\) and positive after, so the vertex is the minimum \((1,-4)\). 3. The graph decreases on \((-1,1)\) and increases on \((1,4)\). Since \(f''(x)=2>0\), it is concave up throughout. 4. The endpoint values are \(f(-1)=0\) and \(f(4)=5\).

a) Vertex \((1,-4)\); x-intercepts \((-1,0)\) and \((3,0)\). b) \(f'(x)=2x-2\) is negative on \((-1,1)\) and positive on \((1,4)\), so the curve decreases and then increases. Since \(f''(x)=2>0\), it is concave up throughout. c) Left endpoint \((-1,0)\); right endpoint \((4,5)\).

- Find where the first derivative equals zero. - Use the sign of the second derivative to classify each critical point. - Compare the y-coordinates of the extrema with the x-axis. - Use the degree and leading coefficient to determine both ends of the graph.

1. The derivative is \(f'(x)=x(x-1)(x-2)\), giving critical numbers \(0,1,2\). The second derivative is \(f''(x)=3x^2-6x+2\). 2. Since \(f''(0)=2>0\), \((0,-4)\) is a local minimum. Since \(f''(1)=-1<0\), \((1,-\frac{15}{4})\) is a local maximum. Since \(f''(2)=2>0\), \((2,-4)\) is a local minimum. 3. The positive leading coefficient and even degree give \(f(x)\to\infty\) as \(x\to\pm\infty\). All three extrema have negative function values, so the graph has one zero to the left of \(0\), none between \(0\) and \(2\), and one to the right of \(2\). There are exactly two real zeros.

1. Local minima: \((0,-4)\) and \((2,-4)\). Local maximum: \((1,-\frac{15}{4})\). 2. Exactly \(2\) real zeros.

- Find the first and second derivatives. - Carefully evaluate the function at each critical number. - What must happen between an extremum below the x-axis and an extremum above it? - Determine the graph's end behavior from its degree and leading coefficient.

1. Differentiate: \(f'(x)=2x(x^2-2)\), giving critical numbers \(x=0\) and \(x=\pm\sqrt{2}\). 2. The second derivative is \(f''(x)=6x^2-4\). Since \(f''(0)=-4<0\), \((0,1.5)\) is a local maximum. Since \(f''(\pm\sqrt{2})=8>0\), \((-\sqrt{2},-0.5)\) and \((\sqrt{2},-0.5)\) are local minima. 3. Since the leading coefficient is positive and the degree is even, \(f(x)\to\infty\) as \(x\to\pm\infty\). The two minima are below the x-axis and the maximum between them is above it. Each of the four monotonic branches therefore crosses the x-axis once, giving exactly four real zeros.

Local maximum: \((0,1.5)\). Local minima: \((-\sqrt{2},-0.5)\) and \((\sqrt{2},-0.5)\). Exactly \(4\) real zeros.

- Use the first and second derivatives to locate and classify extrema. - Substitute an x-coordinate into the original function to find the corresponding y-coordinate. - The tangent slope equals the value of the first derivative at the point of tangency. - Use point-slope form for the tangent line.

1. Differentiate: \(f'(x)=x^2-2x-3\), \(f''(x)=2x-2\), and \(f'''(x)=2\). 2. Solve \(f'(x)=0\): \((x+1)(x-3)=0\), so the critical numbers are \(x=-1\) and \(x=3\). Since \(f''(-1)=-4<0\), \((-1, \frac{11}{3})\) is a local maximum. Since \(f''(3)=4>0\), \((3, -7)\) is a local minimum. 3. Solve \(f''(x)=0\): \(2x-2=0\), so \(x=1\). Because \(f'''(1)=2\ne0\), the graph has an inflection point. Since \(f(1)=-\frac{5}{3}\), the inflection point is \((1, -\frac{5}{3})\). Its slope is \(f'(1)=-4\). 4. Using point-slope form, \(y+\frac{5}{3}=-4(x-1)\), so the tangent line is \(y=-4x+\frac{7}{3}\).

a) Local maximum: \((-1, \frac{11}{3})\); local minimum: \((3, -7)\) b) Inflection point: \((1, -\frac{5}{3})\); slope: \(-4\) c) \(y=-4x+\frac{7}{3}\)

- Use the necessary and sufficient derivative tests for extrema and inflection points. - A tangent line requires both the point of tangency and the slope there. - To find an axis intercept, set the other coordinate equal to zero.

1. Differentiate: \(g'(x)=-3x^2+12x-9\), \(g''(x)=-6x+12\), and \(g'''(x)=-6\). 2. Solve \(g'(x)=0\): \(-3(x-1)(x-3)=0\), so \(x=1\) and \(x=3\). Since \(g''(1)=6>0\), \((1, 0)\) is a local minimum. Since \(g''(3)=-6<0\), \((3, 4)\) is a local maximum. 3. Solve \(g''(x)=0\): \(x=2\). Since \(g'''(2)=-6\ne0\), the graph has an inflection point at \((2, 2)\). 4. The slope at the inflection point is \(g'(2)=3\). Point-slope form gives \(y-2=3(x-2)\), or \(y=3x-4\). 5. Setting \(x=0\) gives the y-intercept \((0, -4)\). Setting \(y=0\) gives \(x=\frac{4}{3}\), so the x-intercept is \((\frac{4}{3}, 0)\).

a) Local minimum: \((1, 0)\); local maximum: \((3, 4)\); inflection point: \((2, 2)\) b) \(y=3x-4\) c) x-intercept: \((\frac{4}{3}, 0)\); y-intercept: \((0, -4)\)

- The beginning of the observation period corresponds to \(t=0\). - Use the sign of the first derivative to find where the water level decreases. - Solve \(h'(t)=0\) to locate interior candidates for extrema of the water level. - Check both critical values and endpoints for an absolute maximum. - To find the fastest decrease, minimize the first derivative on the time interval.

1. Evaluate the function: \(h(0)=200\) and \(h(5)=-125+375-240+200=210\). The levels are \(200\,\text{cm}\) and \(210\,\text{cm}\). 2. The derivative is \(h'(t)=-3t^2+30t-48=-3(t-2)(t-8)\). It is negative on \([0, 2)\) and \((8, 12]\). Therefore, the water level decreases from \(t=0\) to \(t=2\) and from \(t=8\) to \(t=12\). 3. Check the critical values and endpoints: \(h(0)=200\), \(h(2)=156\), \(h(8)=264\), and \(h(12)=56\). The absolute maximum is \(264\,\text{cm}\) at \(t=8\). 4. The fastest decrease occurs where \(h'(t)\) is smallest on \([0, 12]\). The derivative is a downward-opening parabola, so its minimum on the closed interval occurs at an endpoint. Since \(h'(0)=-48\) and \(h'(12)=-120\), the water level is decreasing most rapidly at \(t=12\).

a) \(200\,\text{cm}\) initially and \(210\,\text{cm}\) after \(5\) hours b) From \(t=0\) to \(t=2\), and from \(t=8\) to \(t=12\) c) After \(8\) hours; \(264\,\text{cm}\) d) At \(t=12\) hours

- Treat each formula as a separate graph-analysis problem before matching any properties. - Use the sign of the first derivative for monotonicity and local extrema. - A stationary inflection point needs both a horizontal tangent and a concavity change. - Use algebraic symmetry and factoring only for the symmetry and zero parts of the blueprint.

1. For \(f(x)=2x^2-x^4\), the graph is y-axis symmetric and the zeros are \(-\sqrt2,0,\sqrt2\). Since \(f'(x)=4x(1-x^2)\), \(f\) increases on \(( -\infty,-1)\) and \((0,1)\), and decreases on \((-1,0)\) and \((1,\infty)\). Thus the local maxima are \((-1,1)\) and \((1,1)\), and the local minimum is \((0,0)\). Its inflection points are not stationary. Therefore A and E apply. 2. For \(g(x)=x^3-3x\), the graph is origin-symmetric and the zeros are \(-\sqrt3,0,\sqrt3\). Since \(g'(x)=3(x^2-1)\), \(g\) increases on \(( -\infty,-1)\) and \((1,\infty)\), and decreases on \((-1,1)\). Thus \((-1,2)\) is a local maximum and \((1,-2)\) is a local minimum. Its inflection point at \(x=0\) is not stationary. Therefore B, C, and E apply. 3. Since \(h(x)=(x-1)^3\), its only zero is \(x=1\). The derivative \(h'(x)=3(x-1)^2\) is nonnegative and does not change sign, so \(h\) is strictly increasing on \(\mathbb{R}\) and has no local extremum. Because \(h''(x)=6(x-1)\) changes sign at \(x=1\) and \(h'(1)=0\), \((1,0)\) is a stationary inflection point. Therefore D applies.

(1) Blueprint: y-axis symmetry; zeros \(-\sqrt2,0,\sqrt2\); increasing on \(( -\infty,-1)\), \((0,1)\); decreasing on \((-1,0)\), \((1,\infty)\); local maxima \((-1,1)\), \((1,1)\); local minimum \((0,0)\); no stationary inflection point. Properties A, E. (2) Blueprint: origin symmetry; zeros \(-\sqrt3,0,\sqrt3\); increasing on \(( -\infty,-1)\), \((1,\infty)\); decreasing on \((-1,1)\); local maximum \((-1,2)\); local minimum \((1,-2)\); no stationary inflection point. Properties B, C, E. (3) Blueprint: zero \(1\); strictly increasing on \(\mathbb{R}\); no local extrema; stationary inflection point \((1,0)\). Property D.

- Factor or complete the square before differentiating when that reveals a useful graph feature. - A zero of the first derivative does not automatically create a local extremum; check the sign on both sides. - For a stationary inflection point, verify a concavity change at a horizontal tangent. - Assemble all of the features for one function before matching its statements.

1. For \(f(x)=x^4+1\), \(f'(x)=4x^3\) has the single zero \(x=0\). The function decreases on \(( -\infty,0)\), increases on \((0,\infty)\), and has a local minimum at \((0,1)\). It has no real zeros and no stationary inflection point. Thus only A applies. 2. Since \(g(x)=(x+1)^3\), \(g'(x)=3(x+1)^2\) has the single zero \(x=-1\) but does not change sign. The function is strictly increasing on \(\mathbb{R}\), has no local extremum, and \(g''(x)=6(x+1)\) changes sign at \(-1\). Thus \((-1,0)\) is a stationary inflection point. A, B, and C apply. 3. For \(h(x)=(x-1)^2+4\), \(h'(x)=2(x-1)\) has the single zero \(x=1\). The function decreases on \(( -\infty,1)\), increases on \((1,\infty)\), and has a local minimum at \((1,4)\). It has no real zeros and is symmetric about \(x=1\). Thus A and D apply.

(1) Blueprint: no real zeros; decreasing then increasing at local minimum \((0,1)\); no stationary inflection point. Statement A. (2) Blueprint: zero and stationary inflection point \((-1,0)\); strictly increasing on \(\mathbb{R}\); no local extremum. Statements A, B, C. (3) Blueprint: no real zeros; decreasing on \(( -\infty,1)\), increasing on \((1,\infty)\); local minimum \((1,4)\); symmetry line \(x=1\). Statements A, D.

- Locate the local maximum and local minima. - The number of solutions equals the number of intersections with the horizontal line \(y=a\). - Use the end behavior of the quartic. - Use the extrema and end behavior to compare all horizontal-line cases.

1. The derivative is \(f'(x)=x(x-2)(x+2)\), with critical numbers \(-2,0,2\). The second derivative is \(f''(x)=3x^2-4\). 2. Since \(f''(\pm2)=8>0\), the points \((-2,-1)\) and \((2,-1)\) are local minima. Since \(f''(0)=-4<0\), \((0,3)\) is a local maximum. Also, \(f(x)\to\infty\) as \(x\to\pm\infty\). 3. The number of solutions equals the number of intersections with the horizontal line \(y=a\). There are four when \(-1<a<3\), three when \(a=3\), two when \(a=-1\) or \(a>3\), none when \(a<-1\), and never exactly one.

Four solutions: \(-1<a<3\) Three solutions: \(a=3\) Two solutions: \(a=-1\) or \(a>3\) One solution: no values of \(a\) No solutions: \(a<-1\)

- A maximum concentration occurs at a critical value of the concentration function. - The first derivative is the concentration's rate of change. - “Increasing most rapidly” asks when the first derivative is greatest. - Use only times in the stated interval.

1. The first derivative is \(k'(t)=-3t^2+18t+48=-3(t-8)(t+2)\). The only critical value in \([0, 12]\) is \(t=8\). 2. Since \(k''(t)=-6t+18\) and \(k''(8)=-30<0\), \(t=8\) gives a local maximum. Comparing with the endpoints confirms the absolute maximum: \(k(0)=50\), \(k(8)=498\), and \(k(12)=194\). Thus, the maximum concentration is \(498\,\text{mg/L}\) after \(8\) hours. 3. The concentration increases most rapidly when \(k'(t)\) is largest. Solve \(k''(t)=0\): \(t=3\). Because \(k'''(t)=-6<0\), the first derivative has a maximum there. The rate is \(k'(3)=75\,\frac{\text{mg}}{\text{L}\cdot\text{h}}\).

a) After \(8\) hours; \(498\,\text{mg/L}\) b) After \(3\) hours; \(75\,\frac{\text{mg}}{\text{L}\cdot\text{h}}\)

- Compare the height at its critical times and both ends of the period. - What derivative tells you where the graph changes concavity? - Think about how changing concavity affects the growth rate represented by \(h'\).

1. \(h'(t)=-0.06t^2+1.2t=-0.06t(t-20)\). The only interior critical number is \(t=20\). Compare \(h(0)=20\), \(h(20)=100\), and \(h(25)=82.5\). The greatest modeled height occurs after \(20\) days. 2. \(h''(t)=-0.12t+1.2\) changes from positive to negative at \(t=10\). Thus \((10,60)\) is an inflection point of the height graph. 3. Since the first derivative \(h'\) increases before \(10\) and decreases after \(10\), growth is fastest at this inflection time. The rate is \(h'(10)=6\,\frac{\text{cm}}{\text{day}}\).

a) After \(20\) days. b) Inflection point: \((10,60)\). The growth rate is greatest there, after \(10\) days, at \(6\,\frac{\text{cm}}{\text{day}}\).

- Acceleration is the derivative of velocity. - Find the absolute maximum of velocity by checking critical values and endpoints. - At an inflection point of velocity, interpret the tangent slope as acceleration with the correct units.

1. The first two derivatives are \(v'(t)=-0.3t^2+3.6t\) and \(v''(t)=-0.6t+3.6\). 2. Solve \(v'(t)=0\): \(t=0\) or \(t=12\). Since \(v''(12)=-3.6<0\), \(t=12\) gives a local maximum. Comparing \(v(0)=5\), \(v(12)=91.4\), and \(v(15)=72.5\) confirms that the maximum velocity occurs after \(12\) seconds. 3. Solve \(v''(t)=0\): \(t=6\). Because \(v'''(t)=-0.6<0\), the acceleration \(v'(t)\) reaches its maximum there. The slope is \(v'(6)=10.8\,\frac{\text{m}}{\text{s}^2}\). Thus, after \(6\) seconds, the vehicle has its greatest acceleration.

a) After \(12\) seconds b) The inflection time is \(t=6\) seconds. The tangent slope is \(10.8\,\frac{\text{m}}{\text{s}^2}\), the vehicle's maximum acceleration.

