- Find the inflection-point x-coordinate from the second derivative. - Express the parameter in terms of \(x\). - Substitute into the y-coordinate.

1. The derivatives are \(g_t'(x)=3x^2-6tx\), \(g_t''(x)=6x-6t\), and \(g_t'''(x)=6\). 2. The inflection point satisfies \(6x-6t=0\), so \(x=t\). The nonzero third derivative confirms an inflection point. 3. Substitute \(t=x\) into the original function: \(y=x^3-3x\cdot x^2+1=-2x^3+1\).

\(y=-2x^3+1\)

- What must be held fixed when finding a derivative for a family of functions? - How could you verify that a critical number really gives a minimum? - What information about the family of points should an equation of the locus preserve? - What restrictions do the parameter values place on the set of points?

1. Differentiate: \(f_a'(x)=2x-2a\), so the critical number is \(x=a\). 2. Since \(f_a''(x)=2>0\), this point is a local minimum for every \(a\). 3. Its value is \(f_a(a)=a^2-2a^2+3=3-a^2\). 4. At every minimum \(x=a\) and \(y=3-a^2\). Replacing \(a\) by \(x\) gives the locus \(y=3-x^2\).

a) \(x=a\); \(f_a''(a)=2>0\), so it is a local minimum. b) \(y=3-a^2\). c) \(y=3-x^2\).

- Which critical number satisfies the stated sign restriction on the x-coordinate? - What must be checked before you can classify that point? - How can the set of all resulting extremum points be expressed without a parameter? - Which restriction on the parameter matters for the resulting points?

1. Differentiate: \(f_a'(x)=3x^2-3a^2\) and \(f_a''(x)=6x\). 2. At \(x=a\), \(f_a'(a)=0\) and \(f_a''(a)=6a>0\), so the point is a local minimum. 3. Its y-coordinate is \(f_a(a)=a^3-3a^3=-2a^3\). 4. Since the point has coordinates \((a,-2a^3)\), replace \(a\) with its x-coordinate to get \(y=-2x^3\), for \(x>0\).

a) \(x=a\) is a local minimum, since \(f_a''(a)=6a>0\). b) \(y=-2a^3\). c) \(y=-2x^3\), for \(x>0\).

- Use the second derivative to locate a possible change in concavity. - After finding its x-coordinate, substitute it into the original function to obtain the full point. - Check the sign of the second derivative on the two sides. - Use the derivative to locate the critical numbers for the extrema.

1. Differentiate: \(f_k'(x)=3x^2-6kx\), \(f_k''(x)=6x-6k\), and \(f_k'''(x)=6\). 2. Solve \(f_k''(x)=0\): \(6x-6k=0\), so \(x=k\). Since the second derivative is linear with nonzero slope, this is its only zero and it changes sign there. 3. The function value is \(f_k(k)=2-2k^3\), so the unique inflection point is \((k,2-2k^3)\). 4. Since \(f_k''(x)=6(x-k)\), the graph is concave down for \(x<k\) and concave up for \(x>k\). 5. Solve \(f_k'(x)=3x(x-2k)=0\): the critical numbers are \(x=0\) and \(x=2k\). Because \(f_k''(0)=-6k<0\), \((0,2)\) is a local maximum. Because \(f_k''(2k)=6k>0\), \((2k,2-4k^3)\) is a local minimum.

a) Unique inflection point: \((k,2-2k^3)\). b) Concave down for \(x<k\) and concave up for \(x>k\). c) Local maximum: \((0,2)\). Local minimum: \((2k,2-4k^3)\).

- Locate the change in concavity from the second derivative. - A point needs both an x-coordinate and its function value. - Compare the sign of the second derivative on the two sides of its zero. - Use the critical numbers and the second derivative to classify the extrema.

1. Differentiate: \(h_a'(x)=ax^2-4x\), \(h_a''(x)=2ax-4\), and \(h_a'''(x)=2a\). 2. Solve \(h_a''(x)=0\): \(2ax-4=0\), so \(x=\frac2a\). Since \(a>0\), \(h_a'''(x)=2a\ne0\), so the graph changes concavity there. 3. The function value is \(h_a\left(\frac2a\right)=5-\frac{16}{3a^2}\). Thus the inflection point is \(\left(\frac2a,5-\frac{16}{3a^2}\right)\). 4. For \(x<\frac2a\), \(h_a''(x)<0\), so the graph is concave down. For \(x>\frac2a\), \(h_a''(x)>0\), so it is concave up. 5. The derivative factors as \(h_a'(x)=x(ax-4)\). At \(x=0\), \(h_a''(0)=-4<0\), so there is a local maximum. The other critical number is \(x=\frac4a\), where \(h_a''\left(\frac4a\right)=4>0\), so there is a local minimum.

a) \(\left(\frac2a,5-\frac{16}{3a^2}\right)\) b) Concave down for \(x<\frac2a\); concave up for \(x>\frac2a\). c) Local maximum at \(x=0\); local minimum at \(x=\frac4a\).

- Evaluate the function and its first two derivatives at \(x=0\). - Factor out the greatest common power of \(x\) to find the zeros. - Find the second critical number and express its function value in terms of \(t\). - Set the local minimum value equal to \(-4\).

1. The value is \(g_t(0)=0\), so every graph passes through the origin. Differentiate: \(g_t'(x)=3x(x-2t)\), \(g_t''(x)=6x-6t\). 2. At \(x=0\), \(g_t'(0)=0\) and \(g_t''(0)=-6t<0\). Hence the origin is a local maximum by the second derivative test. 3. The factorization \(g_t(x)=x^2(x-3t)\) gives a double zero at \(0\) and a simple zero at \(3t\). 4. The other critical number is \(x=2t\). Here \(g_t''(2t)=6t>0\), so it is a local minimum. Its value is \(g_t(2t)=-4t^3\). Solving \(-4t^3=-4\) with \(t>0\) gives \(t=1\).

a) Local maximum at \((0,0)\) b) \(x=0\), multiplicity \(2\); \(x=3t\), multiplicity \(1\) c) \(t=1\)

- Factor the function to find its zeros. - Solve the first-derivative equation in terms of \(k\). - Use \(k>0\) when classifying the critical points. - Use the second and third derivatives to locate the inflection point.

1. Factor \(f_k(x)=x(x^2-3k^2)\). The x-intercepts are \((0,0)\) and \((\pm\sqrt{3}k,0)\); the y-intercept is \((0,0)\). 2. Since \(f_k'(x)=3(x^2-k^2)\), the critical numbers are \(x=\pm k\). The second derivative is \(f_k''(x)=6x\). Evaluating gives \(f_k''(-k)=-6k<0\) and \(f_k''(k)=6k>0\), so the local maximum is \((-k,2k^3)\) and the local minimum is \((k,-2k^3)\). 3. The second derivative changes from negative to positive at \(x=0\), and \(f_k(0)=0\). Hence \((0,0)\) is the inflection point.

Intercepts: \((0,0)\) and \((\pm\sqrt{3}k,0)\) Local maximum: \((-k,2k^3)\) Local minimum: \((k,-2k^3)\) Inflection point: \((0,0)\)

- Factor the quartic to find its zeros. - Use the sign of the second derivative to classify the critical points. - Substitute parameterized x-values carefully into the original function. - Use the graph's even symmetry to check paired points.

1. Factor: \(g_a(x)=\frac{1}{4}x^2(x^2-4a^2)\). The intercepts are \((0,0)\) and \((\pm2a,0)\). 2. The derivative is \(g_a'(x)=x(x^2-2a^2)\), with critical numbers \(x=0\) and \(x=\pm\sqrt{2}a\). Since \(g_a''(x)=3x^2-2a^2\), \(g_a''(0)=-2a^2<0\) gives a local maximum at \((0,0)\), and \(g_a''(\pm\sqrt{2}a)=4a^2>0\) gives local minima at \((\pm\sqrt{2}a,-a^4)\). 3. The second derivative is zero at \(x=\pm\sqrt{\frac{2}{3}}a\) and changes sign at each because \(a>0\). The inflection points are \((\pm\sqrt{\frac{2}{3}}a,-\frac{5}{9}a^4)\).