- Look for the greatest common exponential factor. - Use the fact that \(e^{2x}=(e^x)^2\). - Set the first derivative equal to zero for extrema and the second derivative equal to zero for possible inflection points. - Verify each type with the next derivative.

1. Factor the function: \(f(x)=2e^x(3-e^x)\). Since \(e^x>0\), the zero satisfies \(e^x=3\), so \(x=\ln(3)\). 2. Differentiate: \(f'(x)=6e^x-4e^{2x}\), \(f''(x)=6e^x-8e^{2x}\), and \(f'''(x)=6e^x-16e^{2x}\). 3. For an extremum, solve \(f'(x)=2e^x(3-2e^x)=0\). Thus \(e^x=\frac{3}{2}\), so \(x=\ln\left(\frac{3}{2}\right)\). The second derivative there is \(-9<0\), so the point is a local maximum. Its y-coordinate is \(\frac{9}{2}\). 4. For an inflection point, solve \(f''(x)=2e^x(3-4e^x)=0\). Thus \(e^x=\frac{3}{4}\), so \(x=\ln\left(\frac{3}{4}\right)\). The third derivative there is \(-\frac{9}{2}\neq0\), confirming an inflection point. Its y-coordinate is \(\frac{27}{8}\).

a) \(f(x)=2e^x(3-e^x)\); zero at \(x=\ln(3)\). b) Local maximum: \(\left(\ln\left(\frac{3}{2}\right),\frac{9}{2}\right)\). Inflection point: \(\left(\ln\left(\frac{3}{4}\right),\frac{27}{8}\right)\).

- Treat the expression as a quadratic in \(e^x\). - Substitute \(u=e^x\), factor, and then substitute back. - Use the first derivative for extrema and the second derivative for possible inflection points. - Verify the classifications and evaluate the original function.

1. Let \(u=e^x\). Then \(g(x)=u^2-4u+3=(u-1)(u-3)\). Substituting back gives \(g(x)=(e^x-1)(e^x-3)\). 2. The zeros satisfy \(e^x=1\) or \(e^x=3\), so \(x=0\) and \(x=\ln(3)\). 3. Differentiate: \(g'(x)=2e^{2x}-4e^x\), \(g''(x)=4e^{2x}-4e^x\), and \(g'''(x)=8e^{2x}-4e^x\). 4. Solve \(g'(x)=2e^x(e^x-2)=0\). Thus \(x=\ln(2)\). Since \(g''(\ln(2))=8>0\), the point is a local minimum, and \(g(\ln(2))=-1\). 5. Solve \(g''(x)=4e^x(e^x-1)=0\). Thus \(x=0\). Since \(g'''(0)=4\neq0\), the point is an inflection point, and \(g(0)=0\).

a) \(g(x)=(e^x-1)(e^x-3)\); zeros at \(x=0\) and \(x=\ln(3)\). b) Local minimum: \((\ln(2),-1)\). Inflection point: \((0,0)\).

- Even powers suggest y-axis symmetry. - Factor out the greatest common factor to find the zeros. - Use the first and second derivatives to find and classify extrema. - The leading term determines the graph's end behavior.

1. Because \(f(-x)=f(x)\), the graph is symmetric about the y-axis. 2. Factoring gives \(f(x)=x^2(x-2)(x+2)\). The zeros are \(x=0\), with multiplicity \(2\), and \(x=\pm2\), each with multiplicity \(1\). 3. The derivative is \(f'(x)=4x(x^2-2)\), so the critical numbers are \(x=0\) and \(x=\pm\sqrt{2}\). The second derivative is \(f''(x)=12x^2-8\). Thus, \((0, 0)\) is a local maximum, and \((\pm\sqrt{2}, -4)\) are local minima. 4. The leading term is \(x^4\), so \(f(x)\to\infty\) as \(x\to\pm\infty\). The graph passes through the intercepts and extrema, is symmetric about the y-axis, and has both ends rising.

1. The graph is symmetric about the y-axis. 2. The zeros are \(x=0\), with multiplicity \(2\), and \(x=\pm2\), each with multiplicity \(1\). 3. Local maximum: \((0, 0)\); local minima: \((-\sqrt{2}, -4)\) and \((\sqrt{2}, -4)\) 4. \(f(x)\to\infty\) as \(x\to\pm\infty\).

- Substitute \(u=x^2\) to turn the quartic equation into a quadratic. - Remember both square roots when substituting back. - Use the first and second derivatives to find the extrema. - Use symmetry and end behavior to complete the graph description.

1. Set \(f(x)=0\) and multiply by \(-2\): \(x^4-8x^2+12=0\). Let \(u=x^2\). Then \(u^2-8u+12=(u-2)(u-6)=0\), so \(u=2\) or \(u=6\). Therefore, the x-intercepts occur at \(x=\pm\sqrt{2}\) and \(x=\pm\sqrt{6}\). 2. The derivative is \(f'(x)=-2x(x^2-4)\), so the critical numbers are \(x=-2\), \(x=0\), and \(x=2\). Since \(f''(x)=-6x^2+8\), \((0, -6)\) is a local minimum and \((\pm2, 2)\) are local maxima. 3. The graph is symmetric about the y-axis. It passes through the four x-intercepts, the local minimum, and the two local maxima. Since the leading coefficient is negative, both ends fall toward \(-\infty\).

1. The x-intercepts occur at \(x=\pm\sqrt{2}\) and \(x=\pm\sqrt{6}\). 2. Local minimum: \((0, -6)\); local maxima: \((-2, 2)\) and \((2, 2)\) 3. The graph is symmetric about the y-axis, passes through the listed intercepts and extrema, and satisfies \(f(x)\to-\infty\) as \(x\to\pm\infty\).

- Factor out the common exponential term. - Which exponential factor could equal zero? - Check the second derivative at the critical number. - Compare the growth of \(e^x\) and \(e^{2x}\) for large positive and negative inputs.

1. Factoring gives \(f(x)=e^x(4-e^x)\). Since \(e^x>0\), the zero is \(x=\ln(4)\). 2. The first derivative is \(f'(x)=2e^x(2-e^x)\). Its only critical number is \(x=\ln(2)\). 3. Since \(f''(x)=4e^x-4e^{2x}\), \(f''(\ln(2))=8-16=-8<0\), so the point is a local maximum. Its height is \(f(\ln(2))=4\). 4. As \(x\to-\infty\), both exponential terms approach zero, so \(f(x)\to0\). As \(x\to\infty\), the negative \(e^{2x}\) term dominates, so \(f(x)\to-\infty\).

a) \(f(x)=e^x(4-e^x)\); zero \(x=\ln(4)\). b) Local maximum: \((\ln(2),4)\). c) \(\lim_{x\to\infty}f(x)=-\infty\); \(\lim_{x\to-\infty}f(x)=0\).

- Substitute \(u=e^x\) to turn the zero equation into a quadratic. - Use the first derivative for the local extremum. - Use the second derivative for a possible inflection point. - Remember that \(e^x\) is never zero.

1. Let \(u=e^x\). The zero equation becomes \(u^2-5u+4=0\), or \((u-1)(u-4)=0\). Thus \(e^x=1\) or \(e^x=4\), giving \(x=0\) and \(x=\ln(4)\). 2. Differentiate: \(f'(x)=2e^{2x}-5e^x=e^x(2e^x-5)\). The critical number satisfies \(e^x=\frac{5}{2}\), so \(x=\ln\left(\frac{5}{2}\right)\). 3. The second derivative is \(f''(x)=4e^{2x}-5e^x=e^x(4e^x-5)\). At the critical number, \(f''(x)=\frac{25}{2}>0\), so the point is a local minimum. Its y-coordinate is \(-\frac{9}{4}\). 4. A possible inflection point satisfies \(f''(x)=0\), so \(e^x=\frac{5}{4}\) and \(x=\ln\left(\frac{5}{4}\right)\). The third derivative is \(f'''(x)=8e^{2x}-5e^x\), which equals \(\frac{25}{4}\neq0\) there. Therefore, the graph has an inflection point at that x-coordinate.

a) \(x=0\) and \(x=\ln(4)\). b) Local minimum: \(\left(\ln\left(\frac{5}{2}\right),-\frac{9}{4}\right)\). c) Inflection-point x-coordinate: \(x=\ln\left(\frac{5}{4}\right)\).

- Exponential expressions are always positive. - Compare \(f(-x)\) with \(f(x)\). - Apply the chain rule to both exponential terms. - The sign of the second derivative determines concavity.

1. Since \(f(0)=1+1=2\), the y-intercept is \((0,2)\). Both exponential terms are positive for every real \(x\), so their sum cannot be zero. Therefore, there are no x-intercepts. 2. \(f(-x)=e^{-0.2x}+e^{0.2x}=f(x)\), so \(f\) is even. As \(x\to\infty\), the first term grows without bound and the second approaches \(0\), so \(f(x)\to\infty\). By symmetry, \(f(x)\to\infty\) as \(x\to-\infty\). 3. The derivatives are \(f'(x)=0.2e^{0.2x}-0.2e^{-0.2x}\) and \(f''(x)=0.04e^{0.2x}+0.04e^{-0.2x}=0.04f(x)\). Since \(f(x)>0\), \(f''(x)>0\) for all \(x\), so the graph is concave up everywhere. 4. Set \(f'(x)=0\): \(e^{0.2x}=e^{-0.2x}\), so \(x=0\). Because the graph is concave up, \((0,2)\) is a local minimum.

a) Y-intercept: \((0,2)\). There are no x-intercepts because \(e^{0.2x}>0\) and \(e^{-0.2x}>0\). b) \(f\) is even. Also, \(\lim_{x\to\infty}f(x)=\infty\) and \(\lim_{x\to-\infty}f(x)=\infty\). c) \(f''(x)=0.04f(x)>0\), so the graph is concave up for all real \(x\). d) Local minimum: \((0,2)\).

- Compare \(f(-x)\) with \(f(x)\). - Use the product rule and chain rule when differentiating. - The exponential factor is always positive. - Use the sign of the second derivative to determine concavity.

1. Since \(f(-x)=4e^{-(-x)^2/2}=f(x)\), the function is even. Also, \(-x^2/2\to-\infty\) as \(x\to\pm\infty\), so \(f(x)\to0\). 2. The derivatives are \(f'(x)=-4xe^{-x^2/2}\) and \(f''(x)=(4x^2-4)e^{-x^2/2}\). The first derivative is zero only at \(x=0\), and \(f''(0)=-4<0\). Thus \((0,4)\) is a local maximum. 3. Since the exponential factor is positive, \(f''(x)=0\) when \(x^2-1=0\), so the inflection-point x-coordinates are \(x=-1\) and \(x=1\). The second derivative is positive for \(|x|>1\) and negative for \(|x|<1\). Therefore, the graph is concave up on \((-\infty,-1)\cup(1,\infty)\) and concave down on \((-1,1)\).

a) \(f\) is even, and \(\lim_{x\to\pm\infty}f(x)=0\). b) Local maximum: \((0,4)\). c) Inflection-point x-coordinates: \(x=-1\) and \(x=1\). The graph is concave up on \((-\infty,-1)\cup(1,\infty)\) and concave down on \((-1,1)\).

- Compare \(f(-x)\) with \(f(x)\) and \(-f(x)\). - Analyze the quadratic exponent and use the fact that \(e^u\) increases as \(u\) increases. - Inflection points occur where the second derivative changes sign. - Use point-slope form for the tangent line.

1. Since \(f(-x)=2e^{1-(-x)^2/2}=f(x)\), the function is even, so its graph is symmetric about the y-axis. It is not symmetric about the origin. As \(x\to\pm\infty\), the exponent approaches \(-\infty\), so \(f(x)\to0\). 2. The exponent \(1-x^2/2\) has its maximum value \(1\) at \(x=0\). Because the exponential function is strictly increasing, \(f\) is also largest at \(x=0\). Thus the maximum point is \((0,2e)\). 3. Differentiate: \(f'(x)=-2xe^{1-x^2/2}\), and \(f''(x)=(2x^2-2)e^{1-x^2/2}\). Since the exponential factor is positive, the second derivative is zero at \(x=\pm1\). It changes sign at both values, so the inflection points are \((-1,2\sqrt{e})\) and \((1,2\sqrt{e})\). 4. At \(x=1\), \(f(1)=2\sqrt{e}\) and \(f'(1)=-2\sqrt{e}\). Using point-slope form, \(y-2\sqrt{e}=-2\sqrt{e}(x-1)\), or \(y=-2\sqrt{e}x+4\sqrt{e}\).

a) The graph is symmetric about the y-axis, and \(\lim_{x\to\pm\infty}f(x)=0\). b) Maximum: \((0,2e)\). c) Inflection points: \((-1,2\sqrt{e})\) and \((1,2\sqrt{e})\). d) \(y=-2\sqrt{e}x+4\sqrt{e}\).

- Factor the exponential terms after differentiating. - Use the sign change of the first derivative for the extremum. - Use the sign of the second derivative on the intervals separated by its zero. - Compare the two exact x-coordinates only after finding both points.

1. Differentiate and factor: \(f'(x)=4e^x-2e^{2x}=2e^x(2-e^x)\). 2. Since \(e^x>0\), the critical number satisfies \(e^x=2\), so \(x=\ln2\). The derivative changes from positive to negative there, giving a local maximum. Its value is \(f(\ln2)=4\). 3. \(f''(x)=4e^x-4e^{2x}=4e^x(1-e^x)\). It is positive for \(x<0\) and negative for \(x>0\), so \(f\) is concave up on \(( -\infty,0)\) and concave down on \((0,\infty)\). The inflection point is \((0,3)\). 4. Since \(0<\ln2\), the inflection point occurs to the left of the local maximum.

a) \(f'(x)=2e^x(2-e^x)\). b) Local maximum: \((\ln2,4)\). c) Concave up on \(( -\infty,0)\), concave down on \((0,\infty)\); inflection point \((0,3)\). d) The inflection point occurs first because \(0<\ln2\).

- Identify which factor can equal zero. - Compare linear growth with exponential growth or decay. - Use the product and chain rules. - An inflection point occurs where the second derivative changes sign. - Use the point and slope to write the tangent line.