Intercepts: \((0,0)\) and \((\pm2a,0)\) Local maximum: \((0,0)\) Local minima: \((\pm\sqrt{2}a,-a^4)\) Inflection points: \((\pm\sqrt{\frac{2}{3}}a,-\frac{5}{9}a^4)\)

- Use the second derivative to find the inflection point in terms of the parameter. - Write both coordinates as expressions involving \(a\). - Solve the \(x\)-coordinate equation for \(a\). - Substitute to eliminate the parameter. - Use the restriction \(a>0\) to determine the domain of the locus.

1. Differentiate: \(g_a'(x)=3x^2+6ax\) and \(g_a''(x)=6x+6a\). 2. Solve \(g_a''(x)=0\): \(x=-a\). Since \(g_a'''(x)=6\ne0\), this value gives an inflection point. 3. Its \(y\)-coordinate is \(g_a(-a)=(-a)^3+3a(-a)^2-2a=2a^3-2a\). 4. From \(x=-a\), obtain \(a=-x\). Substitute into the \(y\)-coordinate: \(y=2(-x)^3-2(-x)=-2x^3+2x\). 5. Because \(a>0\), the inflection-point coordinate satisfies \(x=-a<0\).

\(y=-2x^3+2x\) for \(x<0\)

- Find the critical numbers in terms of \(a\). - Compute the corresponding function values. - Eliminate \(a\) in both cases and compare the equations. - Apply the restriction \(a\ne0\).

1. Differentiate: \(g_a'(x)=3(x^2-a^2)\), so the critical numbers are \(x=a\) and \(x=-a\). Since \(a\ne0\), they are distinct and neither is zero. 2. The second derivative is \(g_a''(x)=6x\). Thus \(g_a''(a)=6a\) and \(g_a''(-a)=-6a\). At the negative critical number the test gives a local maximum; at the positive critical number it gives a local minimum, regardless of the sign of \(a\). 3. At \(x=a\), \(g_a(a)=2-2a^3\), giving \(y=2-2x^3\). At \(x=-a\), \(g_a(-a)=2+2a^3\); substituting \(a=-x\) gives the same equation. 4. Because \(a\ne0\), the locus is \(y=2-2x^3\) for \(x\ne0\).

\(y=2-2x^3\), for \(x\ne0\)

- Factor the function before finding its zeros. - The sign of \(k\) affects the second derivative at each critical number. - Use the first two derivatives to find and classify the local extrema. - Substitute each x-coordinate into the original function to find the point coordinates.

1. Factor \(f_k(x)=x^2(x-3k)\). The zeros are \(x=0\), multiplicity \(2\), and \(x=3k\), multiplicity \(1\). 2. The derivatives are \(f_k'(x)=3x(x-2k)\), \(f_k''(x)=6(x-k)\), and \(f_k'''(x)=6\). The critical numbers are \(0\) and \(2k\). 3. The second derivative values are \(f_k''(0)=-6k\) and \(f_k''(2k)=6k\). For \(k>0\), the point \((0,0)\) is a local maximum and \((2k,-4k^3)\) is a local minimum. For \(k<0\), their types reverse. 4. Since \(f_k''(k)=0\) and \(f_k'''(k)=6\ne0\), the inflection point is \((k,-2k^3)\).

Zeros: \(x=0\), multiplicity \(2\); \(x=3k\), multiplicity \(1\) For \(k>0\): local maximum \((0,0)\); local minimum \((2k,-4k^3)\) For \(k<0\): local minimum \((0,0)\); local maximum \((2k,-4k^3)\) Inflection point: \((k,-2k^3)\)

- Factor before finding the zeros. - Substitute the given point to solve for the parameter. - Use the first derivative for critical points and the second derivative for classification. - Track how the sign of the parameter affects the second derivative.

1. Factor \(g_a(x)=ax^2(x-2)(x+2)\). Since \(a\ne0\), the intercepts are \((-2,0)\), \((0,0)\), and \((2,0)\). 2. From \(g_a(1)=6\), \(-3a=6\), so \(a=-2\). 3. The derivative is \(g_a'(x)=4ax(x^2-2)\), giving critical numbers \(0\) and \(\pm\sqrt{2}\), independent of \(a\). 4. The second derivative is \(g_a''(x)=a(12x^2-8)\). Evaluating gives \(g_a''(0)=-8a\) and \(g_a''(\pm\sqrt{2})=16a\). If \(a>0\), \(x=0\) is a local maximum and \(x=\pm\sqrt{2}\) are minima. If \(a<0\), the types reverse.

a) X-intercepts: \((-2,0)\), \((0,0)\), \((2,0)\) b) \(a=-2\) c) Critical-number x-values: \(0\), \(\pm\sqrt{2}\). For \(a>0\): maximum at \(0\), minima at \(\pm\sqrt{2}\). For \(a<0\): minimum at \(0\), maxima at \(\pm\sqrt{2}\).

- Factor the first derivative to find the critical numbers. - How does the sign of the second derivative depend on \(k\)? - Find the inflection point before eliminating the parameter. - Does the restriction \(k\ne0\) rule out a point on the resulting curve?

1. The derivatives are \(f_k'(x)=3x(x-2k)\) and \(f_k''(x)=6x-6k\). 2. The critical numbers are \(x=0\) and \(x=2k\), giving points \((0,2)\) and \((2k,2-4k^3)\). 3. Since \(f_k''(0)=-6k\) and \(f_k''(2k)=6k\), for \(k>0\) the first point is a local maximum and the second is a local minimum. For \(k<0\) these classifications reverse. 4. The second derivative changes sign at \(x=k\), giving the inflection point \((k,2-2k^3)\). Eliminating \(k\) gives \(y=2-2x^3\), with \(x\ne0\).

a) For \(k>0\), maximum \((0,2)\) and minimum \((2k,2-4k^3)\); for \(k<0\), the classifications reverse. b) \(y=2-2x^3\), with \(x\ne0\).

- Factor the first derivative. - Check how the second derivative at each critical number depends on the sign of \(a\). - Eliminate the parameter from the inflection-point coordinates. - Keep the restriction on \(a\) in the final locus.

1. Differentiate: \(f_a'(x)=3(x-a)(x-3a)\), \(f_a''(x)=6x-12a\). 2. The critical points are \((a,4a^3)\) and \((3a,0)\). Because \(f_a''(a)=-6a\) and \(f_a''(3a)=6a\), when \(a>0\) the first is a local maximum and the second a local minimum. For \(a<0\), they switch roles. 3. The second derivative changes sign at \(x=2a\). The inflection point is \((2a,2a^3)\). 4. With \(x=2a\), its y-coordinate is \(2(x/2)^3=\frac{1}{4}x^3\). Since \(a\ne0\), the point \((0,0)\) is not attained.

a) For \(a>0\), maximum \((a,4a^3)\) and minimum \((3a,0)\); for \(a<0\), the classifications reverse. b) Inflection point \((2a,2a^3)\) lies on \(y=\frac{1}{4}x^3\); \((0,0)\) is not attained.

- Find the derivative and solve for the critical number. - Evaluate the second derivative at that number. - What condition does a point on \(y=5\) satisfy? - Where is the y-intercept, and what gives the tangent slope there?

1. Differentiate: \(f_a'(x)=1-e^{a-x}\), \(f_a''(x)=e^{a-x}\). 2. From \(f_a'(x)=0\), \(e^{a-x}=1\), so the only critical number is \(x=a\). 3. Since \(f_a''(a)=1>0\), the point is a local minimum, at \((a,a+1)\). 4. Requiring \(a+1=5\) gives \(a=4\). 5. At the y-intercept, \(f_a'(0)=1-e^a\). Setting \(1-e^a=1-e\) gives \(a=1\).

a) Local minimum: \((a,a+1)\) b) \(a=4\) c) \(a=1\)

- The exponential factor in the first derivative is never zero. - Evaluate the second derivative at the critical number. - A point on the x-axis has height zero. - Parallel tangents and lines have the same slope.