1. Since the exponential factor is never zero, \(f(x)=0\) only when \(x-2=0\). Thus the zero is \(x=2\). 2. As \(x\to\infty\), exponential decay dominates the linear factor, so \(f(x)\to0\). As \(x\to-\infty\), \(x-2\to-\infty\) and \(e^{-0.5x}\to\infty\), so \(f(x)\to-\infty\). 3. The first derivative is \(f'(x)=(2-0.5x)e^{-0.5x}\). It is zero at \(x=4\). The second derivative is \(f''(x)=(0.25x-1.5)e^{-0.5x}\), and \(f''(4)=-0.5e^{-2}<0\). Therefore, the local maximum is \((4,2e^{-2})\). 4. The second derivative is zero only at \(x=6\), and its linear factor changes sign there. Thus the unique inflection point is \((6,4e^{-3})\). The slope there is \(f'(6)=-e^{-3}\). Using point-slope form gives \(y-4e^{-3}=-e^{-3}(x-6)\), or \(y=-e^{-3}x+10e^{-3}\).

a) \(x=2\). b) \(\lim_{x\to\infty}f(x)=0\), and \(\lim_{x\to-\infty}f(x)=-\infty\). c) Local maximum: \((4,2e^{-2})\). d) Inflection point: \((6,4e^{-3})\). Tangent line: \(y=-e^{-3}x+10e^{-3}\).

- An exponential factor is never zero. - Use the product and chain rules. - A simple zero of the second derivative produces a sign change here. - Compare linear growth with exponential growth or decay.

1. Since the exponential factor is never zero, \(f(x)=0\) only when \(2x=0\). Thus the x-intercept is \((0,0)\). 2. Differentiate: \(f'(x)=(2-2x)e^{1-x}\). The derivative is zero at \(x=1\). The second derivative is \(f''(x)=(2x-4)e^{1-x}\), and \(f''(1)=-2<0\). Therefore, the local maximum is \((1,2)\). 3. The second derivative is zero only at \(x=2\). Its linear factor changes sign there, so this is the unique inflection point. Since \(f(2)=4e^{-1}\), the point is \(\left(2,\frac{4}{e}\right)\). 4. As \(x\to\infty\), exponential decay dominates linear growth, so \(f(x)\to0\). As \(x\to-\infty\), the linear factor is negative and unbounded while \(e^{1-x}\to\infty\), so \(f(x)\to-\infty\).

a) \((0,0)\). b) Local maximum: \((1,2)\). c) Inflection point: \(\left(2,\frac{4}{e}\right)\). d) \(\lim_{x\to\infty}f(x)=0\), and \(\lim_{x\to-\infty}f(x)=-\infty\).

- Find where the denominator is zero and use the numerator discriminant to check for x-intercepts. - Rewrite the numerator using \((x-2)^2\). - Use the sign of the second derivative to determine concavity. - An inflection point must be a point on the graph where concavity changes.

1. The denominator is zero at \(x=2\), so the domain is \(\mathbb{R}\setminus\{2\}\). 2. The numerator has discriminant \((-4)^2-4\cdot 1\cdot 7=-12<0\), so it has no real zeros. Therefore, the graph has no x-intercepts. 3. Since \((x-2)^2+3=x^2-4x+7\), \(f(x)=x-2+\frac{3}{x-2}\). 4. The vertical asymptote is \(x=2\), and the slant asymptote is \(y=x-2\). 5. For \(x<2\), \((x-2)^3<0\), so \(f''(x)<0\), and the graph is concave down. For \(x>2\), \(f''(x)>0\), and the graph is concave up. 6. Although concavity differs across \(x=2\), that value is not in the domain. Also, \(f''(x)\) is never zero. Therefore, the graph has no inflection points.

a) Domain: \(\mathbb{R}\setminus\{2\}\); no x-intercepts b) Vertical asymptote: \(x=2\); slant asymptote: \(y=x-2\) c) Concave down on \((-\infty, 2)\), concave up on \((2, \infty)\), and no inflection points

- Check whether the denominator can equal zero. - Compare \(f(-x)\) with \(f(x)\), and compare the leading terms for end behavior. - Use the sign of the first derivative to classify the critical point. - Set the second derivative equal to zero and verify a sign change.

1. Since \(x^2+2>0\) for every real \(x\), the domain is \(\mathbb{R}\). Setting the numerator equal to zero gives \(8-2x^2=0\), so the x-intercepts are \((-2, 0)\) and \((2, 0)\). 2. Since \(f(-x)=f(x)\), the graph is symmetric about the y-axis. The numerator and denominator have the same degree, so \(\lim_{x\to\pm\infty}f(x)=-2\). 3. The first derivative is \(f'(x)=\frac{-24x}{(x^2+2)^2}\). It is positive for \(x<0\) and negative for \(x>0\), so the graph has a local maximum at \((0, 4)\). 4. The second derivative is \(f''(x)=\frac{72x^2-48}{(x^2+2)^3}=\frac{24(3x^2-2)}{(x^2+2)^3}\). It is zero when \(x=\pm\sqrt{\frac{2}{3}}=\pm\frac{\sqrt{6}}{3}\), and its sign changes at both values. 5. Substitution gives \(f\left(\pm\frac{\sqrt{6}}{3}\right)=\frac{5}{2}\). Therefore, the inflection points are \(\left(-\frac{\sqrt{6}}{3}, \frac{5}{2}\right)\) and \(\left(\frac{\sqrt{6}}{3}, \frac{5}{2}\right)\).

a) Domain: \(\mathbb{R}\); x-intercepts: \((-2, 0)\) and \((2, 0)\) b) Symmetric about the y-axis; \(\lim_{x\to\pm\infty}f(x)=-2\) c) Local maximum at \((0, 4)\) d) Inflection points: \(\left(-\frac{\sqrt{6}}{3}, \frac{5}{2}\right)\) and \(\left(\frac{\sqrt{6}}{3}, \frac{5}{2}\right)\)

- Compare \(h(-x)\) with \(h(x)\). - Use the first derivative to find and classify extrema. - Factor the numerator of the second derivative. - Verify that the second derivative changes sign at each candidate.

1. Since \(h(-x)=-h(x)\), the graph is symmetric about the origin. Because the denominator's degree is greater than the numerator's degree, \(\lim_{x\to\pm\infty}h(x)=0\), so the horizontal asymptote is \(y=0\). 2. The first derivative is \(h'(x)=\frac{3-x^2}{(x^2+3)^2}\). It is zero at \(x=\pm\sqrt{3}\). 3. The derivative changes from negative to positive at \(x=-\sqrt{3}\), giving a local minimum at \(\left(-\sqrt{3}, -\frac{\sqrt{3}}{6}\right)\). It changes from positive to negative at \(x=\sqrt{3}\), giving a local maximum at \(\left(\sqrt{3}, \frac{\sqrt{3}}{6}\right)\). 4. The second derivative is \(h''(x)=\frac{2x^3-18x}{(x^2+3)^3}=\frac{2x(x-3)(x+3)}{(x^2+3)^3}\). 5. The denominator is always positive, and the numerator has three distinct zeros at \(x=-3\), \(x=0\), and \(x=3\). Its sign changes at each zero, so all three values give inflection points. 6. Evaluating \(h\) gives the points \(\left(-3, -\frac{1}{4}\right)\), \((0, 0)\), and \(\left(3, \frac{1}{4}\right)\).

a) Symmetric about the origin; horizontal asymptote: \(y=0\) b) Local minimum at \(\left(-\sqrt{3}, -\frac{\sqrt{3}}{6}\right)\); local maximum at \(\left(\sqrt{3}, \frac{\sqrt{3}}{6}\right)\) c) Inflection points: \(\left(-3, -\frac{1}{4}\right)\), \((0, 0)\), and \(\left(3, \frac{1}{4}\right)\)

- Evaluate the model at \(t=0\) and examine the exponential term as \(t\) becomes large. - Half the population corresponds to \(N(t)=1000\). - Differentiate using the chain rule. - A maximum growth rate corresponds to a change in concavity of the original function.

1. \(N(0)=\frac{2000}{40}=50\), so \(50\) people initially know the message. Since \(e^{-0.4t}\to 0\), \(\lim_{t\to\infty}N(t)=2000\), meaning the model approaches the city's full population. 2. Set \(N(t_0)=1000\). Then \(1+39e^{-0.4t_0}=2\), so \(e^{0.4t_0}=39\). Therefore, \(t_0=\frac{\ln 39}{0.4}\approx 9.16\) hours. 3. Differentiate: \(N'(t)=\frac{31{,}200e^{-0.4t}}{(1+39e^{-0.4t})^2}\). Thus, \(N'(0)=\frac{31{,}200}{1600}=19.5\) people per hour. 4. To verify the graph feature directly, differentiate again: \(N''(t)=\frac{12{,}480e^{-0.4t}(39e^{-0.4t}-1)}{(1+39e^{-0.4t})^3}\). At \(t=t_0\), \(39e^{-0.4t_0}=1\), so \(N''(t_0)=0\). For \(t<t_0\), \(N''(t)>0\), and for \(t>t_0\), \(N''(t)<0\). Thus the graph changes from concave up to concave down at \(t_0\), so \(t_0\) is an inflection point.

a) \(N(0)=50\); \(\lim_{t\to\infty}N(t)=2000\) b) \(t_0=\frac{\ln 39}{0.4}\approx 9.16\,\text{hours}\) c) \(N'(0)=19.5\) people per hour d) An inflection point, because the first derivative has a maximum there

- Use the signs of the first and second derivatives. - Analyze the intervals separated by the excluded domain value. - A vertical asymptote occurs when the denominator approaches zero while the numerator remains nonzero. - Check the denominators of the derivative formulas.

1. The derivatives are \(f'(x)=\frac{-3}{(x+1)^2}\) and \(f''(x)=\frac{6}{(x+1)^3}\). 2. Since \((x+1)^2>0\) on the domain, \(f'(x)<0\). Therefore, \(f\) is strictly decreasing on \((-\infty,-1)\) and on \((-1,\infty)\). 3. For \(x<-1\), \((x+1)^3<0\), so \(f''(x)<0\) and the graph is concave down. For \(x>-1\), \(f''(x)>0\) and the graph is concave up. 4. The denominator of \(f\) is zero at \(x=-1\), while the numerator is \(3\neq0\), so \(x=-1\) is a vertical asymptote. The derivatives have nonzero constant numerators and denominators containing powers of \(x+1\), so their magnitudes also become unbounded as \(x\to-1\). Thus \(x=-1\) is the common vertical asymptote.

1. Strictly decreasing on \((-\infty,-1)\) and \((-1,\infty)\). Concave down on \((-\infty,-1)\); concave up on \((-1,\infty)\). 2. The graphs of \(f\), \(f'\), and \(f''\) all have vertical asymptote \(x=-1\).

- A product is zero when at least one factor is zero. - Use the product and chain rules. - Show that the first derivative has only one zero by analyzing the second derivative. - Examine the logarithm near the left endpoint of the domain.

1. The product is zero when \(x-2=0\) or \(\ln(x-1)=0\). Both conditions give \(x=2\), so this is the only zero. 2. Differentiate using the product rule: \(g'(x)=\ln(x-1)+\frac{x-2}{x-1}\). At \(x=2\), both terms are zero, so \(x=2\) is a critical number. 3. The second derivative is \(g''(x)=\frac{1}{x-1}+\frac{1}{(x-1)^2}>0\) for every \(x>1\). Therefore, \(g'\) is strictly increasing and has no other zero. Since \(g''(2)=2>0\), the second derivative test confirms that \((2,0)\) is a local minimum. 4. As \(x\to1^+\), \(x-2\to-1\) and \(\ln(x-1)\to-\infty\), so \(g(x)\to\infty\). Thus \(x=1\) is a vertical asymptote. As \(x\to\infty\), \(g(x)\to\infty\). Since \(g(x)/x\to\infty\), there is no horizontal or slant asymptote.

a) Zero: \(x=2\). Local minimum: \((2,0)\). b) \(\lim_{x\to1^+}g(x)=\infty\) and \(\lim_{x\to\infty}g(x)=\infty\). Vertical asymptote: \(x=1\). There are no horizontal or slant asymptotes.

- The logarithm determines the domain. - Compare quadratic and logarithmic growth as \(x\to\infty\). - Set the first derivative equal to zero. - Use the second derivative to classify the extremum.

1. Since \(\ln(x)\) requires \(x>0\), the domain is \((0,\infty)\). 2. As \(x\to0^+\), \(-2\ln(x)\to\infty\), so \(f(x)\to\infty\). As \(x\to\infty\), the quadratic term dominates the logarithmic term, so \(f(x)\to\infty\). 3. The derivatives are \(f'(x)=x-\frac{2}{x}\) and \(f''(x)=1+\frac{2}{x^2}\). 4. Set \(f'(x)=0\): \(x^2=2\). The domain gives \(x=\sqrt{2}\). 5. Since \(f''(\sqrt{2})=2>0\), the second derivative test gives a local minimum at \(\left(\sqrt{2},1-\ln(2)\right)\).

a) Domain: \((0,\infty)\). Also, \(\lim_{x\to0^+}f(x)=\infty\) and \(\lim_{x\to\infty}f(x)=\infty\). b) Local minimum: \((\sqrt{2},1-\ln(2))\).

- The logarithm determines the domain. - Compare logarithmic and power growth. - Apply the quotient rule carefully. - Use the second derivative to classify the critical point.

1. The domain is \((0,\infty)\). As \(x\to0^+\), \(\ln(x)\to-\infty\) while \(x^2\to0^+\), so \(g(x)\to-\infty\). As \(x\to\infty\), the power in the denominator grows faster than the logarithm, so \(g(x)\to0\). 2. The quotient rule gives \(g'(x)=\frac{1-2\ln(x)}{x^3}\). The second derivative is \(g''(x)=\frac{6\ln(x)-5}{x^4}\). 3. Set \(g'(x)=0\): \(\ln(x)=\frac{1}{2}\), so \(x=\sqrt{e}\). 4. Since \(g''(\sqrt{e})=-\frac{2}{e^2}<0\), the second derivative test gives a local maximum at \(\left(\sqrt{e},\frac{1}{2e}\right)\).

a) Domain: \((0,\infty)\). Also, \(\lim_{x\to0^+}g(x)=-\infty\) and \(\lim_{x\to\infty}g(x)=0\). b) Local maximum: \(\left(\sqrt{e},\frac{1}{2e}\right)\).

- A horizontal tangent has slope zero. - The argument of a logarithm must be positive. - Determine what happens when the logarithm's argument approaches \(0^+\). - Use the second derivative to classify the critical point.

1. Differentiate using the chain rule: \(f_k'(x)=\frac{k-2x}{kx-x^2}\). 2. A horizontal tangent at \(x=2\) requires \(f_k'(2)=0\). Therefore, \(k-4=0\), so \(k=4\). 3. For \(k=4\), the logarithm requires \(4x-x^2=x(4-x)>0\). Thus the maximal domain is \((0, 4)\). 4. As \(x\to0^+\) or \(x\to4^-\), the logarithm's argument approaches \(0^+\), so the function approaches \(-\infty\). The vertical asymptotes are \(x=0\) and \(x=4\). 5. For \(k=4\), \(f_4'(2)=0\) and \(f_4''(2)=-\frac{1}{2}<0\). The second derivative test gives a local maximum.

a) \(k=4\). b) Domain: \((0, 4)\); vertical asymptotes: \(x=0\) and \(x=4\). c) The point is a local maximum.

- Find the zeros of the first derivative. - Check critical values and endpoints for the maximum. - Use the sign of the first derivative for increasing and decreasing intervals. - The second derivative describes how the growth rate changes.

1. The derivatives are \(f'(t)=-3t(t-8)\) and \(f''(t)=-6t+24\). The critical values in the interval are \(t=0\) and \(t=8\). 2. Since \(f''(8)=-24<0\), the population has a local maximum at \(t=8\). Comparing the endpoints gives \(f(0)=60\), \(f(8)=316\), and \(f(12)=60\), so the maximum population is \(316\) insects after \(8\) weeks. 3. The first derivative is positive for \(0<t<8\) and negative for \(8<t<12\). Therefore, the population increases on \([0, 8]\) and decreases on \([8, 12]\). 4. Solving \(f''(t)=0\) gives \(t=4\). At this inflection time, the growth rate \(f'(t)\) reaches its maximum. Thus, the population is increasing most rapidly after \(4\) weeks.

a) After \(8\) weeks b) \(316\) insects c) Increasing on \([0, 8]\); decreasing on \([8, 12]\) d) At \(t=4\), the population is increasing most rapidly.

- Factor the first derivative to find all critical numbers. - What do the values of the second derivative at those numbers tell you? - Look for a change in the sign of the second derivative to find the inflection point.

1. Differentiate: \(f'(x)=3(x-1)(x-3)\), so the critical numbers are \(x=1\) and \(x=3\). 2. Since \(f''(x)=6x-12\), \(f''(1)=-6<0\) gives a local maximum at \((1,4)\), while \(f''(3)=6>0\) gives a local minimum at \((3,0)\). 3. The second derivative is negative for \(x<2\) and positive for \(x>2\), so the concavity changes at \(x=2\). Since \(f(2)=2\), the inflection point is \((2,2)\).

a) Critical numbers: \(x=1\) and \(x=3\). Local maximum: \((1,4)\); local minimum: \((3,0)\). b) Inflection point: \((2,2)\). The second derivative changes from negative to positive.

- Factor the first derivative to find the critical numbers. - Evaluate the second derivative at each critical number. - Can factoring the original function show that it is never negative? - Where does the second derivative change sign?

1. The derivative is \(f'(x)=4x(x-1)(x-2)\), so the critical numbers are \(x=0\), \(x=1\), and \(x=2\). 2. The second derivative is \(f''(x)=12x^2-24x+8\). Since \(f''(0)=8>0\) and \(f''(2)=8>0\), \((0,0)\) and \((2,0)\) are local minima. Since \(f''(1)=-4<0\), \((1,1)\) is a local maximum. 3. Factor: \(f(x)=x^2(x-2)^2\geq0\), and both minima have value \(0\). Thus both are also absolute minima. 4. Solving \(f''(x)=0\) gives \(x=1\pm\frac{\sqrt{3}}{3}\). The second derivative changes sign across each value, and the function value there is \(\frac{4}{9}\). Hence the inflection points are \(\left(1-\frac{\sqrt{3}}{3},\frac{4}{9}\right)\) and \(\left(1+\frac{\sqrt{3}}{3},\frac{4}{9}\right)\).

a) Critical numbers: \(x=0,1,2\). Local minima: \((0,0)\) and \((2,0)\); local maximum: \((1,1)\). b) Yes. \(f(x)=x^2(x-2)^2\geq0\), and each local minimum has value \(0\). c) Inflection points: \(\left(1-\frac{\sqrt{3}}{3},\frac{4}{9}\right)\) and \(\left(1+\frac{\sqrt{3}}{3},\frac{4}{9}\right)\).

- How does replacing \(x\) with \(-x\) change this polynomial? - Set the first derivative equal to zero and evaluate the second derivative at each solution. - Compare the coordinates of the maximum and minimum. - Does the second derivative change sign at the center of symmetry?

1. Since \(g(-x)=-g(x)\), the graph is symmetric about the origin. 2. Differentiate: \(g'(x)=\frac{3}{4}x^2-3\). The critical numbers are \(x=-2\) and \(x=2\). 3. Since \(g''(x)=\frac{3}{2}x\), \(g''(-2)=-3<0\) gives a local maximum at \((-2,4)\), while \(g''(2)=3>0\) gives a local minimum at \((2,-4)\). The extrema are images of each other under a half-turn about the origin. 4. The second derivative changes from negative to positive at \(x=0\), so \((0,0)\) is an inflection point.

a) \(g(-x)=-g(x)\), so the graph is symmetric about the origin. b) Critical numbers: \(x=-2\), \(x=2\). Local maximum: \((-2,4)\); local minimum: \((2,-4)\). They are symmetric about the origin. c) Inflection point: \((0,0)\).

- Does the expression change when \(x\) is replaced by \(-x\)? - Find the zeros of the first derivative. - Evaluate the second derivative at the critical numbers. - How does the negative leading term affect the graph far to the left and right?

1. Replacing \(x\) by \(-x\) does not change \(f(x)\), so \(f\) is even and the graph is symmetric about the y-axis. 2. The derivative is \(f'(x)=2x(3-x^2)\), so the critical numbers are \(x=0\) and \(x=\pm\sqrt{3}\). 3. Since \(f''(x)=6-6x^2\), \(f''(0)=6>0\) gives a local minimum at \((0,0)\). Also, \(f''(\pm\sqrt{3})=-12<0\) gives local maxima at \(\left(\pm\sqrt{3},\frac{9}{2}\right)\). 4. The function tends to \(-\infty\) at both ends of the real line. Its only critical values are \(0\) and \(\frac{9}{2}\), so the two local maxima are also absolute maxima.

a) \(f(-x)=f(x)\), so the graph is symmetric about the y-axis. b) Critical numbers: \(x=0\), \(x=\pm\sqrt{3}\). Local minimum: \((0,0)\); local maxima: \(\left(\pm\sqrt{3},\frac{9}{2}\right)\). c) Yes. The two maxima both have value \(\frac{9}{2}\), which is the greatest function value.

- Read zeros and multiplicities from the factored form. - Differentiate and solve for the critical numbers. - Evaluate the second derivative at each critical number. - How does a local maximum with height zero behave near the x-axis?

1. The zeros are \(x=-2\), with multiplicity \(2\), and \(x=1\), with multiplicity \(1\). 2. Expanding gives \(f(x)=x^3+3x^2-4\), so \(f'(x)=3x(x+2)\). The critical numbers are \(x=-2\) and \(x=0\). 3. Since \(f''(x)=6x+6\), \(f''(-2)=-6<0\) gives a local maximum at \((-2,0)\), and \(f''(0)=6>0\) gives a local minimum at \((0,-4)\). 4. Because the function has a local maximum of \(0\) at \(x=-2\), it touches the x-axis there and stays below it on both sides near that point.

a) Zeros: \(x=-2\) (multiplicity \(2\)) and \(x=1\) (multiplicity \(1\)). The even-multiplicity zero is \(x=-2\). b) Critical numbers: \(x=-2\) and \(x=0\). Local maximum: \((-2,0)\); local minimum: \((0,-4)\). c) The graph touches the x-axis at \((-2,0)\) rather than crossing it.

- Factor the first derivative to find the critical numbers. - Evaluate the second derivative at each critical number. - Where does the second derivative change sign? - Compare the midpoint of the two extrema with the inflection point.

1. Differentiate: \(f'(x)=x^2-2x-3=(x-3)(x+1)\). The critical numbers are \(x=-1\) and \(x=3\). 2. Since \(f''(x)=2x-2\), \(f''(-1)=-4<0\) gives the local maximum \(\left(-1,\frac{16}{3}\right)\), and \(f''(3)=4>0\) gives the local minimum \(\left(3,-\frac{16}{3}\right)\). 3. The second derivative changes from negative to positive at \(x=1\), and \(f(1)=0\). Thus \((1,0)\) is the inflection point. 4. The midpoint of the two extrema is \(\left(\frac{-1+3}{2},\frac{16/3-16/3}{2}\right)=(1,0)\). A half-turn about the inflection point swaps the extrema.

a) Local maximum: \(\left(-1,\frac{16}{3}\right)\); local minimum: \(\left(3,-\frac{16}{3}\right)\). b) Inflection point: \((1,0)\). c) The extrema are symmetric about the inflection point: it is their midpoint.

- Find the critical numbers from the factored first derivative. - Evaluate the second derivative at each one. - Can the function be written as the negative of a square? - At which values does the second derivative change sign?

1. Differentiate: \(g'(x)=-\frac{1}{2}x(x^2-4)\). The critical numbers are \(x=-2\), \(x=0\), and \(x=2\). 2. Since \(g''(x)=-\frac{3}{2}x^2+2\), \(g''(0)=2>0\) gives a local minimum at \((0,-2)\), while \(g''(\pm2)=-4<0\) gives local maxima at \((-2,0)\) and \((2,0)\). 3. Since \(g(x)=-\frac{1}{8}(x^2-4)^2\leq0\) for all \(x\), both maxima of height \(0\) are absolute maxima. 4. Solving \(g''(x)=0\) gives \(x=\pm\frac{2}{\sqrt{3}}\). The second derivative changes sign at both values, and \(g\left(\pm\frac{2}{\sqrt{3}}\right)=-\frac{8}{9}\). The inflection points are \(\left(-\frac{2}{\sqrt{3}},-\frac{8}{9}\right)\) and \(\left(\frac{2}{\sqrt{3}},-\frac{8}{9}\right)\).

a) Critical numbers: \(x=-2,0,2\). Local minimum: \((0,-2)\); local maxima: \((-2,0)\), \((2,0)\). b) Both maxima are absolute because \(g(x)=-\frac{1}{8}(x^2-4)^2\leq0\). c) Inflection points: \(\left(-\frac{2}{\sqrt{3}},-\frac{8}{9}\right)\), \(\left(\frac{2}{\sqrt{3}},-\frac{8}{9}\right)\).

- How many zeros can a cubic or quadratic polynomial have? - Which derivatives locate possible extrema and inflection points? - Check the second derivative at each critical number. - Does the second derivative change sign at both of its zeros?

1. The first derivative of a fourth-degree polynomial has degree \(3\), so there can be at most \(3\) distinct local extrema. Its second derivative has degree \(2\), so there can be at most \(2\) distinct inflection points. 2. Here, \(f'(x)=\frac{1}{2}x(x^2-4)\), with critical numbers \(x=-2,0,2\). Since \(f''(x)=\frac{3}{2}x^2-2\), \(f''(\pm2)=4>0\) gives local minima at \(x=-2,2\), and \(f''(0)=-2<0\) gives a local maximum at \(x=0\). There are \(3\) extrema. 3. Solving \(f''(x)=0\) gives \(x=\pm\frac{2}{\sqrt{3}}\). The second derivative changes sign at each, so both are inflection-point x-coordinates. There are \(2\) inflection points. This function reaches both upper bounds.

a) At most \(3\) local extrema and \(2\) inflection points. b) Local minima at \(x=-2\) and \(x=2\); local maximum at \(x=0\). Yes, the function has \(3\) extrema. c) Inflection-point x-coordinates: \(x=-\frac{2}{\sqrt{3}}\) and \(x=\frac{2}{\sqrt{3}}\). Yes, there are \(2\).

- Factor the first derivative to find the critical numbers. - Use the second derivative at each one to classify the extrema. - What degree is the second derivative of a cubic? - Does a nonconstant linear function always change sign at its zero?

1. The derivative is \(f'(x)=x^2-2x-3=(x-3)(x+1)\), so the critical numbers are \(x=-1\) and \(x=3\). 2. Since \(f''(x)=2x-2\), \(f''(-1)=-4<0\) gives a local maximum at \(\left(-1,\frac{32}{3}\right)\), and \(f''(3)=4>0\) gives a local minimum at \((3,0)\). 3. The second derivative changes from negative to positive at \(x=1\), and \(f(1)=\frac{16}{3}\). Thus the inflection point is \(\left(1,\frac{16}{3}\right)\). 4. Every polynomial of degree exactly \(3\) has a nonconstant linear second derivative. It has exactly one real zero, with a sign change there, so every cubic has exactly one inflection point.

a) Critical numbers: \(x=-1\) and \(x=3\). Local maximum: \(\left(-1,\frac{32}{3}\right)\); local minimum: \((3,0)\). b) Inflection point: \(\left(1,\frac{16}{3}\right)\). A cubic's second derivative is a nonconstant linear function, so it has exactly one sign-changing zero.

- Differentiate and find the critical numbers. - Evaluate the second derivative at each one. - Can the polynomial be written as a square? - Where does the second derivative change sign?

1. The derivative is \(f'(x)=4x(x^2-9)\), so the critical numbers are \(x=-3,0,3\). 2. Since \(f''(x)=12x^2-36\), \(f''(0)=-36<0\) gives a local maximum at \((0,81)\), and \(f''(\pm3)=72>0\) gives local minima at \((-3,0)\) and \((3,0)\). 3. Since \(f(x)=(x^2-9)^2\geq0\) everywhere, both local minima are also absolute minima. 4. The second derivative changes sign at \(x=\pm\sqrt{3}\); there \(f(x)=36\). Thus the inflection points are \((-\sqrt{3},36)\) and \((\sqrt{3},36)\).

a) Critical numbers: \(x=-3,0,3\). Local maximum: \((0,81)\); local minima: \((-3,0)\) and \((3,0)\). b) Yes. \(f(x)=(x^2-9)^2\geq0\), and both minima equal \(0\). c) Inflection points: \((-\sqrt{3},36)\) and \((\sqrt{3},36)\).

- Find the critical numbers from the first derivative. - Evaluate the second derivative at those numbers. - What happens when \(x\) is replaced by \(-x\)? - Where does the second derivative change sign?

1. The derivative is \(f'(x)=x(x^2-5)\), so the critical numbers are \(x=0\) and \(x=\pm\sqrt{5}\). 2. Since \(f''(x)=3x^2-5\), \(f''(0)=-5<0\) gives a local maximum at \((0,4)\), while \(f''(\pm\sqrt{5})=10>0\) gives local minima at \(\left(-\sqrt{5},-\frac{9}{4}\right)\) and \(\left(\sqrt{5},-\frac{9}{4}\right)\). 3. Because \(f(-x)=f(x)\), the graph is symmetric about the y-axis. The two minima have opposite x-coordinates and equal heights, and the maximum lies on the symmetry axis. 4. The second derivative changes sign at \(x=\pm\sqrt{\frac{5}{3}}\), where \(f(x)=\frac{19}{36}\). Thus the inflection points are \(\left(-\sqrt{\frac{5}{3}},\frac{19}{36}\right)\) and \(\left(\sqrt{\frac{5}{3}},\frac{19}{36}\right)\).

a) Critical numbers: \(x=0\) and \(x=\pm\sqrt{5}\). Local maximum: \((0,4)\); local minima: \(\left(-\sqrt{5},-\frac{9}{4}\right)\) and \(\left(\sqrt{5},-\frac{9}{4}\right)\). b) The graph is symmetric about the y-axis; the minima are mirror images and the maximum lies on the axis. c) Inflection points: \(\left(-\sqrt{\frac{5}{3}},\frac{19}{36}\right)\) and \(\left(\sqrt{\frac{5}{3}},\frac{19}{36}\right)\).