1. The first derivative is \(g_k'(x)=e^x(k-e^x)\). Its only critical number is \(x=\ln(k)\). 2. Since \(g_k''(x)=ke^x-2e^{2x}\), \(g_k''(\ln(k))=-k^2<0\). Thus \(g_k\) has a local maximum at \(\left(\ln(k),\frac{k^2}{2}-2\right)\). 3. Requiring \(\frac{k^2}{2}-2=0\) and using \(k>0\) gives \(k=2\). 4. At \(x=0\), the tangent slope is \(g_k'(0)=k-1\). Setting \(k-1=3\) gives \(k=4\).

a) Local maximum: \(\left(\ln(k),\frac{k^2}{2}-2\right)\) b) \(k=2\) c) \(k=4\)

- Apply both the product and chain rules. - The exponential factor is always positive. - Solve each resulting quadratic equation. - Use \(k>0\) to verify that the square roots are real.

1. Apply the product and chain rules: \(f_k'(x)=(-x^2+2x+k)e^{-x}\) and \(f_k''(x)=(x^2-4x+2-k)e^{-x}\). 2. Since \(e^{-x}>0\), the critical numbers satisfy \(x^2-2x-k=0\). 3. Solving gives \(x=1\pm\sqrt{1+k}\). 4. The x-coordinates of inflection points satisfy \(x^2-4x+2-k=0\). 5. Solving gives \(x=2\pm\sqrt{2+k}\). Because \(k>0\), the two roots are real and distinct. The quadratic factor in \(f_k''\) changes sign at each root, so both are x-coordinates of inflection points.

Critical numbers: \(x=1\pm\sqrt{1+k}\) Inflection-point x-coordinates: \(x=2\pm\sqrt{2+k}\)

- The nonzero parameter does not affect where the derivative is zero. - Factor the cubic expression before solving it. - Check the sign changes required for local extrema and inflection points. - Decide whether the requested x-values depend on \(a\).

1. Differentiate: \(g_a'(x)=a(1-x^2)e^{-x^2/2}\). 2. Since \(a\neq0\) and the exponential factor is positive, the critical numbers satisfy \(1-x^2=0\). Thus, \(x=\pm1\). 3. Differentiate again: \(g_a''(x)=a(x^3-3x)e^{-x^2/2}\). 4. The x-coordinates of inflection points satisfy \(x(x^2-3)=0\), so \(x=0\) and \(x=\pm\sqrt{3}\). 5. The relevant derivative factors change sign at each listed value, confirming the local extrema and inflection points.

Critical numbers: \(x=-1\) and \(x=1\) Inflection-point x-coordinates: \(x=-\sqrt{3}\), \(x=0\), and \(x=\sqrt{3}\)

- Check whether \(g_k(-x)=g_k(x)\). - The symmetry may suggest where to look for a critical point. - Use the second derivative at that point to check for a minimum. - What must the minimum's height be for it to lie on the x-axis?

1. Since \(g_k(-x)=k(e^{-x}+e^x)-k^2=g_k(x)\), each graph is symmetric about the y-axis. 2. The first derivative is \(g_k'(x)=k(e^x-e^{-x})\). Its only critical number is \(x=0\). 3. The second derivative is \(g_k''(x)=k(e^x+e^{-x})\), so \(g_k''(0)=2k>0\). The second derivative test gives a local minimum. 4. Its height is \(g_k(0)=2k-k^2\). Setting \(k(2-k)=0\) and using \(k>0\) gives \(k=2\).

a) \(g_k(-x)=g_k(x)\), so the graph is symmetric about the y-axis. b) \(k=2\)

- Compare linear growth with exponential decay. - Set the first derivative equal to zero and check whether \(c\) remains in the resulting equation. - The greatest decrease occurs where the first derivative is minimal. - Substitute the maximum time into the original model.

1. Since exponential decay dominates linear growth, \(h_c(t)=\frac{c(t+40)}{e^{0.02t}}\to0\) as \(t\to\infty\). 2. Differentiate: \(h_c'(t)=ce^{-0.02t}(0.2-0.02t)\). The derivative is zero at \(t=10\), independent of \(c\), and changes from positive to negative there. 3. The second derivative is \(h_c''(t)=ce^{-0.02t}(0.0004t-0.024)\). It is zero at \(t=60\). At that time the first derivative reaches its minimum, so the concentration is decreasing most rapidly. 4. Use the maximum at \(t=10\): \(c(50)e^{-0.2}=500\). Thus, \(c=10e^{0.2}\approx12.21\,\text{mg}/(\text{L}\cdot\text{h})\).

a) \(0\,\text{mg/L}\) b) \(t=10\) hours c) \(t=60\) hours d) \(c=10e^{0.2}\approx12.21\,\text{mg}/(\text{L}\cdot\text{h})\)

- Use the leading term to determine end behavior. - Inflection points occur where the second derivative changes sign. - A point on the y-axis has x-coordinate zero. - Evaluate the first derivative at the inflection point to find the tangent slope.

1. The leading term is \(\frac13kx^3\). Therefore, when \(k>0\), \(f_k(x)\to\infty\) as \(x\to\infty\) and \(f_k(x)\to-\infty\) as \(x\to-\infty\). When \(k<0\), these directions reverse. 2. Differentiate twice: \(f_k'(x)=kx^2+2(k+2)x+12\), \(f_k''(x)=2kx+2k+4\). Setting the second derivative equal to zero gives \(2kx+2k+4=0\), so \(x=-1-\frac2k\). Since \(f_k'''(x)=2k\ne0\), this critical point of the first derivative is an inflection point. 3. For the inflection point to lie on the y-axis, \(-1-\frac2k=0\), which gives \(k=-2\). Then \(f_{-2}(x)=-\frac23x^3+12x\), so \(f_{-2}(0)=0\), and the inflection point is \((0, 0)\). Finally, \(f_{-2}'(x)=-2x^2+12\), so the tangent slope at the origin is \(f_{-2}'(0)=12\).

a) For \(k>0\), \(f_k(x)\to\infty\) as \(x\to\infty\) and \(f_k(x)\to-\infty\) as \(x\to-\infty\). For \(k<0\), \(f_k(x)\to-\infty\) as \(x\to\infty\) and \(f_k(x)\to\infty\) as \(x\to-\infty\). b) \(x=-1-\frac2k\) c) \(k=-2\); \(W_{-2}=(0, 0)\); tangent slope \(12\).

- Look for an input that makes the function value independent of \(k\). - Use the product and chain rules to find the first two derivatives. - Eliminate \(k\) from the coordinates of the local maxima.

1. At \(x=0\), \(f_k(0)=0\) for every \(k\), so \((0, 0)\) is common to all graphs. If two distinct parameter values \(k\) and \(m\) give the same function value at \(x\ne0\), then \(e^{1-kx}=e^{1-mx}\), which implies \((k-m)x=0\), a contradiction. Thus, there is no other common point. 2. Differentiate: \(f_k'(x)=(1-kx)e^{1-kx}\) and \(f_k''(x)=(k^2x-2k)e^{1-kx}\). The first derivative is zero at \(x=\frac1k\). Since \(f_k''\left(\frac1k\right)=-k<0\), the point is a local maximum, with \(H=\left(\frac1k, \frac1k\right)\). The second derivative is zero at \(x=\frac2k\). Its sign changes there, so the inflection point is \(W=\left(\frac2k, \frac{2}{ke}\right)\). 3. Every local maximum has equal x- and y-coordinates, so all local maxima lie on \(y=x\).

a) \((0, 0)\) b) \(H=\left(\frac1k, \frac1k\right)\) and \(W=\left(\frac2k, \frac{2}{ke}\right)\) c) \(y=x\)

- Determine where \(\sqrt{x}\) is defined. - Use the zero-product property. - Differentiate and solve for an interior critical number. - What does the second derivative tell you at that number? - Rewrite \(a\sqrt{a}\) using exponents.