- Factor the first derivative to locate the critical numbers. - Evaluate the second derivative at each critical number. - Compare the extremum heights with the x-axis. - What does continuity guarantee between function values of opposite signs?

1. The derivative is \(f'(x)=x^2-x-2=(x-2)(x+1)\). The critical numbers are \(x=-1\) and \(x=2\). 2. Since \(f''(x)=2x-1\), \(f''(-1)=-3<0\) gives a local maximum at \(\left(-1,\frac{13}{6}\right)\), and \(f''(2)=3>0\) gives a local minimum at \(\left(2,-\frac{7}{3}\right)\). 3. The cubic has a positive leading coefficient, so \(f(x)\to-\infty\) as \(x\to-\infty\) and \(f(x)\to\infty\) as \(x\to\infty\). The local maximum is above the x-axis and the local minimum below it. 4. By continuity, there is at least one zero in each of \((-\infty,-1)\), \((-1,2)\), and \((2,\infty)\). A cubic has at most three real zeros, so there is exactly one in each interval.

a) Local maximum: \(\left(-1,\frac{13}{6}\right)\); local minimum: \(\left(2,-\frac{7}{3}\right)\). b) The maximum is positive, the minimum is negative, and the end behavior has opposite signs. Continuity gives a zero in each of \((-\infty,-1)\), \((-1,2)\), and \((2,\infty)\), so there are exactly three distinct real zeros.

- Check whether the denominator can be zero and compare \(f(-x)\) with \(f(x)\). - Compare the numerator and denominator degrees for end behavior. - Use the first derivative for extrema and the second derivative for inflection points. - Verify sign changes at all derivative zeros.

1. Since \(x^2+4>0\) for every real \(x\), the domain is \(\mathbb{R}\). 2. Since \(f(-x)=-f(x)\), the graph is symmetric about the origin. The x- and y-intercepts are both \((0, 0)\). 3. The denominator has greater degree than the numerator, so \(\lim_{x\to\pm\infty}f(x)=0\). The horizontal asymptote is \(y=0\). 4. The first derivative is \(f'(x)=\frac{16-4x^2}{(x^2+4)^2}\). It is zero at \(x=\pm 2\). The derivative changes from negative to positive at \(x=-2\), giving a local minimum at \((-2, -1)\), and from positive to negative at \(x=2\), giving a local maximum at \((2, 1)\). 5. The second derivative is \(f''(x)=\frac{8x(x^2-12)}{(x^2+4)^3}\). Its zeros are \(x=-2\sqrt{3}\), \(x=0\), and \(x=2\sqrt{3}\), and its sign changes at each one. 6. The inflection points are \(\left(-2\sqrt{3}, -\frac{\sqrt{3}}{2}\right)\), \((0, 0)\), and \(\left(2\sqrt{3}, \frac{\sqrt{3}}{2}\right)\).

Domain: \(\mathbb{R}\); symmetric about the origin; intercept: \((0, 0)\); horizontal asymptote: \(y=0\); local minimum: \((-2, -1)\); local maximum: \((2, 1)\); inflection points: \(\left(-2\sqrt{3}, -\frac{\sqrt{3}}{2}\right)\), \((0, 0)\), and \(\left(2\sqrt{3}, \frac{\sqrt{3}}{2}\right)\)

- Check whether the denominator can be zero and compare \(f(-x)\) with \(f(x)\). - Use the numerator for x-intercepts and leading coefficients for the horizontal asymptote. - Evaluate the second derivative at the intercept x-values and verify sign changes. - Use the first and second derivatives to classify the critical point.

1. Since \(x^2+3>0\) for every real \(x\), the domain is \(\mathbb{R}\). Since \(f(-x)=f(x)\), the graph is symmetric about the y-axis. 2. The numerator is zero at \(x=\pm 1\), so the x-intercepts are \((-1, 0)\) and \((1, 0)\). Equal numerator and denominator degrees give the horizontal asymptote \(y=1\). 3. The first derivative is \(f'(x)=\frac{8x}{(x^2+3)^2}\), and the second derivative is \(f''(x)=\frac{24(1-x^2)}{(x^2+3)^3}\). 4. The second derivative is zero at \(x=\pm 1\). Its denominator is positive, and the numerator changes sign at both values, so \((-1, 0)\) and \((1, 0)\) are inflection points. 5. The first derivative is zero only at \(x=0\). Since \(f''(0)>0\), the graph has a local minimum at \(\left(0, -\frac{1}{3}\right)\).

1) Domain: \(\mathbb{R}\); symmetric about the y-axis 2) x-intercepts: \((-1, 0)\) and \((1, 0)\); horizontal asymptote: \(y=1\) 3) Both x-intercepts are inflection points. 4) Local minimum: \(\left(0, -\frac{1}{3}\right)\)

- Treat the square root as a power and apply the chain rule. - Simplify the second derivative before examining its sign. - What must happen to concavity for an inflection point to exist? - Look at the sign of the first derivative on either side of a critical point.

1. By the chain rule, \(f'(x)=\frac{x}{\sqrt{x^2+5}}\). 2. Differentiate again: \(f''(x)=\frac{5}{(x^2+5)^{3/2}}\). 3. Because the denominator is positive for every real \(x\), \(f''(x)>0\) everywhere. The graph is concave up on \(( -\infty,\infty )\) and is never concave down. 4. Since the concavity never changes, there are no inflection points. 5. The derivative \(f'(x)\) is negative when \(x<0\), zero at \(x=0\), and positive when \(x>0\). Thus the graph decreases and then increases, so it has a local (and absolute) minimum at \((0,\sqrt{5})\).

a) \(f'(x)=\frac{x}{\sqrt{x^2+5}}\), \(f''(x)=\frac{5}{(x^2+5)^{3/2}}\) b) Concave up for all real \(x\); nowhere concave down c) No inflection points; the second derivative is always positive. d) Local and absolute minimum at \((0,\sqrt{5})\), with \(f'\) changing from negative to positive.

- Rewrite everything in powers of \(2^x\) before solving the zero and critical-number equations. - The signs of the first and second derivatives determine the monotonicity and concavity pieces of the blueprint. - Use the limiting behavior to decide how the far-left and far-right pieces of the graph must continue. - Order the significant x-values only after computing each feature.

1. Rewrite \(4^x=(2^x)^2\). Then \(f(x)=2^x(2^x-1)\), so the only zero is \(x=0\). 2. \(f'(x)=\ln(2)2^x(2\cdot2^x-1)\). Thus the only critical number is \(x=-1\); \(f'\) is negative before \(-1\) and positive after it, so the point \((-1,-\tfrac14)\) is a local minimum. 3. \(f''(x)=(\ln2)^2 2^x(4\cdot2^x-1)\). It changes sign at \(x=-2\), giving the inflection point \((-2,-\tfrac{3}{16})\). The graph is concave down on \(( -\infty,-2)\) and concave up on \((-2,\infty)\). 4. As \(x\to\infty\), \(f(x)\to\infty\). As \(x\to-\infty\), \(f(x)\to0\) from below. 5. Therefore the blueprint is: approach \(y=0\) from below, concave down and decreasing; pass through the inflection point at \(x=-2\); continue decreasing concave up to the local minimum at \(x=-1\); then increase concave up through the zero at \(x=0\) and toward \(\infty\).

a) Zero: \((0,0)\). b) Local minimum: \((-1,-\tfrac14)\). c) Inflection point: \((-2,-\tfrac{3}{16})\); concave down on \(( -\infty,-2)\), concave up on \((-2,\infty)\). d) \(f(x)\to\infty\) as \(x\to\infty\); \(f(x)\to0^-\) as \(x\to-\infty\). e) Decreasing on \(( -\infty,-1)\), increasing on \((-1,\infty)\); left-to-right landmarks: inflection at \(x=-2\), minimum at \(x=-1\), zero at \(x=0\).

- Positivity of both exponential terms settles the zero question immediately. - Compare the two positive exponential terms in the first derivative to locate the turning point. - The sign of the second derivative tells whether the graph ever changes concavity. - A graph blueprint should connect the two ends through the extremum using the derivative information.

1. Both terms are positive for every real \(x\), so \(g(x)>0\) and there are no zeros. 2. \(g'(x)=\ln(2)(2^x-4\cdot2^{-x})\). Solving \(g'(x)=0\) gives \(x=1\). The derivative is negative for \(x<1\) and positive for \(x>1\), so \((1,4)\) is a local minimum. 3. \(g''(x)=(\ln2)^2(2^x+4\cdot2^{-x})>0\) for every real \(x\), so the graph is concave up everywhere. 4. As \(x\to\infty\), the term \(2^x\) makes \(g(x)\to\infty\). As \(x\to-\infty\), the term \(4\cdot2^{-x}\) makes \(g(x)\to\infty\). 5. Thus the graph descends from \(\infty\), remains concave up, reaches its minimum at \((1,4)\), and then rises to \(\infty\), staying above the x-axis throughout.

a) No zeros because \(g(x)>0\) for all \(x\). b) Local minimum: \((1,4)\). c) Concave up on \(\mathbb{R}\). d) \(g(x)\to\infty\) as \(x\to\pm\infty\). e) Decreasing and concave up on \(( -\infty,1)\); minimum \((1,4)\); increasing and concave up on \((1,\infty)\).

- The argument of a logarithm must be positive. - Use the first derivative to locate critical numbers and the second derivative to classify them. - A constant sign for the second derivative means the concavity does not change. - Examine the function as \(x\) approaches the excluded value.

1. The logarithm requires \(x^2>0\), so the domain is \(\mathbb{R}\setminus\{0\}\). 2. Differentiate: \(f'(x)=\frac{2}{x}-x+2\) and \(f''(x)=-\frac{2}{x^2}-1\). 3. Solve \(f'(x)=0\): \(\frac{2-x^2+2x}{x}=0\), so \(x^2-2x-2=0\). The critical numbers are \(x=1\pm\sqrt{3}\). 4. Since \(f''(x)<0\) everywhere in the domain, both critical points are local maxima. Using \(x^2=2x+2\) at either critical number, their coordinates are \(\left(1+\sqrt{3},\ln(4+2\sqrt{3})+\sqrt{3}\right)\) and \(\left(1-\sqrt{3},\ln(4-2\sqrt{3})-\sqrt{3}\right)\). 5. The second derivative is never zero and is negative on both domain intervals. Thus the graph is concave down throughout its domain and has no inflection points. 6. The point \(x=0\) is not in the domain, and \(f(x)\to-\infty\) as \(x\to0^-\) or \(x\to0^+\). The vertical asymptote separates the graph into two branches, so the two maxima are not connected by a continuous portion of the graph.

a) \(\mathbb{R}\setminus\{0\}\). b) Local maxima at \(\left(1+\sqrt{3},\ln(4+2\sqrt{3})+\sqrt{3}\right)\) and \(\left(1-\sqrt{3},\ln(4-2\sqrt{3})-\sqrt{3}\right)\). c) There are no inflection points. d) The vertical asymptote \(x=0\) separates the graph into two branches, so no connected part of the graph must pass through a minimum or an inflection point between the maxima.

- Which factor of the second derivative determines where its sign changes? - Compare the sign of that factor between and outside its roots. - How can the value of the second derivative at a critical point classify the extremum?

1. The derivatives are \(f'(x)=(x^2+2x-3)e^x\) and \(f''(x)=(x^2+4x-1)e^x\). 2. The exponential factor is positive. The quadratic \(x^2+4x-1\) is zero at \(x=-2-\sqrt{5}\) and \(x=-2+\sqrt{5}\), positive outside those roots, and negative between them. Thus \(f\) is concave up on \((-\infty,-2-\sqrt{5})\) and \((-2+\sqrt{5},\infty)\), and concave down on \((-2-\sqrt{5},-2+\sqrt{5})\). 3. Since the second derivative changes sign at both roots, both x-coordinates give inflection points. 4. Factor \(f'(x)=(x+3)(x-1)e^x\). At \(x=-3\), \(f''(-3)=-4e^{-3}<0\), giving a local maximum at \((-3,6e^{-3})\). At \(x=1\), \(f''(1)=4e>0\), giving a local minimum at \((1,-2e)\).

a) Concave up on \((-\infty,-2-\sqrt{5})\) and \((-2+\sqrt{5},\infty)\); concave down on \((-2-\sqrt{5},-2+\sqrt{5})\). b) \(x=-2-\sqrt{5}\) and \(x=-2+\sqrt{5}\). c) Local maximum: \((-3,6e^{-3})\); local minimum: \((1,-2e)\).

- Use a double-angle identity for \(\sin(x)\cos(x)\). - A product is zero when at least one factor is zero. - Rewrite the derivative equation as a quadratic equation in \(\cos(x)\). - Use higher derivatives when the second derivative test is inconclusive.

1. Expanding gives \(f(x)=\sin(x)+\sin(x)\cos(x)=\sin(x)+\frac{1}{2}\sin(2x)\). 2. The product is zero when \(\sin(x)=0\) or \(1+\cos(x)=0\). On \([0,2\pi]\), the zeros are \(x=0,\pi,2\pi\). 3. Since \(f'(x)=\cos(x)+\cos(2x)\), use \(\cos(2x)=2\cos^2(x)-1\) to obtain \(2\cos^2(x)+\cos(x)-1=0\). Thus \(\cos(x)=\frac{1}{2}\) or \(\cos(x)=-1\), giving critical numbers \(x=\frac{\pi}{3},\pi,\frac{5\pi}{3}\). 4. The second derivative is \(f''(x)=-\sin(x)-2\sin(2x)\). At \(x=\frac{\pi}{3}\), it is \(-\frac{3\sqrt{3}}{2}<0\), giving a local maximum at \(\left(\frac{\pi}{3},\frac{3\sqrt{3}}{4}\right)\). At \(x=\frac{5\pi}{3}\), it is \(\frac{3\sqrt{3}}{2}>0\), giving a local minimum at \(\left(\frac{5\pi}{3},-\frac{3\sqrt{3}}{4}\right)\). 5. At \(x=\pi\), \(f'(\pi)=0\) and \(f''(\pi)=0\), so the second derivative test is inconclusive. But \(f'''(\pi)=-3\ne0\), so concavity changes there; \((\pi,0)\) is a stationary inflection point, not a local extremum.

a) \(f(x)=\sin(x)+\frac{1}{2}\sin(2x)\). b) \(x=0,\pi,2\pi\). c) Local maximum: \(\left(\frac{\pi}{3},\frac{3\sqrt{3}}{4}\right)\); local minimum: \(\left(\frac{5\pi}{3},-\frac{3\sqrt{3}}{4}\right)\); stationary inflection: \((\pi,0)\).

- Compare symmetry, intercepts, end behavior, or extrema for part a. - Use the leading term to determine the end behavior. - Substitute \(u=x^2\) to solve the quartic equation. - Use the first and second derivatives to find and classify extrema.