1. The square root requires \(x\geq0\), so the domain is \([0,\infty)\). The zeros are \(x=0\) and \(x=a\). 2. For \(x>0\), differentiate: \(h_a'(x)=\frac{a-3x}{2\sqrt{x}}\). 3. The only interior critical number is \(x=\frac{a}{3}\). Since \(h_a''(x)=\frac{-3x-a}{4x\sqrt{x}}\), we have \(h_a''\left(\frac{a}{3}\right)=-\frac{3}{2\sqrt{a/3}}<0\). The second derivative test gives a local maximum. 4. Its y-coordinate is \(\left(a-\frac{a}{3}\right)\sqrt{\frac{a}{3}}=\frac{2a\sqrt{3a}}{9}\). The maximum point is \(\left(\frac{a}{3}, \frac{2a\sqrt{3a}}{9}\right)\). 5. Setting \(\frac{2a\sqrt{3a}}{9}=2\) gives \(a^{3/2}=3^{3/2}\), so \(a=3\).

a) Domain: \([0,\infty)\); zeros: \(x=0\) and \(x=a\). b) Local maximum: \(\left(\frac{a}{3}, \frac{2a\sqrt{3a}}{9}\right)\). c) \(a=3\).

- Examine the function as \(x\) approaches the left endpoint of its domain. - A local extremum can occur where the first derivative is zero. - Use the second derivative to classify the extremum. - Analyze each coordinate of the minimum as a function of \(k\).

1. As \(x\to0^+\), \(\frac{x}{k}\to0\) and \(-\ln(x)\to\infty\). Therefore, \(f_k(x)\to\infty\), so \(x=0\) is a vertical asymptote. 2. Differentiate: \(f_k'(x)=\frac{1}{k}-\frac{1}{x}\). Setting the derivative equal to zero gives \(x=k\). 3. The second derivative is \(f_k''(x)=\frac{1}{x^2}\). Since \(f_k''(k)=\frac{1}{k^2}>0\), the second derivative test gives a local minimum. 4. Its y-coordinate is \(f_k(k)=1-\ln(k)\). Thus, the local minimum is \((k, 1-\ln(k))\). 5. As \(k\) increases, the x-coordinate increases and \(1-\ln(k)\) decreases. The minimum moves to the right and downward.

a) \(x=0\). b) \((k, 1-\ln(k))\). c) It moves to the right and downward as \(k\) increases.

- Find the critical numbers from the first derivative. - Use the second derivative to classify them. - Analyze the sign of the first derivative between its zeros. - Use the parameter restriction \(t>0\) in part c.

1. Differentiate: \(f_t'(x)=x^2-t^2\) and \(f_t''(x)=2x\). 2. The critical numbers satisfy \(x^2-t^2=0\), so \(x=-t\) and \(x=t\). 3. Since \(f_t''(-t)=-2t<0\), there is a local maximum at \(\left(-t,5+\frac{2}{3}t^3\right)\). Since \(f_t''(t)=2t>0\), there is a local minimum at \(\left(t,5-\frac{2}{3}t^3\right)\). 4. The derivative \((x-t)(x+t)\) is positive on \(( -\infty,-t )\) and \((t,\infty)\), and negative on \((-t,t)\). Thus, \(f_t\) increases on the first and third intervals and decreases on the middle interval. 5. A local extremum at \(x=4\) requires \(4=t\) or \(4=-t\). Since \(t>0\), \(t=4\).

a) Local maximum: \(\left(-t,5+\frac{2}{3}t^3\right)\); local minimum: \(\left(t,5-\frac{2}{3}t^3\right)\). b) Increasing on \((-\infty,-t)\) and \((t,\infty)\); decreasing on \((-t,t)\). c) \(t=4\).

- Find the positive local-minimum x-coordinate and the positive inflection-point x-coordinate separately. - Form the ratio only after expressing both coordinates in terms of \(a\). - Simplify the radicals.

1. The derivatives are \(g_a'(x)=4ax^3-4x=4x(ax^2-1)\) and \(g_a''(x)=12ax^2-4\). 2. The positive critical number is \(x_{\min}=\frac{1}{\sqrt{a}}\). Because \(g_a''\left(\frac{1}{\sqrt{a}}\right)=8>0\), the second derivative test gives a local minimum. 3. The positive solution of \(g_a''(x)=0\) is \(x_{\mathrm{infl}}=\frac{1}{\sqrt{3a}}\). The second derivative changes sign there, so it is an inflection-point x-coordinate. 4. Therefore, \(\frac{x_{\min}}{x_{\mathrm{infl}}}=\frac{1/\sqrt{a}}{1/\sqrt{3a}}=\sqrt{3}\), which is independent of \(a\).

\(\frac{x_{\min}}{x_{\mathrm{infl}}}=\sqrt{3}\).

- Factor the function and retain multiplicity. - Use the second derivative to find the inflection point, then eliminate \(a\). - Use the first derivative to make \(x=4\) critical and the second derivative to classify it.

1. Factor \(f_a(x)=x^2(a-x)\). Thus \(x=0\) is a double zero and \(x=a\) is a zero. 2. The derivatives are \(f_a'(x)=2ax-3x^2\) and \(f_a''(x)=2a-6x\). The second derivative changes sign at \(x=\frac{a}{3}\), so the inflection point has \(x=\frac{a}{3}\) and \(y=\frac{2a^3}{27}\). Since \(a=3x\), substitution gives \(y=2x^3\). 3. For \(x=4\) to be critical, \(f_a'(4)=8a-48=0\), so \(a=6\). Then \(f_6''(4)=12-24=-12<0\), so the point is a local maximum.

1) \(x=0\) is a double zero; \(x=a\) is a zero. 2) \(y=2x^3\) 3) \(a=6\); local maximum

- Compare \(g_a(-x)\) with \(g_a(x)\). - Find the critical numbers and use the second derivative test. - Substitute the critical numbers into the original function. - Apply the restriction \(a>0\) when solving for \(a\).

1. Since \(g_a(-x)=(-x)^4-2a(-x)^2=g_a(x)\), every graph is symmetric about the y-axis. 2. The derivative is \(g_a'(x)=4x^3-4ax=4x(x^2-a)\). For \(a>0\), the critical numbers are \(x=0\) and \(x=\pm\sqrt{a}\). 3. Since \(g_a''(x)=12x^2-4a\), \(g_a''(\pm\sqrt{a})=8a>0\), giving local minima at \(\left(-\sqrt{a},-a^2\right)\) and \(\left(\sqrt{a},-a^2\right)\). 4. Setting \(-a^2=-25\) and using \(a>0\) gives \(a=5\).

a) \(g_a(-x)=g_a(x)\), so every graph is symmetric about the y-axis. b) \(\left(-\sqrt{a},-a^2\right)\) and \(\left(\sqrt{a},-a^2\right)\). c) \(a=5\).

- When does a quadratic have no two distinct real zeros? - Which derivative do you use to locate possible inflection points? - What must the y-coordinate of a point on the x-axis equal? - Factor the equation in \(k\) before solving it.

1. Differentiate: \(g_k'(x)=x^2-2kx+6\). Local extrema require the derivative to change sign at a zero. The quadratic has no two distinct real zeros when its discriminant is nonpositive: \(D=(-2k)^2-4(1)(6)=4k^2-24\le0\). Thus \(k^2\le6\), so \(-\sqrt{6}\le k\le\sqrt{6}\). At the boundary values, the derivative has a double zero but does not change sign, so there is still no local extremum. 2. The second derivative is \(g_k''(x)=2x-2k\). Solving \(g_k''(x)=0\) gives \(x=k\). Since \(g_k'''(x)=2\ne0\), this value gives an inflection point. 3. Evaluate the function at \(x=k\): \(g_k(k)=\frac{1}{3}k^3-k^3+6k=-\frac{2}{3}k^3+6k\). Therefore, \(W_k=\left(k, -\frac{2}{3}k^3+6k\right)\). 4. For the point to lie on the x-axis, solve \(-\frac{2}{3}k^3+6k=0\). Factoring gives \(k\left(-\frac{2}{3}k^2+6\right)=0\), so \(k=-3\), \(k=0\), or \(k=3\).

a) \(-\sqrt{6}\le k\le\sqrt{6}\) b) \(W_k=\left(k, -\frac{2}{3}k^3+6k\right)\) c) \(k\in\{-3, 0, 3\}\)

- Factor the first derivative to find the critical numbers. - Use the sign of the second derivative and the sign of \(k\) to classify the extrema. - Find the inflection point from the second derivative. - Check separately what happens when \(k=0\).