1. The graph shown cannot represent \(f\). For example, \(f\) is even and therefore symmetric about the y-axis, while the displayed graph is symmetric about \(x=1\). Also, \(f(0)=6\), but the displayed graph has y-intercept \((0, 1)\). 2. The positive leading coefficient and even degree give \(f(x)\to\infty\) as \(x\to\pm\infty\). 3. Solving \(\frac{1}{2}x^4-4x^2+6=0\) and substituting \(u=x^2\) gives \(u^2-8u+12=0\). Thus, \(u=2\) or \(u=6\), so the zeros are \(x=\pm\sqrt{2}\approx\pm1.41\) and \(x=\pm\sqrt{6}\approx\pm2.45\). 4. The derivative is \(f'(x)=2x(x^2-4)\), so the critical numbers are \(x=-2, 0, 2\). Since \(f''(x)=6x^2-8\), \((0, 6)\) is a local maximum and \((-2, -2)\) and \((2, -2)\) are local minima.

a) Possible reasons include: \(f\) is symmetric about the y-axis, but the displayed graph is symmetric about \(x=1\); and \(f(0)=6\), but the displayed graph has y-intercept \((0, 1)\). b) 1. \(f(x)\to\infty\) as \(x\to\pm\infty\) 2. \(x=\pm\sqrt{2}\approx\pm1.41\) and \(x=\pm\sqrt{6}\approx\pm2.45\) 3. Local maximum: \((0, 6)\); local minima: \((-2, -2)\) and \((2, -2)\)

- Find the first and second derivatives. - Use the second derivative to classify the critical points. - How can a third derivative check the inflection point? - Compare the exact values with the displayed graph's scale.

1. The derivatives are \(f'(x)=-\frac{3}{4}x^2+3\), \(f''(x)=-\frac{3}{2}x\), and \(f'''(x)=-\frac{3}{2}\). 2. Solving \(f'(x)=0\) gives \(x=\pm2\). Since \(f''(-2)=3>0\) and \(f''(2)=-3<0\), \((-2,-4)\) is a local minimum and \((2,4)\) is a local maximum. 3. Solving \(f''(x)=0\) gives \(x=0\). Since the third derivative is nonzero, the inflection point is \((0,0)\). 4. The displayed graph has extrema at approximately \((-2,-3)\) and \((2,3)\), not at \((-2,-4)\) and \((2,4)\). Therefore, it does not represent \(f\) correctly.

a) Local minimum: \((-2,-4)\); local maximum: \((2,4)\); inflection point: \((0,0)\) b) No. The displayed graph has extremum y-values \(-3\) and \(3\), but \(f\) has extremum y-values \(-4\) and \(4\).

- Set the first derivative equal to \(0\) to find critical numbers. - Use the second derivative to classify each extremum. - Set the second derivative equal to \(0\) to find possible concavity changes. - Verify each value with the third derivative or a sign change.

1. The derivatives are \(f'(x)=-\frac{2}{5}x^3+\frac{8}{5}x\), \(f''(x)=-\frac{6}{5}x^2+\frac{8}{5}\), and \(f'''(x)=-\frac{12}{5}x\). 2. Solving \(f'(x)=0\) gives \(x=0\) and \(x=\pm2\). Since \(f''(0)=\frac{8}{5}>0\) and \(f''(\pm2)=-\frac{16}{5}<0\), the graph has a local minimum at \((0,1)\) and local maxima at \((-2,\frac{13}{5})\) and \((2,\frac{13}{5})\). 3. Solving \(f''(x)=0\) gives \(x=\pm\frac{2\sqrt{3}}{3}\). The third derivative is nonzero at both values, so these are the concavity-change x-values.

a) Local maxima: \((-2,\frac{13}{5})\) and \((2,\frac{13}{5})\); local minimum: \((0,1)\) b) \(x=\pm\frac{2\sqrt{3}}{3}\)

- Compare the end behavior of a positive-leading-coefficient cubic with each graph. - Factor \(f\) to identify its zeros and their multiplicities. - Check whether \(f\) is even or odd. - Find the critical numbers and compare the locations and classifications of the extrema.

1. Graph A cannot represent \(f\). The function has end behavior \(f(x)\to\infty\) as \(x\to\infty\) and \(f(x)\to-\infty\) as \(x\to-\infty\), while Graph A has the opposite end behavior. Also, \(f(x)=x^2(x-3)<0\) for every \(x<0\), but Graph A lies above the x-axis there. In addition, \(f\) has a local maximum at \((0, 0)\) and a local minimum at \((2, -4)\), while Graph A reverses those classifications. 2. Graph B cannot represent \(f\). Graph B is symmetric about the origin, but \(f\) is neither even nor odd. Graph B crosses the x-axis at the origin, while \(f(x)=x^2(x-3)\) has a double zero there and only touches the axis. Also, the local maximum of \(f\) is at the origin, not at a negative x-value as shown in Graph B.

Graph A: It has the wrong end behavior; it lies above the x-axis for negative \(x\), although \(f(x)<0\) there; and it reverses the local maximum and local minimum. Graph B: It has origin symmetry, but \(f\) has no such symmetry; it crosses rather than touches the x-axis at \(x=0\); and its local maximum is at the wrong x-value.

- Compare the graph’s right-end behavior with each exponential factor. - Use the graph’s x-intercept to eliminate choices. - A local maximum has a horizontal tangent. - Concavity is determined by the sign of the second derivative. - Differentiate the selected function to verify both statements.

1. The graph approaches the x-axis as \(x\to\infty\), so option A is not possible because \((x+2)e^x\to\infty\). 2. The graph has an x-intercept at \(x=-2\), which rules out option B. Therefore, option C is the matching function: \(f(x)=(x+2)e^{-x}\). 3. Differentiate: \(f'(x)=-(x+1)e^{-x}\). Thus \(f'(-1)=0\), so statement (1) is true. This also agrees with the local maximum shown at \(x=-1\). 4. The second derivative is \(f''(x)=xe^{-x}\). For \(x<0\), the exponential factor is positive and \(x\) is negative, so \(f''(x)<0\). The graph is concave down, not concave up, so statement (2) is false.

a) (C) \(f_3(x)=(x+2)e^{-x}\). b) (1) True. \(f'(-1)=0\). (2) False. Since \(f''(x)=xe^{-x}<0\) for \(x<0\), the graph is concave down there.

- Rewrite the expression as a quotient to analyze its end behavior. - Apply the product and chain rules. - Use the first and second derivatives to locate and classify the local maximum. - Verify a change in concavity at the candidate inflection point. - The exponential factor is never zero.

1. Rewrite \(f(x)=\frac{x+1}{e^x}\). Exponential growth dominates linear growth, so \(\lim_{x\to\infty}f(x)=0\). Equivalently, L’Hôpital’s rule gives \(\lim_{x\to\infty}\frac{1}{e^x}=0\). 2. Differentiate: \(f'(x)=-xe^{-x}\). The only critical number is \(x=0\). Since \(f''(x)=(x-1)e^{-x}\) and \(f''(0)=-1<0\), the graph has a local maximum at \((0, 1)\). 3. The second derivative is zero at \(x=1\). Since \(f'''(x)=(2-x)e^{-x}\) and \(f'''(1)=e^{-1}\ne0\), the graph has an inflection point at \(\left(1, \frac{2}{e}\right)\). 4. Since \(e^{-x}\ne0\), \((x+1)e^{-x}=0\) only when \(x=-1\).

a) \(\lim_{x\to\infty}f(x)=0\). b) \((0, 1)\). c) \(\left(1, \frac{2}{e}\right)\). d) \(x=-1\).

- A maximum occurs at a critical point of the concentration function. - In part b, set the function equal to the given concentration. - The equation in part b has two solutions; choose the one after the maximum. - Interpret the slope of the graph as the rate at which concentration changes.

1. Differentiate: \(C'(t) = \frac{10 - 10t^2}{(t^2 + 1)^2}\). 2. Solve \(C'(t) = 0\). Since \(t \ge 0\), the relevant critical time is \(t = 1\). Also, \(C''(t) = \frac{20t(t^2 - 3)}{(t^2 + 1)^3}\), and \(C''(1) = -5 < 0\), so the concentration is maximized at \(t = 1\). The maximum is \(C(1) = 5\,\text{mg/L}\). 3. Solve \(\frac{10t}{t^2 + 1} = 4\). This gives \(2t^2 - 5t + 2 = 0\), with solutions \(t = \frac{1}{2}\) and \(t = 2\). The time after the maximum is \(t = 2\,\text{h}\). 4. At \(t = \sqrt{3}\), the concavity changes and \(C'(t)\) reaches its minimum value. In context, the concentration is decreasing at its fastest rate at that time.

a) The maximum concentration is \(5\,\text{mg/L}\) at \(t = 1\,\text{h}\). b) After the maximum, the concentration is \(4\,\text{mg/L}\) at \(t = 2\,\text{h}\). c) At \(t = \sqrt{3}\,\text{h}\), the concentration is decreasing most rapidly.

- Calculate each y-intercept and find the zeros. - Differentiate each function and use the derivative sign to locate extrema. - Compare end behavior in both directions. - Match the resulting features with the displayed graphs.

1. For \(f(x)=(x-2)e^x\), \(f(0)=-2\), and the only zero is \(x=2\). Also, \(f'(x)=(x-1)e^x\), so \(f\) decreases for \(x<1\), increases for \(x>1\), and has a local minimum at \(x=1\). Since \(f(x)\to0\) as \(x\to-\infty\), these features match graph \(A\). 2. For \(g(x)=3-e^x\), \(g(0)=2\), \(g'(x)=-e^x<0\), so the function is strictly decreasing, and \(g(x)\to3\) as \(x\to-\infty\). These features match graph \(B\). 3. For \(h(x)=xe^{-x}\), the graph passes through \((0, 0)\). Since \(h'(x)=(1-x)e^{-x}\), it increases for \(x<1\), decreases for \(x>1\), and has a local maximum at \(x=1\). Also, \(h(x)\to0\) as \(x\to\infty\). These features match graph \(C\).

Graph \(A\): function 1; graph \(B\): function 2; graph \(C\): function 3.

- Factor the polynomial to find its zeros. - Use the first and second derivatives to find the local maximum. - Set the second derivative equal to \(0\) for possible inflection points. - Use the graph's symmetry to check paired answers.

1. Since \(f(x)=\frac{1}{4}(x^2-4)^2\), the zeros are \(x=-2\) and \(x=2\), each with multiplicity \(2\). Thus, \(N_1=(-2, 0)\) and \(N_2=(2, 0)\). 2. The derivatives are \(f'(x)=x^3-4x\), \(f''(x)=3x^2-4\), and \(f'''(x)=6x\). The second-derivative test gives a local maximum at \(H=(0, 4)\). 3. Solving \(f''(x)=0\) gives \(x=\pm\frac{2\sqrt{3}}{3}\). At either value, \(f(x)=\frac{16}{9}\), and the third derivative is nonzero. Therefore, \(W_1=(-\frac{2\sqrt{3}}{3}, \frac{16}{9})\) and \(W_2=(\frac{2\sqrt{3}}{3}, \frac{16}{9})\).

\(N_1=(-2, 0)\), \(N_2=(2, 0)\), \(H=(0, 4)\), \(W_1=(-\frac{2\sqrt{3}}{3}, \frac{16}{9})\), and \(W_2=(\frac{2\sqrt{3}}{3}, \frac{16}{9})\)

- Factor the polynomial to find zeros and multiplicities. - Use the first and second derivatives for extrema. - Use the second derivative for the inflection point. - Compare each coordinate with the graph.

1. Factoring gives \(f(x)=x(x-3)^2\). Therefore, \(x=0\) is a simple zero and \(x=3\) is a double zero. 2. The derivatives are \(f'(x)=3(x-1)(x-3)\), \(f''(x)=6x-12\), and \(f'''(x)=6\). The second-derivative test gives a local maximum at \(H=(1, 4)\) and a local minimum at \(T=(3, 0)\). 3. Solving \(f''(x)=0\) gives \(x=2\), and \(f(2)=2\). Thus, \(W=(2, 2)\). All points agree with the graph.

Zeros: \((0, 0)\), simple, and \((3, 0)\), double; local maximum: \(H=(1, 4)\); local minimum: \(T=(3, 0)\); inflection point: \(W=(2, 2)\) All points agree with the graph.

- Factor the polynomial to find x-intercepts and multiplicities. - Use the first and second derivatives to find and classify extrema. - Use the second derivative to find the inflection point. - Compare the exact coordinates with the graph.

1. Since \(f(x)=x^2(x-3)\), the x-intercepts are \((0, 0)\), with multiplicity \(2\), and \((3, 0)\), with multiplicity \(1\). The y-intercept is also \((0, 0)\). 2. The derivatives are \(f'(x)=3x(x-2)\), \(f''(x)=6x-6\), and \(f'''(x)=6\). The second-derivative test gives a local maximum at \((0, 0)\) and a local minimum at \((2, -4)\). 3. The inflection point occurs at \(x=1\), where \(f(1)=-2\). Thus, the inflection point is \((1, -2)\). 4. These exact points agree with the diagram, so the graph is correct.

Yes, the graph is correct. Intercepts: \((0, 0)\), a double x-intercept and the y-intercept; \((3, 0)\), a simple x-intercept Local maximum: \((0, 0)\); local minimum: \((2, -4)\) Inflection point: \((1, -2)\)

- Evaluate the function at \(x=0\). - Factor the cubic to find zeros and multiplicities. - Use the first and second derivatives to find and classify extrema. - Compare the exact values with the graph's scale.

1. The y-intercept is \((0, 2)\). Factoring gives \(f(x)=-(x+1)^2(x-2)\), so the x-intercepts are \((-1, 0)\), with multiplicity \(2\), and \((2, 0)\), with multiplicity \(1\). 2. The derivatives are \(f'(x)=-3x^2+3\) and \(f''(x)=-6x\). The critical numbers are \(x=-1\) and \(x=1\). The second-derivative test gives a local minimum at \((-1, 0)\) and a local maximum at \((1, 4)\). 3. The displayed graph has y-intercept \((0, 1.6)\) and a local maximum near \((1, 3.2)\). These do not match the exact values, so the graph is incorrect.

No. Intercepts: \((0, 2)\), \((-1, 0)\) with multiplicity \(2\), and \((2, 0)\) Local minimum: \((-1, 0)\); local maximum: \((1, 4)\)

- Use even symmetry to check paired points. - Factor the polynomial to find its zeros. - Set the first derivative to zero. - Use the second derivative to classify the extrema.

1. The function is even. Factoring gives \(f(x)=\frac{1}{2}x^2(x^2-4)\), so the x-intercepts are \((0,0)\), a double zero, and \((\pm2,0)\). The y-intercept is \((0,0)\). 2. Since \(f'(x)=2x(x^2-2)\), the critical numbers are \(0\) and \(\pm\sqrt{2}\). With \(f''(x)=6x^2-4\), we have \(f''(0)=-4<0\) and \(f''(\pm\sqrt{2})=8>0\). Thus the local maximum is \((0,0)\) and the local minima are \((\pm\sqrt{2},-2)\). 3. These features agree with the graph, so it is correct.

Yes. Intercepts: \((0,0)\), a double x-intercept and the y-intercept; \((\pm2,0)\). Local maximum: \((0,0)\); local minima: \((\pm\sqrt{2},-2)\).