1. Differentiate: \(f_k'(x)=3x^2-12kx+9k^2=3(x-k)(x-3k)\), and \(f_k''(x)=6x-12k\). 2. The critical numbers are \(x=k\) and \(x=3k\), with values \(f_k(k)=4k^3\) and \(f_k(3k)=0\). 3. If \(k>0\), then \(f_k''(k)=-6k<0\) and \(f_k''(3k)=6k>0\). Thus \((k,4k^3)\) is a local maximum and \((3k,0)\) is a local minimum. 4. If \(k<0\), the signs reverse, so \((k,4k^3)\) is a local minimum and \((3k,0)\) is a local maximum. 5. If \(k=0\), the critical numbers coincide at \(x=0\) and \(f_0''(0)=0\), so the second derivative test is inconclusive. Since \(f_0(x)=x^3\) increases through \(0\), there is no local extremum; the origin is a stationary inflection point. 6. The equation \(f_k''(x)=0\) gives \(x=2k\). Since \(f_k'''(x)=6\ne0\), this is always an inflection point, with coordinates \((2k,2k^3)\).

For \(k>0\): local maximum \((k,4k^3)\); local minimum \((3k,0)\). For \(k<0\): local minimum \((k,4k^3)\); local maximum \((3k,0)\). For \(k=0\): no local extrema; stationary inflection point \((0,0)\). Inflection point for all \(k\): \((2k,2k^3)\).

- Test the endpoint \(x=0\) for a common point. - Differentiate for \(x>0\) and solve the critical-number equation. - Eliminate \(k\) from the minimum coordinates. - Include the attainable x-values in the locus.

1. Since \(f_k(0)=0\), every graph passes through \((0,0)\). 2. For \(x>0\), \(f_k'(x)=1-\frac{k}{2\sqrt{x}}\). Setting this equal to zero gives \(x=\frac{k^2}{4}\), the only interior critical number. 3. Since \(f_k''(x)=\frac{k}{4x^{3/2}}\) and \(f_k''\left(\frac{k^2}{4}\right)=\frac{2}{k^2}>0\), the second derivative test gives a local minimum. Its point is \(\left(\frac{k^2}{4},-\frac{k^2}{4}\right)\). At the endpoint \(x=0\), nearby values are negative, so the endpoint is not a local minimum. 4. The minimum coordinates satisfy \(y=-x\). Since \(k>0\), every positive \(x\) occurs and \(x=0\) does not. The locus is \(y=-x\) for \(x>0\).

a) \((0,0)\). b) \(\left(\frac{k^2}{4},-\frac{k^2}{4}\right)\). c) \(y=-x\) for \(x>0\).

- Find and classify both critical points. - Substitute the minimum x-coordinate into the function. - Eliminate the parameter and preserve its domain restriction.

1. The derivatives are \(f_a'(x)=3x(x-2a)\) and \(f_a''(x)=6x-6a\). 2. The critical numbers are \(x=0\) and \(x=2a\). Since \(a>0\), \(f_a''(0)=-6a<0\) and \(f_a''(2a)=6a>0\). Thus the local minimum occurs at \(x=2a\). 3. Its y-coordinate is \(f_a(2a)=4-4a^3\), so the minimum is \((2a,4-4a^3)\). 4. Since \(a=\frac{x}{2}\), the locus is \(y=4-\frac{1}{2}x^3\). Because \(a>0\), the domain of the locus is \(x>0\).

a) \((2a,4-4a^3)\). b) \(y=4-\frac{1}{2}x^3\) for \(x>0\).

- Use the second derivative to locate the inflection point. - Verify with the third derivative. - Eliminate the parameter from the coordinates.

1. The derivatives are \(g_k'(x)=x^2+2kx+k^2-1\), \(g_k''(x)=2x+2k\), and \(g_k'''(x)=2\). 2. The inflection point satisfies \(2x+2k=0\), so \(x=-k\). The nonzero third derivative confirms an inflection point. 3. Its y-coordinate is \(g_k(-k)=-\frac{1}{3}k^3+k\). Thus the inflection point is \(\left(-k, -\frac{1}{3}k^3+k\right)\). 4. Since \(k=-x\), substitution gives \(y=\frac{1}{3}x^3-x\).

a) \(\left(-k, -\frac{1}{3}k^3+k\right)\) b) \(y=\frac{1}{3}x^3-x\)

- Find both critical numbers and evaluate the second derivative there. - Express the parameter using the moving critical point's x-coordinate. - Check that the fixed extremum also lies on the resulting curve.

1. The derivative is \(f_a'(x)=3x(x-2a)\), so the critical numbers are \(x=0\) and \(x=2a\). 2. Since \(f_a''(x)=6x-6a\), \(f_a''(0)=-6a\ne0\) and \(f_a''(2a)=6a\ne0\). Both are local extrema; for \(a>0\), \((0,2)\) is a maximum and the moving point is a minimum, with the types reversed when \(a<0\). 3. The fixed extremum is \((0,2)\). The moving extremum has \(x=2a\), so \(a=\frac{x}{2}\). Substituting gives \(y=x^3-3\left(\frac{x}{2}\right)x^2+2=2-\frac{1}{2}x^3\). 4. For \(a\ne0\), the moving points cover every \(x\ne0\), while the fixed extremum gives \(x=0\). Thus the whole curve \(y=2-\frac{1}{2}x^3\) is attained.

For \(a>0\), \((0,2)\) is a local maximum and \((2a,2-4a^3)\) is a local minimum; for \(a<0\), their types reverse. The locus of all extrema is \(y=2-\frac{1}{2}x^3\), for all real \(x\).

- Apply the product and chain rules. - The exponential factor cannot equal zero. - Solve the critical-point condition for \(k\), then substitute. - Check the second derivative at the critical point.

1. The product and chain rules give \(f_k'(x)=e^{0.5x}(0.5k-0.5x-1)\). 2. Since the exponential factor is positive, a critical point satisfies \(0.5k-0.5x-1=0\), or \(k=x+2\). 3. Substituting into the family gives \(y=(x+2-x)e^{0.5x}=2e^{0.5x}\). 4. The second derivative is \(f_k''(x)=e^{0.5x}(0.25k-0.25x-1)\). At each critical number, \(k=x+2\), so \(f_k''(x)=-0.5e^{0.5x}<0\). Every point on the curve is a local maximum of its corresponding family member.

The locus is \(y=2e^{0.5x}\), for all real \(x\); every point is a local maximum of the corresponding function.

- Set the first derivative equal to zero and use the nonzero critical number. - Eliminate the parameter between the extremum coordinates. - Use the second derivative to find the inflection point. - Simplify the ratio and check whether the parameter remains.

1. Differentiate: \(f_k'(x)=2kx-3x^2=x(2k-3x)\). The nonzero critical number is \(x_{\mathrm{ext}}=\frac{2k}{3}\), with height \(\frac{4k^3}{27}\). Since \(f_k''\left(\frac{2k}{3}\right)=-2k\), the point is a local maximum for \(k>0\) and a local minimum for \(k<0\). 2. From \(x=\frac{2k}{3}\), obtain \(k=\frac{3x}{2}\). Substitution into \(y=\frac{4k^3}{27}\) gives \(y=\frac{1}{2}x^3\). Since \(k\ne0\), the locus has \(x\ne0\). 3. Since \(f_k''(x)=2k-6x\), the inflection-point x-coordinate is \(x_{\mathrm{infl}}=\frac{k}{3}\). The third derivative is \(-6\ne0\), confirming a concavity change. Therefore, \(\frac{x_{\mathrm{ext}}}{x_{\mathrm{infl}}}=\frac{2k/3}{k/3}=2\).

1. \(\left(\frac{2k}{3},\frac{4k^3}{27}\right)\); local maximum for \(k>0\), local minimum for \(k<0\). 2. \(y=\frac{1}{2}x^3\) for \(x\ne0\). 3. \(x_{\mathrm{infl}}=\frac{k}{3}\) and \(\frac{x_{\mathrm{ext}}}{x_{\mathrm{infl}}}=2\).

- Check how the parameter term behaves at a common point. - Factor the function to analyze its zeros. - Use the first and second derivatives to classify the extrema. - Eliminate \(k\) between the coordinates of the local minimum.

1. Since \(f_k(0)=0\), all graphs pass through the origin. At any fixed \(x\ne0\), the term \(-kx\) changes as \(k\) changes, so no other common point exists. 2. Factoring gives \(f_k(x)=x(x^2-k)\). The three distinct real zeros \(0\) and \(\pm\sqrt{k}\) exist exactly when \(k>0\). 3. The derivative is \(f_k'(x)=3x^2-k\). For \(k>0\), its critical numbers are \(x=\pm\sqrt{\frac{k}{3}}\). Because \(f_k''(x)=6x\), \(f_k''\left(\sqrt{\frac{k}{3}}\right)=6\sqrt{\frac{k}{3}}>0\) gives a local minimum at \(\left(\sqrt{\frac{k}{3}},-\frac{2k}{3}\sqrt{\frac{k}{3}}\right)\). 4. At the local minimum, \(k=3x^2\), so \(y=-\frac{2k}{3}x=-2x^3\). Since \(k>0\), the locus has \(x>0\).

a) The origin is the only common point. b) \(k>0\). c) \(\left(\sqrt{\frac{k}{3}},-\frac{2k}{3}\sqrt{\frac{k}{3}}\right)\). d) \(y=-2x^3\) for \(x>0\).

- Relate the multiplicity of a zero to how the graph meets the x-axis. - Use the product rule and factor the derivative. - Use the second derivative to classify each extremum. - A point on the y-axis has x-coordinate \(0\).

1. Since \(x=a\) is a zero of even multiplicity and \(a\ne-1\), the graph is tangent to the x-axis there. 2. Differentiate: \(g_a'(x)=2(x-a)(x+1)+(x-a)^2=(x-a)(3x+2-a)\), with critical numbers \(x=a\) and \(x=\frac{a-2}{3}\). 3. The extremum points are \(P_1=(a,0)\) and \(P_2=\left(\frac{a-2}{3},\frac{4}{27}(a+1)^3\right)\). Since \(g_a''(x)=6x+2-4a\), \(g_a''(a)=2(a+1)\) and \(g_a''\left(\frac{a-2}{3}\right)=-2(a+1)\). For \(a>-1\), \(P_1\) is a minimum and \(P_2\) a maximum; for \(a<-1\), the types reverse. 4. When \(a>-1\), the maximum is \(P_2\), which lies on the y-axis when \(a=2\), giving \((0,4)\). For \(a<-1\), the maximum is \(P_1\), whose x-coordinate cannot be zero. Thus \(a=2\) is the only solution.

a) The graph is tangent at \(x=a\) because this is a zero of even multiplicity. b) \(P_1=(a,0)\) and \(P_2=\left(\frac{a-2}{3},\frac{4}{27}(a+1)^3\right)\). For \(a>-1\), \(P_1\) is a minimum and \(P_2\) a maximum; for \(a<-1\), the types reverse. c) \(a=2\), giving maximum \((0,4)\).

- Factor the first derivative to find the critical numbers. - Treat the cases \(k>0\), \(k<0\), and \(k=0\) separately. - Eliminate \(k\) using the critical-number equation. - Check directly what happens when the second derivative test is inconclusive.

1. Differentiate: \(f_k'(x)=x(x^2-2k)\) and \(f_k''(x)=3x^2-2k\). 2. For \(k>0\), the nonzero critical numbers are \(x=\pm\sqrt{2k}\). There \(f_k''=4k>0\), so both are local minima. Eliminating \(k=\frac{x^2}{2}\) gives \(y=\frac{1}{4}x^4-\frac{x^4}{2}=-\frac{1}{4}x^4\). 3. For \(k<0\), \(f_k''(0)=-2k>0\), so the origin is a local minimum. For \(k=0\), \(f_0''(0)=0\), and the second derivative test is inconclusive; however, \(f_0(x)=\frac{1}{4}x^4\geq0\), so the origin is also a local minimum. 4. Every nonzero x-value on \(y=-\frac{1}{4}x^4\) is attained by choosing \(k=\frac{x^2}{2}>0\), and the origin is attained for \(k\leq0\). Thus the locus is the entire curve \(y=-\frac{1}{4}x^4\).

\(y=-\frac{1}{4}x^4\) for all real \(x\).

- A graph is tangent to the x-axis when both the function value and derivative are zero. - Find both critical numbers and evaluate the second derivative at each. - Eliminate the parameter using the x-coordinate of the other extremum.

1. Differentiate: \(f_a'(x)=3(x-a)(x+a)\), so the critical numbers are \(x=a\) and \(x=-a\). Since \(f_a''(x)=6x\), \(f_a''(a)=6a\ne0\) and \(f_a''(-a)=-6a\ne0\), so both are local extrema. For \(a>0\), the point at \(a\) is a minimum and the point at \(-a\) a maximum; the types reverse for \(a<0\). 2. At \(x=a\), \(f_a(a)=0\) and \(f_a'(a)=0\), so the graph is tangent to the x-axis at \((a,0)\). 3. The other extremum occurs at \(x=-a\), with height \(f_a(-a)=4a^3\). Its coordinates are \((-a,4a^3)\). 4. Since \(a=-x\), the locus is \(y=4(-x)^3=-4x^3\), with \(x\ne0\) because \(a\ne0\).

a) The graph is tangent to the x-axis at \((a,0)\), which is a local minimum for \(a>0\) and a local maximum for \(a<0\). b) The other extremum is \((-a,4a^3)\); its locus is \(y=-4x^3\) for \(x\ne0\).

- A local extremum must satisfy a first-derivative condition. - Use the second derivative to classify it. - Count the roots of the second derivative and check whether its sign changes.

1. The first derivative is \(f_k'(x)=x^3+kx+2\). Requiring \(f_k'(-1)=0\) gives \(-1-k+2=0\), so \(k=1\). 2. The second derivative is \(f_k''(x)=3x^2+k\). Since \(f_1''(-1)=4>0\), the point is a local minimum. 3. Inflection points require a sign change in \(f_k''(x)=3x^2+k\). If \(k<0\), the equation \(3x^2+k=0\) has two distinct roots, and the second derivative changes sign at both, so there are two inflection points. 4. If \(k>0\), the second derivative is always positive, so there are no inflection points. If \(k=0\), its only zero is \(x=0\), but there is no sign change. Therefore, there are no inflection points for \(k\ge0\).

a) \(k=1\); local minimum b) \(k<0\): two inflection points; \(k\ge0\): no inflection points

- Count the real roots of the first derivative using its discriminant. - Set the second derivative equal to zero. - Substitute the inflection x-coordinate into the original function.

1. The first three derivatives are \(g_a'(x)=x^2+2ax+4\), \(g_a''(x)=2x+2a\), and \(g_a'''(x)=2\). 2. Critical points satisfy \(x^2+2ax+4=0\). This quadratic has two distinct real roots when its discriminant is positive: \(4a^2-16>0\), or \(|a|>2\). At those roots, the second derivative is nonzero, so both are local extrema. 3. An inflection point occurs where \(g_a''(x)=0\). Thus \(2x+2a=0\), so \(x=-a\). Because \(g_a'''(x)=2\ne0\), this is an inflection point. 4. Its y-coordinate is \(g_a(-a)=-\frac{1}{3}a^3+a^3-4a=\frac{2}{3}a^3-4a\).

a) \(a<-2\) or \(a>2\) b) \(\left(-a, \frac{2}{3}a^3-4a\right)\)

- Factor out the common exponential factor. - Solve equations of the form \(e^x=c\) with logarithms. - Use the second derivative test to classify the extremum. - Substitute the critical x-coordinate into the original function.

1. Factor: \(f_a(x)=e^x(e^x-a)\). Since \(e^x\ne0\), the zero occurs at \(x=\ln(a)\). 2. The first derivative is \(f_a'(x)=e^x(2e^x-a)\). The critical-number equation gives \(x=\ln\left(\frac{a}{2}\right)\). 3. The second derivative is \(f_a''(x)=4e^{2x}-ae^x\), and \(f_a''\left(\ln\left(\frac{a}{2}\right)\right)=\frac{a^2}{2}>0\). Thus the critical point is a local minimum. 4. Its y-coordinate is \(\left(\frac{a}{2}\right)^2-a\left(\frac{a}{2}\right)=-\frac{a^2}{4}\).

Zero: \(x=\ln(a)\). Local minimum: \(\left(\ln\left(\frac{a}{2}\right),-\frac{a^2}{4}\right)\).