- Check \(f(0)\) before doing further calculations. - Find the zeros of the first derivative; if the first and second derivatives are both \(0\), check the third derivative. - Compute the exact stationary-point coordinates and compare them with the graph.

1. The y-intercept is \((0, 2)\), but the displayed graph has y-intercept \((0, 4)\). This already shows that the graph is incorrect. 2. The derivative is \(f'(x)=x^2(x-3)\). Thus, the critical numbers are \(x=0\) and \(x=3\). 3. At \(x=3\), \(f''(3)=9>0\), so there is a local minimum at \((3, -\frac{19}{4})\). At \(x=0\), the first and second derivatives are \(0\), while \(f'''(0)=-6\ne0\), so \((0, 2)\) is a stationary inflection point. 4. These points do not match the displayed graph, so the graph is incorrect.

No. Y-intercept and stationary inflection point: \((0, 2)\) Local minimum: \((3, -\frac{19}{4})\)

- Use even symmetry to check the intercepts. - Think of the equation for the zeros as a quadratic in \(x^2\). - Find the critical points and check the sign of the second derivative.

1. The function is even, and its y-intercept is \((0,-5)\). 2. Let \(u=x^2\). Solving \(-u^2+6u-5=0\) gives \(u=1\) or \(u=5\). Thus the x-intercepts are \((\pm1,0)\) and \((\pm\sqrt{5},0)\). 3. Since \(f'(x)=-4x(x^2-3)\), the critical numbers are \(0\) and \(\pm\sqrt{3}\). With \(f''(x)=-12x^2+12\), we have \(f''(0)=12>0\) and \(f''(\pm\sqrt{3})=-24<0\). Therefore, there is a local minimum at \((0,-5)\) and local maxima at \((\pm\sqrt{3},4)\). 4. The graph agrees with these features, so it is correct.

Yes. Intercepts: \((0,-5)\), \((\pm1,0)\), and \((\pm\sqrt{5},0)\). Local minimum: \((0,-5)\); local maxima: \((\pm\sqrt{3},4)\).

- Find the critical numbers from the first derivative and classify them with the second derivative. - A local minimum on the x-axis is a multiple zero. - Set the second derivative equal to \(0\) to locate the inflection point. - Compare the exact critical numbers and inflection-point location with the graph.

1. The y-intercept is \((0,4)\). The derivative is \(f'(x)=3x(x-2)\), so the critical numbers are \(x=0\) and \(x=2\). 2. Since \(f''(0)=-6<0\), \((0,4)\) is a local maximum. Since \(f''(2)=6>0\), \((2,0)\) is a local minimum. 3. Because the local minimum lies on the x-axis, \(x=2\) is a double zero. Factoring gives \(f(x)=(x-2)^2(x+1)\), so the other x-intercept is \((-1,0)\). 4. Solving \(f''(x)=6x-6=0\) gives \(x=1\). Since \(f(1)=2\), the inflection point is \((1,2)\). 5. The displayed graph places the local minimum at \(x=3\), not \(x=2\), and its inflection point is also misplaced. Therefore, it is incorrect.

No. Intercepts: \((0,4)\), \((-1,0)\), and \((2,0)\), where \(x=2\) is a double zero Local maximum: \((0,4)\); local minimum: \((2,0)\) Inflection point: \((1,2)\)

- Count the x-axis crossings. - Use the polynomial degree to bound the number of zeros. - Identify the graph's local maximum and local minimum. - Use the sign and degree of the leading term for end behavior. - Relate the two extrema to the shape of the quadratic derivative.

1. The graph crosses the x-axis three times, so \(f\) has three real zeros. A degree-\(3\) polynomial can have at most three real zeros because a nonzero polynomial cannot have more zeros than its degree. 2. The graph has two local extrema. Since the leading term is \(-\frac{1}{2}x^3\), \(f(x)\to-\infty\) as \(x\to\infty\), and \(f(x)\to\infty\) as \(x\to-\infty\). 3. The derivative is a quadratic with two distinct zeros at the extrema. Its vertex lies between those zeros, where the derivative changes from decreasing to increasing or vice versa. Therefore, the second derivative changes sign there, producing an inflection point between the extrema.

a) Three real zeros; a cubic has at most three real zeros. b) Two local extrema; \(f(x)\to-\infty\) as \(x\to\infty\), and \(f(x)\to\infty\) as \(x\to-\infty\). c) The quadratic derivative has its vertex between its two zeros, so the second derivative changes sign there.

- Factor the first derivative to find critical numbers. - Use the second derivative to classify the local minimum. - What happens to the first two derivatives at the special inflection point? - What extra feature distinguishes a stationary inflection point from a nonstationary one?

1. The derivatives are \(f'(x)=x(x-3)^2\), \(f''(x)=3(x-1)(x-3)\), and \(f'''(x)=6x-12\). 2. Since \(f'(0)=0\) and \(f''(0)=9>0\), the local minimum is \((0,2)\). 3. At \(x=3\), \(f'(3)=f''(3)=0\) and \(f'''(3)=6\ne0\). Also, \(f(3)=\frac{35}{4}\). Thus \((3,\frac{35}{4})\) is a stationary inflection point. 4. A stationary inflection point also satisfies \(f'(x_0)=0\). A nonhorizontal inflection tangent has \(f'(x_0)\ne0\).

a) Local minimum: \((0,2)\) b) Stationary inflection point: \((3,\frac{35}{4})\) c) \(f'(x_0)=0\)

- Factor first to get the intercepts efficiently. - Use the zeros and signs of \(f'\) to organize the monotonicity pieces. - Use the sign of \(f''\) to organize the concavity pieces and locate the inflection point. - The leading cubic term determines the two ends of the blueprint.

1. Factoring gives \(f(x)=\frac12x(x^2-12)\). The x-intercepts are \((0,0)\) and \((\pm2\sqrt3,0)\). 2. \(f'(x)=\frac32x^2-6\), so the critical numbers are \(x=\pm2\). Since \(f''(x)=3x\), the second derivative test gives a local maximum at \((-2,8)\) and a local minimum at \((2,-8)\). 3. The second derivative changes sign at \(x=0\), so \((0,0)\) is the inflection point. 4. The graph increases on \(( -\infty,-2)\), decreases on \((-2,2)\), and increases on \((2,\infty)\). It is concave down on \(( -\infty,0)\) and concave up on \((0,\infty)\). 5. Because the leading term is positive cubic, the left end goes to \(-\infty\) and the right end goes to \(\infty\). These facts determine the requested blueprint.

a) Intercepts: \((0,0)\), \((\pm2\sqrt3,0)\). Local maximum: \((-2,8)\). Local minimum: \((2,-8)\). Inflection point: \((0,0)\). b) Increasing on \(( -\infty,-2)\), decreasing on \((-2,2)\), increasing on \((2,\infty)\); concave down for \(x<0\), concave up for \(x>0\). Left-to-right: left end down → maximum \((-2,8)\) → inflection \((0,0)\) → minimum \((2,-8)\) → right end up.

- Multiplicity at a zero can help predict whether the graph crosses with a horizontal tangent. - A critical number where \(f'\) does not change sign is not a local extremum. - Split the real line at the zeros of \(f''\) to get the concavity intervals. - Assemble the blueprint in increasing x-order rather than treating the features as unrelated facts.

1. Factoring gives \(f(x)=x^3(x-4)\). Thus \(x=0\) is a zero of multiplicity \(3\), and \(x=4\) is a simple zero. Since the leading term is \(x^4\), \(f(x)\to\infty\) as \(x\to\pm\infty\). 2. \(f'(x)=4x^2(x-3)\). The derivative is negative for \(x<3\) except for the non-sign-changing zero at \(x=0\), and positive for \(x>3\). Therefore, \((3,-27)\) is the only local extremum, a local minimum. 3. \(f''(x)=12x(x-2)\). It is positive for \(x<0\), negative for \(0<x<2\), and positive for \(x>2\). Thus the inflection points are \((0,0)\) and \((2,-16)\). Because \(f'(0)=0\), \((0,0)\) is a stationary inflection point. 4. The graph decreases and is concave up until \((0,0)\), continues decreasing and becomes concave down until \((2,-16)\), continues decreasing and becomes concave up until the minimum \((3,-27)\), then increases concave up and crosses again at \((4,0)\).

a) Zeros: \(x=0\) (multiplicity \(3\)) and \(x=4\); both ends go to \(\infty\). b) Only local extremum: local minimum \((3,-27)\). c) Inflection points \((0,0)\) and \((2,-16)\); \((0,0)\) is stationary. d) Decreasing on \(( -\infty,3)\), increasing on \((3,\infty)\); concave up on \(( -\infty,0)\) and \((2,\infty)\), concave down on \((0,2)\). Significant order: stationary inflection \((0,0)\), inflection \((2,-16)\), minimum \((3,-27)\), zero \((4,0)\).

- Compare the equation’s symmetry with the graph. - Evaluate \(f(0)\). - Find the critical numbers from the first derivative. - Compare the required extremum at \(x=0\) with the graph.

1. The function is even, so its graph must be symmetric about the y-axis. The displayed graph is shifted to the right and lacks this symmetry. 2. The y-intercept is \(f(0)=2\), but the displayed graph crosses the y-axis at \((0,-1)\). 3. The derivative is \(f'(x)=4x(x^2-2)\), so \(x=0\) is a critical number. Since \(f''(0)=-8<0\), the graph of \(f\) must have a local maximum at \((0,2)\). The displayed graph has no extremum at \(x=0\).

The graph is incorrect because it is not symmetric about the y-axis, its y-intercept is \(-1\) instead of \(2\), and it has no local maximum at \((0, 2)\).

- Evaluate each function at \(x=0\) and find any zeros. - Differentiate each function and use the sign of the derivative to determine monotonicity and extrema. - Compare horizontal asymptotes and end behavior. - Check for symmetry about the y-axis.

1. For \(f_1(x)=e^x-2\), the y-intercept is \(-1\), the horizontal asymptote is \(y=-2\) as \(x\to-\infty\), and \(f_1'(x)=e^x>0\), so the function is strictly increasing. These features match graph a). 2. For \(f_2(x)=2e^{-x}\), the y-intercept is \(2\), \(f_2(x)\to0\) as \(x\to\infty\), and \(f_2'(x)=-2e^{-x}<0\), so the function is strictly decreasing. These features match graph b). 3. For \(f_3(x)=4e^{-0.2x^2}\), the function is even, and \(f_3'(x)=-1.6xe^{-0.2x^2}\). Thus, it increases for \(x<0\), decreases for \(x>0\), and has a maximum value of \(4\) at \(x=0\). These features match graph c). 4. For \(f_4(x)=(x-1)e^x\), the zero is \(x=1\), the y-intercept is \(-1\), and \(f_4'(x)=xe^x\). Thus, it decreases for \(x<0\), increases for \(x>0\), and has a local minimum at \(x=0\). These features match graph d).

a) \(f_1\); b) \(f_2\); c) \(f_3\); d) \(f_4\)

- Which derivative identifies a maximum of the concentration? - How can you locate where the rate of change is most negative? - Use the point-slope form of the tangent line at \(t=6\). - For part d, return to the original exponential model rather than the temporary linear continuation.

1. Differentiate using the product and chain rules: \(c'(t)=8e^{-0.25t}-2te^{-0.25t}=(8-2t)e^{-0.25t}\). 2. Since \(e^{-0.25t}>0\), \(c'(t)=0\) when \(8-2t=0\), so \(t=4\). The derivative changes from positive to negative there, so the concentration is maximal. \(c(4)=32e^{-1}\approx11.77\,\text{mg/L}\). 3. The concentration decreases most rapidly where \(c'\) is smallest, which occurs where \(c''(t)=0\). Differentiating again gives \(c''(t)=(0.5t-4)e^{-0.25t}\). Thus, \(t=8\), and the sign change of \(c''\) confirms a minimum of \(c'\). 4. At \(t=6\), \(c(6)=48e^{-1.5}\) and \(c'(6)=-4e^{-1.5}\). The tangent line is \(L(t)=48e^{-1.5}-4e^{-1.5}(t-6)\). 5. Set \(L(t)=0\): \(48-4(t-6)=0\), so \(t=18\). This is \(12\) additional hours after the intervention. 6. Returning to the original exponential model, exponential decay dominates the linear factor, so \(\lim_{t\to\infty}8te^{-0.25t}=0\). The modeled medication concentration approaches \(0\,\text{mg/L}\) over time.

a) The maximum occurs at \(t=4\) hours and is approximately \(11.77\,\text{mg/L}\). b) The concentration is decreasing most rapidly at \(t=8\) hours. c) It reaches \(0\) after \(12\) additional hours, or \(18\) hours after the medication was taken. d) For the original exponential model, the concentration approaches \(0\,\text{mg/L}\).

- Compare \(g(-x)\) with \(g(x)\). - Symmetric tangent points have opposite slopes and equal y-intercepts. - Set the tangent-line equations equal to find their intersection. - Find where the second derivative changes sign.

1. Since \(g(-x)=4e^{-(-x)^2/4}=g(x)\), the graph is symmetric about the y-axis. 2. The derivative is \(g'(x)=-2xe^{-x^2/4}\). At \(x=2\), \(g(2)=\frac{4}{e}\) and \(g'(2)=-\frac{4}{e}\). Therefore, the tangent line is \(y=-\frac{4}{e}(x-2)+\frac{4}{e}=-\frac{4}{e}x+\frac{12}{e}\). By symmetry, the tangent at \(x=-2\) has slope \(\frac{4}{e}\) and equation \(y=\frac{4}{e}x+\frac{12}{e}\). 3. Set the two line equations equal. This gives \(x=0\), and then \(y=\frac{12}{e}\). Their intersection is \(\left(0,\frac{12}{e}\right)\). 4. As \(x\to\pm\infty\), the exponent approaches \(-\infty\), so \(g(x)\to0\). The second derivative is \(g''(x)=(x^2-2)e^{-x^2/4}\). It changes sign at \(x=\pm\sqrt{2}\). The inflection points are \(\left(-\sqrt{2},\frac{4}{\sqrt{e}}\right)\) and \(\left(\sqrt{2},\frac{4}{\sqrt{e}}\right)\).

a) \(g(-x)=g(x)\), so the graph is symmetric about the y-axis. b) At \(x=2\): \(y=-\frac{4}{e}x+\frac{12}{e}\). At \(x=-2\): \(y=\frac{4}{e}x+\frac{12}{e}\). c) \(\left(0,\frac{12}{e}\right)\). d) \(\lim_{x\to\pm\infty}g(x)=0\). Inflection points: \(\left(-\sqrt{2},\frac{4}{\sqrt{e}}\right)\) and \(\left(\sqrt{2},\frac{4}{\sqrt{e}}\right)\).

- What derivative condition identifies a local extremum? - How do the first and second derivatives describe slope and concavity? - Check whether every calculated point lies in the modeled interval. - What does it mean geometrically for a horizontal line to pass through a unique maximum?