- Compare two functions with different parameter values to find common points. - Use the product rule and chain rule. - Factor the first derivative before solving. - Eliminate \(k\) from the maximum coordinates.

1. Every graph passes through \((0,0)\). For \(x\ne0\), equality of two members with distinct parameters would require \(e^{-k_1x}=e^{-k_2x}\), which is impossible unless \(k_1=k_2\). The origin is the only common point. 2. Differentiate: \(f_k'(x)=x(2-kx)e^{-kx}\). The critical numbers are \(x=0\) and \(x=\frac{2}{k}\). 3. The second derivative is \(f_k''(x)=(k^2x^2-4kx+2)e^{-kx}\). Since \(f_k''(0)=2>0\), \((0,0)\) is a local minimum. Since \(f_k''\left(\frac{2}{k}\right)=-2e^{-2}<0\), the local maximum is \(\left(\frac{2}{k},\frac{4}{k^2e^2}\right)\). 4. At the maximum, \(x=\frac{2}{k}>0\), so \(k=\frac{2}{x}\). Substitution gives the locus \(y=\frac{x^2}{e^2}\) for \(x>0\).

a) The origin is the only common point. b) Local minimum: \((0,0)\); local maximum: \(\left(\frac{2}{k},\frac{4}{k^2e^2}\right)\). c) \(y=\frac{x^2}{e^2}\) for \(x>0\).

- Differentiate the product twice, simplifying after each derivative. - The exponential factor is always positive, so another factor controls the sign of the second derivative. - An inflection point requires a change in concavity, not merely a zero of the second derivative. - Eliminate the parameter from the inflection-point coordinates to obtain the locus.

1. By the product rule, \(g_a'(x)=(a-1-x)e^x\). 2. Differentiate again: \(g_a''(x)=(a-2-x)e^x\). 3. Since \(e^x>0\), the sign of \(g_a''(x)\) is the sign of \(a-2-x\). Thus \(g_a''(x)>0\) when \(x<a-2\) and \(g_a''(x)<0\) when \(x>a-2\). 4. The concavity changes at \(x=a-2\), so the inflection point is \(\left(a-2,2e^{a-2}\right)\). 5. At an inflection point, let its x-coordinate be \(x=a-2\). Then its y-coordinate is \(2e^x\), so the locus is \(y=2e^x\).

a) \(g_a''(x)=(a-2-x)e^x\) b) Inflection point: \(\left(a-2,2e^{a-2}\right)\); concave up on \(( -\infty,a-2)\) and concave down on \((a-2,\infty)\) c) \(y=2e^x\)

- The exponential factor never equals zero. - Use the product rule and the second derivative test. - Eliminate \(k\) from the minimum coordinates. - Set two family members equal to test for common points.

1. Since \(e^x\ne0\), the zero is \(x=k\). The derivative is \(f_k'(x)=e^x(x-k+1)\), so the critical number is \(x=k-1\). Also \(f_k''(x)=e^x(x-k+2)\), and \(f_k''(k-1)=e^{k-1}>0\). Thus the local minimum is \((k-1,-e^{k-1})\). 2. At the minimum, \(x=k-1\), so \(k=x+1\); substituting into the height gives \(y=-e^x\) for all real \(x\). 3. If two distinct family members shared a point, then \((x-k)e^x=(x-m)e^x\). Since \(e^x\ne0\), this would give \(k=m\). Thus distinct graphs have no common points.

1. Zero: \(x=k\); local minimum: \((k-1,-e^{k-1})\). 2. \(y=-e^x\) for all real \(x\). 3. Distinct graphs have no common points.

- Use the domain condition when solving for a zero. - Apply the product rule and the derivative of \(\ln(x)\). - Express the maximum y-coordinate using its x-coordinate. - Compare two family members to test for common points.

1. Since \(x>0\), \(f_k(x)=0\) reduces to \(k-\ln(x)=0\), so the zero is \(x=e^k\). 2. By the product rule, \(f_k'(x)=x(2k-1-2\ln(x))\). Because \(x>0\), its only zero is at \(x=e^{k-1/2}\). 3. The second derivative is \(f_k''(x)=2k-3-2\ln(x)\), so at the critical number \(f_k''=-2<0\). The local maximum is \(\left(e^{k-1/2},\frac{1}{2}e^{2k-1}\right)\). 4. Since \(x^2=e^{2k-1}\), the locus is \(y=\frac{1}{2}x^2\) for \(x>0\). 5. If distinct family members shared a point, then \(x^2(k_1-\ln(x))=x^2(k_2-\ln(x))\). This would imply \(x=0\), which is outside the domain, or \(k_1=k_2\). Thus there are no common points.

a) \(x=e^k\). b) Local maximum: \(\left(e^{k-1/2},\frac{1}{2}e^{2k-1}\right)\); locus: \(y=\frac{1}{2}x^2\) for \(x>0\). c) Distinct graphs have no common points.

- Set the first derivative equal to zero. - Use the sign of the second derivative to classify the critical point. - Replace the parameter with the maximum x-coordinate. - A common point must have a function value independent of \(a\).

1. Differentiate: \(f_a'(x)=\frac{a}{x}-1\). Its only zero is \(x=a\), and \(f_a''(a)=-\frac{1}{a}<0\), so there is a local maximum at \((a,a\ln(a)-a)\). 2. Replacing \(a\) with the maximum x-coordinate gives the locus \(y=x\ln(x)-x\) for \(x>0\). 3. A point common to all graphs must have \(\ln(x)=0\), so \(x=1\). Then \(f_a(1)=-1\), giving the common point \((1,-1)\).

a) \((a,a\ln(a)-a)\). b) \(y=x\ln(x)-x\) for \(x>0\). c) \((1,-1)\).

- Use the product rule and chain rule for successive derivatives. - Set the second derivative equal to zero. - Verify that the candidate is an inflection point. - Compare the x- and y-coordinates of the inflection point.

1. By the product and chain rules, \(f_t'(x)=(1+tx)e^{tx}\) and \(f_t''(x)=(2t+t^2x)e^{tx}\). 2. Since \(e^{tx}>0\), the inflection-point equation gives \(2t+t^2x=0\), so \(x=-\frac{2}{t}\). 3. The third derivative is \(f_t'''(x)=(3t^2+t^3x)e^{tx}\). At \(x=-\frac{2}{t}\), it equals \(t^2e^{-2}>0\), confirming an inflection point. 4. The y-coordinate is \(f_t\left(-\frac{2}{t}\right)=-\frac{2}{te^2}\). Thus the inflection point is \(\left(-\frac{2}{t}, -\frac{2}{te^2}\right)\). 5. Since \(y=\frac{1}{e^2}x\), all inflection points lie on the line \(y=\frac{x}{e^2}\).

a) \(\left(-\frac{2}{t}, -\frac{2}{te^2}\right)\) b) \(y=\frac{x}{e^2}\)

- Use the chain rule when differentiating \(e^{2x}\). - Solve the exponential equation using logarithms. - Use the critical-point equation to simplify the y-coordinate. - An inflection point requires a change in concavity.

1. The derivatives are \(g_a'(x)=1-2ae^{2x}\) and \(g_a''(x)=-4ae^{2x}\). 2. The only critical number satisfies \(e^{2x}=\frac{1}{2a}\), so \(x=-\frac{1}{2}\ln(2a)\). At this number, \(g_a''(x)=-2<0\), proving that it is a local maximum. 3. At the maximum, \(ae^{2x}=\frac{1}{2}\), so the ordinate is \(y=x-\frac{1}{2}\). Since \(a>0\) covers every real critical x-coordinate, the locus is the whole line. 4. Because \(g_a''(x)=-4ae^{2x}<0\) everywhere, concavity never changes and there are no inflection points.

a) Local maximum: \(\left(-\frac{1}{2}\ln(2a),-\frac{1}{2}\ln(2a)-\frac{1}{2}\right)\). b) \(y=x-\frac{1}{2}\) for all real \(x\). c) The second derivative is always negative, so there are no inflection points.

- Use the product rule and chain rule. - The exponential factor is always positive. - Use the second derivative test to classify the critical point. - Eliminate \(a\) from the maximum coordinates.