1. Differentiate: \(f'(x)=(x-0.1x^2)e^{-0.2x}=x(1-0.1x)e^{-0.2x}\). 2. The critical numbers are \(x=0\) and \(x=10\). The derivative is positive on \((0,10)\) and negative on \((10,15)\), so the maximum occurs at \(x=10\). 3. The maximum height is \(f(10)=50e^{-2}\approx 6.77\,\text{m}\). 4. The second derivative is \(f''(x)=(0.02x^2-0.4x+1)e^{-0.2x}\). Its zeros satisfy \(x^2-20x+50=0\), so \(x=10\pm 5\sqrt{2}\). 5. The left inflection point is \(10-5\sqrt{2}\approx 2.93\), and the right one is \(10+5\sqrt{2}\approx 17.07\). Thus there is no inflection point with \(x>5\) in \([0,15]\); the next change in concavity would occur beyond the modeled barrier. 6. The crest at \(x=10\) is the unique maximum in \([0,15]\). Therefore, every other point to its right has height less than \(50e^{-2}\), so the horizontal ray does not intersect the barrier again in the modeled interval.

a) The maximum height is \(50e^{-2}\,\text{m}\approx 6.77\,\text{m}\) at \(x=10\,\text{m}\). b) There is no inflection point with \(x>5\) in \([0,15]\). The calculated right inflection point is \(x=10+5\sqrt{2}\approx 17.07\), outside the modeled interval; the interval contains the other inflection point at \(x=10-5\sqrt{2}\approx 2.93\). c) The horizontal ray does not intersect the barrier again to the right of the crest within \([0,15]\).

- Use the first derivative to locate the summit and the second derivative to study how slope changes. - The steepest descent corresponds to the minimum value of the first derivative. - Model the line of sight as the line segment from the observer to the summit. - How does concavity determine whether a curve lies above or below a chord?

1. Differentiate: \(g'(x)=(-0.5x-1)e^{-0.5x}\) and \(g''(x)=0.25x e^{-0.5x}\). 2. Since the exponential factor is positive, \(g'(x)=0\) when \(-0.5x-1=0\), so \(x=-2\). The derivative changes from positive to negative, so the summit is \((-2,2e)\approx(-2,5.44)\). 3. The steepest downward slope occurs where \(g'(x)\) is smallest. Since \(g''(x)<0\) for \(x<0\) and \(g''(x)>0\) for \(x>0\), \(g'\) has its minimum at \(x=0\). There, \(g'(0)=-1\). 4. The line from \(P=(-4,0)\) to the summit \((-2,2e)\) has slope \(m=\frac{2e}{2}=e\). 5. The profile slope at \(P\) is \(g'(-4)=e^2>e\). Also, \(g''(x)<0\) on \([-4,-2]\), so the graph is concave down and lies above the chord joining \(P\) to the summit. The terrain blocks the line of sight, so the observer cannot see the summit.

a) The summit is \((-2,2e)\approx(-2,5.44)\). b) The steepest downward slope occurs at \(x=0\), where the slope is \(-1\). c) No. The terrain initially rises more steeply than the line of sight, and the concave-down profile lies above that line segment.

- Start by evaluating the model at launch and interpreting its units. - Use the first derivative to find when sales stop rising and start falling. - A difference such as \(S(t)-S(20)\) may be positive or negative. - Consider what happens to the exponential term as time grows. - How is an inflection point related to a maximum or minimum of the rate of change?

1. At launch, \(S(0)=900\), so sales are \(900\) bikes per week. 2. \(S'(t)=(160-20t)e^{-0.1t}\) changes from positive to negative at \(t=8\). Thus weekly sales peak after \(8\) weeks at \(S(8)=2000e^{-0.8}+500\approx1398.66\), or \(1399\) bikes per week. 3. \(h(t)\) compares weekly sales at time \(t\) with sales at week \(20\). Its sign shows whether sales are higher or lower. 4. Because \((200t+400)e^{-0.1t}\to0\), the modeled long-run sales level is \(500\) bikes per week. 5. \(S''(t)=(2t-36)e^{-0.1t}\) changes from negative to positive at \(t=18\). The inflection point is \((18,500+4000e^{-1.8})\). Since the first derivative \(S'\) decreases before \(18\) and increases afterward, sales are falling most rapidly at that time.

a) \(900\) bikes per week. b) After \(8\) weeks; about \(1399\) bikes per week. c) \(h(t)\) is weekly sales at time \(t\) minus weekly sales at week \(20\). d) \(500\) bikes per week. e) Inflection point: \((18,500+4000e^{-1.8})\). At \(t=18\) weeks, \(S'\) is smallest, so weekly sales are decreasing most rapidly.

- Start by evaluating the model at the beginning of the time interval. - Compare the sign of the first derivative before and after its zero. - For the long run, consider what an exponential-decay factor does to the linear factor. - Find where the second derivative changes sign, and connect that change to the first derivative.

1. \(c(0)=5\), so the initial concentration is \(5\,\text{mg/L}\). 2. \(c'(t)=(1-0.5t)e^{-0.25t}\) changes from positive to negative at \(t=2\). The maximum is \(c(2)=1+8e^{-0.5}\approx5.85\,\text{mg/L}\). 3. Since \((2t+4)e^{-0.25t}\to0\), the model approaches \(1\,\text{mg/L}\). 4. \(c''(t)=(0.125t-0.75)e^{-0.25t}\) changes from negative to positive at \(t=6\). Thus the inflection point is \((6,1+16e^{-1.5})\). 5. The derivative \(c'\) decreases before \(t=6\) and increases afterward, so the most rapid concentration decrease occurs at \(t=6\) hours.

a) \(5\,\text{mg/L}\). b) After \(2\) hours; approximately \(5.85\,\text{mg/L}\). c) \(1\,\text{mg/L}\). d) Inflection point: \((6,1+16e^{-1.5})\); the concentration is decreasing most rapidly at \(t=6\) hours.

- Use the zero-product property for the x-intercepts. - Expand or use the product rule to differentiate. - A double zero often corresponds to a horizontal tangent. - Use the second derivative to classify the critical points.

1. The y-intercept is \((0,-4)\). From the factored form, the x-intercepts are \((-2,0)\), \((1,0)\), and \((2,0)\); \(x=1\) is a double zero. 2. Expanding gives \(f(x)=x^4-2x^3-3x^2+8x-4\), and \(f'(x)=(x-1)(4x^2-2x-8)\). Thus the critical numbers are \(x=1\) and \(x=\frac{1\pm\sqrt{33}}{4}\). 3. The second derivative at \(x=1\) is \(f''(1)=-6<0\), giving a local maximum at \((1,0)\). 4. At \(x=\frac{1-\sqrt{33}}{4}\), \(f''(x)=\frac{33+3\sqrt{33}}{2}>0\); at \(x=\frac{1+\sqrt{33}}{4}\), \(f''(x)=\frac{33-3\sqrt{33}}{2}>0\). Both are local minima, with coordinates \(\left(\frac{1-\sqrt{33}}{4},-\frac{207+33\sqrt{33}}{32}\right)\) and \(\left(\frac{1+\sqrt{33}}{4},-\frac{207-33\sqrt{33}}{32}\right)\).

a) Intercepts: \((0,-4)\), \((-2,0)\), \((1,0)\), \((2,0)\). b) Local maximum: \((1,0)\). Local minima: \(\left(\frac{1-\sqrt{33}}{4},-\frac{207+33\sqrt{33}}{32}\right)\) and \(\left(\frac{1+\sqrt{33}}{4},-\frac{207-33\sqrt{33}}{32}\right)\).

- Test whether the function is odd. - Look for a way to factor the polynomial before finding zeros and multiplicities. - The derivative has only even powers. How might that simplify solving for its zeros? - Use origin symmetry to check paired extrema.

1. Since \(f(-x)=-f(x)\), the graph is symmetric about the origin. Factoring gives \(f(x)=x(x^2-1)^2=x(x-1)^2(x+1)^2\). Thus \(x=0\) is a simple zero and \(x=\pm1\) are double zeros. 2. The derivative is \(f'(x)=5x^4-6x^2+1=(x^2-1)(5x^2-1)\). The critical numbers are \(x=\pm1\) and \(x=\pm\frac{\sqrt{5}}{5}\). 3. Since \(f''(x)=20x^3-12x\), we have \(f''(-1)=-8<0\) and \(f''(1)=8>0\). Thus \((-1,0)\) is a local maximum and \((1,0)\) is a local minimum. 4. Also, \(f''\left(-\frac{\sqrt{5}}{5}\right)=\frac{8\sqrt{5}}{5}>0\) and \(f''\left(\frac{\sqrt{5}}{5}\right)=-\frac{8\sqrt{5}}{5}<0\). The remaining local minimum is \(\left(-\frac{\sqrt{5}}{5},-\frac{16\sqrt{5}}{125}\right)\), and the remaining local maximum is \(\left(\frac{\sqrt{5}}{5},\frac{16\sqrt{5}}{125}\right)\).

a) Symmetric about the origin. Zeros: \(x=0\), multiplicity \(1\); \(x=\pm1\), multiplicity \(2\). b) Local maxima: \((-1,0)\) and \(\left(\frac{\sqrt{5}}{5},\frac{16\sqrt{5}}{125}\right)\). Local minima: \((1,0)\) and \(\left(-\frac{\sqrt{5}}{5},-\frac{16\sqrt{5}}{125}\right)\).

- Use the sign of the given second derivative to classify the critical point. - Remember that \(\sqrt{x^2}=|x|\) when evaluating limits at infinity. - Complete the square inside the radical to identify the intercepts of the slant asymptotes.

1. The first derivative is \(f^{\prime}(x)=\frac{4x+2}{\sqrt{4x^2+4x+10}}\). Setting the numerator equal to zero gives \(x=-\frac{1}{2}\). Since \(f(-\frac{1}{2})=3\) and \(f^{\prime\prime}(x)>0\), the graph has a local and absolute minimum at \((-\frac{1}{2}, 3)\). 2. The expression \(4x^2+4x+10\) is positive for every real \(x\), so \(f^{\prime\prime}(x)>0\) everywhere. The concavity never changes, so there are no inflection points. 3. Divide numerator and denominator of \(f^{\prime}\) by \(|x|\). For \(x\to\infty\), the limit is \(2\); for \(x\to-\infty\), the limit is \(-2\). 4. Complete the square: \(f(x)=\sqrt{(2x+1)^2+9}\). As \(x\to\infty\), \(f(x)-(2x+1)\to0\), so one asymptote is \(y=2x+1\). As \(x\to-\infty\), \(f(x)-(-2x-1)\to0\), so the other is \(y=-2x-1\).

1. Absolute minimum at \((-\frac{1}{2}, 3)\) 2. \(f^{\prime\prime}(x)>0\) for all \(x\), so there are no inflection points. 3. \(\lim_{x\to\infty}f^{\prime}(x)=2\) and \(\lim_{x\to-\infty}f^{\prime}(x)=-2\) 4. \(y=2x+1\) and \(y=-2x-1\)

- After differentiating twice, which factor controls the sign of the second derivative? - Where does the graph switch between concave up and concave down? - What does the sign of the second derivative tell you at a critical point?

1. Differentiate: \(f'(x)=\left(\frac{1}{2}x^2+2x-2\right)e^{x/2}\) and \(f''(x)=\left(\frac{1}{4}x^2+2x+1\right)e^{x/2}\). 2. The exponential factor is positive. The quadratic factor of \(f''\) has roots \(x=-4\pm2\sqrt{3}\). It is positive outside these roots and negative between them. Thus the graph is concave up on \((-\infty,-4-2\sqrt{3})\) and \((-4+2\sqrt{3},\infty)\), and concave down on \((-4-2\sqrt{3},-4+2\sqrt{3})\). 3. The sign changes give inflection points \(\left(-4-2\sqrt{3},8(3+2\sqrt{3})e^{-2-\sqrt{3}}\right)\) and \(\left(-4+2\sqrt{3},8(3-2\sqrt{3})e^{-2+\sqrt{3}}\right)\). 4. Setting \(f'(x)=0\) gives \(x=-2\pm2\sqrt{2}\). At \(x=-2-2\sqrt{2}\), \(f''<0\), so there is a local maximum at \(\left(-2-2\sqrt{2},8(1+\sqrt{2})e^{-1-\sqrt{2}}\right)\). At \(x=-2+2\sqrt{2}\), \(f''>0\), so there is a local minimum at \(\left(-2+2\sqrt{2},8(1-\sqrt{2})e^{-1+\sqrt{2}}\right)\).

a) Concave up on \((-\infty,-4-2\sqrt{3})\) and \((-4+2\sqrt{3},\infty)\); concave down on \((-4-2\sqrt{3},-4+2\sqrt{3})\). b) \(\left(-4-2\sqrt{3},8(3+2\sqrt{3})e^{-2-\sqrt{3}}\right)\) and \(\left(-4+2\sqrt{3},8(3-2\sqrt{3})e^{-2+\sqrt{3}}\right)\). c) Local maximum: \(\left(-2-2\sqrt{2},8(1+\sqrt{2})e^{-1-\sqrt{2}}\right)\); local minimum: \(\left(-2+2\sqrt{2},8(1-\sqrt{2})e^{-1+\sqrt{2}}\right)\).

- Compare the domains and look for vertical asymptotes. - Identify zeros and y-intercepts. - Differentiate each candidate to locate critical numbers and determine increasing or decreasing intervals. - Analyze end behavior in both directions and check symmetry.

1. For \(f_4(x)=(x^2-1)e^{-x}\), the zeros are \(x=-1\) and \(x=1\), and the y-intercept is \(-1\). Also, \(f_4'(x)=(-x^2+2x+1)e^{-x}\), so its critical numbers are \(x=1\pm\sqrt{2}\). Together with \(f_4(x)\to0\) as \(x\to\infty\), these features match graph \(G_1\). 2. For \(f_3(x)=\frac{e^x}{x}\), the domain excludes \(x=0\), giving a vertical asymptote there. Since \(f_3'(x)=\frac{e^x(x-1)}{x^2}\), the function decreases for \(x<1\) on each part of its domain and increases for \(x>1\). It approaches \(0\) from below as \(x\to-\infty\) and grows without bound as \(x\to\infty\). These features match graph \(G_2\). 3. For \(f_1(x)=2e^{-x^2}\), \(f_1'(x)=-4xe^{-x^2}\). Thus, the function increases for \(x<0\), decreases for \(x>0\), and has a maximum at \((0,2)\). It is even and approaches \(0\) as \(x\to\pm\infty\), matching graph \(G_3\). 4. For \(f_2(x)=x^2e^{-x}\), \(f_2'(x)=x(2-x)e^{-x}\). The function is nonnegative, has a double zero at \(x=0\), and has a local maximum at \(x=2\). None of the displayed graphs has these features, so \(f_2\) is unused.

\(G_1\): \(f_4(x)=(x^2-1)e^{-x}\); \(G_2\): \(f_3(x)=\frac{e^x}{x}\); \(G_3\): \(f_1(x)=2e^{-x^2}\); unused: \(f_2(x)=x^2e^{-x}\)