1. By the product rule, \(f_a'(x)=(1-a-x)e^{-x}\). Since \(e^{-x}>0\), the only critical number is \(x=1-a\). 2. The second derivative is \(f_a''(x)=(x+a-2)e^{-x}\), and \(f_a''(1-a)=-e^{a-1}<0\), so this point is a local maximum. 3. Its y-coordinate is \(f_a(1-a)=e^{a-1}\), giving the maximum \((1-a,e^{a-1})\). 4. Since \(x=1-a\), \(a=1-x\). Substitution gives the locus \(y=e^{-x}\), for all real \(x\).

a) \((1-a,e^{a-1})\). b) \(y=e^{-x}\) for all real \(x\).

- Set the first derivative equal to zero. - Use the one-to-one property of the exponential function. - Evaluate the second derivative at the critical number. - Replace \(\ln(k)\) using the minimum x-coordinate.

1. Differentiate: \(g_k'(x)=ke^x-1\). Its only zero satisfies \(e^x=\frac{1}{k}\), so \(x=-\ln(k)\). 2. Since \(g_k''(x)=ke^x\), we have \(g_k''(-\ln(k))=1>0\). Therefore, this is the unique local minimum. 3. Its y-coordinate is \(g_k(-\ln(k))=1+\ln(k)\). Since \(x=-\ln(k)\), the locus is \(y=1-x\) for all real \(x\).

a) The unique local minimum is \((-\ln(k),1+\ln(k))\). b) \(y=1-x\) for all real \(x\).

- Solve the first-derivative equation by combining the exponential terms. - Check when the resulting right side is positive. - Use the second derivative test for the extremum. - For inflection points, solve the second-derivative equation and verify with the third derivative.

1. The derivatives are \(f_a'(x)=e^x-ae^{-x}\) and \(f_a''(x)=e^x+ae^{-x}\). 2. The critical-number equation is \(e^{2x}=a\). If \(a>0\), its only solution is \(x=\frac{1}{2}\ln(a)\); if \(a<0\), there is no solution. 3. For \(a>0\), the second derivative at the critical number is \(2\sqrt{a}>0\), giving a local minimum at \(\left(\frac{1}{2}\ln(a),2\sqrt{a}\right)\). 4. For \(a<0\), \(f_a''(x)=0\) is equivalent to \(e^{2x}=-a\), which has the unique solution \(x=\frac{1}{2}\ln(-a)\). 5. The third derivative is \(f_a'''(x)=e^x-ae^{-x}\). At this x-coordinate it equals \(2\sqrt{-a}\ne0\), so concavity changes and the graph has exactly one inflection point.

1. For \(a>0\): one local extremum; for \(a<0\): no local extrema. 2. For \(a>0\), the point \(\left(\frac{1}{2}\ln(a),2\sqrt{a}\right)\) is a local minimum. 3. For \(a<0\), exactly one inflection point at \(x=\frac{1}{2}\ln(-a)\).

- Find the first three derivatives. - Solve the first- and second-derivative equations separately on \(x>0\). - Track how the sign of \(k\) affects whether each equation has a real solution. - Use logarithm properties to simplify the y-coordinate.

1. The first three derivatives are \(g_k'(x)=\frac{k}{x}+2x\), \(g_k''(x)=-\frac{k}{x^2}+2\), and \(g_k'''(x)=\frac{2k}{x^3}\). 2. A critical point satisfies \(2x^2=-k\). Since \(x>0\), this has exactly one solution when \(k<0\): \(x=\sqrt{-\frac{k}{2}}\). At this point, \(g_k''(x)=4>0\), so it is a local minimum. 3. An inflection point satisfies \(2x^2=k\). This has exactly one positive solution when \(k>0\): \(x=\sqrt{\frac{k}{2}}\). Since \(g_k'''(x)\ne0\) there, the concavity changes. 4. The y-coordinate is \(k\ln\left(\sqrt{\frac{k}{2}}\right)+\frac{k}{2}=\frac{k}{2}\left(\ln\left(\frac{k}{2}\right)+1\right)\).

1. For \(k<0\), there is exactly one local minimum. For \(k>0\), there is exactly one inflection point. 2. For \(k>0\), the inflection point is \(\left(\sqrt{\frac{k}{2}}, \frac{k}{2}\left(\ln\left(\frac{k}{2}\right)+1\right)\right)\).

- Compare \(f_k(-x)\) with \(f_k(x)\). - A natural logarithm is zero when its argument is \(1\). - Use the quotient rule for the second derivative. - Tangency to the x-axis requires both a zero function value and zero slope.

1. Since \(f_k(-x)=f_k(x)\), each graph is symmetric about the y-axis. Also, \(x^2+k\to\infty\) as \(x\to\pm\infty\), so \(f_k(x)\to\infty\). 2. Zeros satisfy \(x^2+k=1\), or \(x^2=1-k\). Thus there are two zeros when \(0<k<1\), one zero at \(x=0\) when \(k=1\), and no zeros when \(k>1\). 3. The second derivative is \(f_k''(x)=\frac{2(k-x^2)}{(x^2+k)^2}\). It is zero at \(x=\pm\sqrt{k}\), and its sign changes at both points. Therefore, there are exactly two inflection points. 4. Tangency at the origin requires \(f_k(0)=0\) and \(f_k'(0)=0\). The first condition gives \(\ln(k)=0\), so \(k=1\); the derivative condition is then also satisfied.

1. Symmetric about the y-axis; \(f_k(x)\to\infty\) as \(x\to\pm\infty\) 2. \(0<k<1\): two zeros; \(k=1\): one zero; \(k>1\): no zeros 3. \(x=\pm\sqrt{k}\) 4. \(k=1\)

- The argument of a logarithm must be positive. - Examine the function as the argument approaches \(0^+\). - Find the critical point from the first derivative. - Study the sign of the second derivative on the entire domain.

1. The logarithm requires \(a-x^2>0\), so the maximal domain is \((-\sqrt{a},\sqrt{a})\). As \(x\to(-\sqrt{a})^{+}\) or \(x\to(\sqrt{a})^{-}\), the argument approaches \(0^+\), so \(g_a(x)\to-\infty\). Thus, \(x=\pm\sqrt{a}\) are vertical asymptotes. 2. The first derivative is \(g_a'(x)=\frac{-2x}{a-x^2}\), which is zero only at \(x=0\). 3. The second derivative is \(g_a''(x)=-\frac{2(a+x^2)}{(a-x^2)^2}\), which is negative throughout the domain. Therefore, \((0, \ln(a))\) is a local and absolute maximum. 4. Because the second derivative is always negative and never zero, the graph has no inflection points.

1. Domain: \((-\sqrt{a},\sqrt{a})\); \(g_a(x)\to-\infty\) at both endpoints 2. Maximum: \((0, \ln(a))\) 3. No inflection points

- Rewrite the rational expression before differentiating. - Use the first and second derivative tests. - Eliminate \(a\) from each extremum's coordinates. - Keep track of the sign restrictions implied by \(a>0\).

1. Rewrite \(f_a(x)=x+2a+\frac{4a^2}{x}\). Then \(f_a'(x)=1-\frac{4a^2}{x^2}\) and \(f_a''(x)=\frac{8a^2}{x^3}\). 2. The critical-number equation gives \(x=\pm2a\). Since \(a>0\), \(f_a''(2a)=\frac{1}{a}>0\) and \(f_a''(-2a)=-\frac{1}{a}<0\). 3. Thus the local minimum is \((2a,6a)\), and the local maximum is \((-2a,-2a)\). 4. For the minima, \(x=2a>0\), so \(y=6a=3x\). Their locus is \(y=3x\) for \(x>0\). 5. For the maxima, \(x=-2a<0\), and \(y=-2a=x\). Their locus is \(y=x\) for \(x<0\).

a) Local minimum: \((2a,6a)\); local maximum: \((-2a,-2a)\). b) \(y=3x\) for \(x>0\). c) \(y=x\) for \(x<0\).

- Find and classify both critical numbers. - Use \(k>0\) when applying the second derivative test. - Replace \(k\) using the maximum x-coordinate. - Include the x-values implied by \(k>0\).

1. Differentiate: \(f_k'(x)=x^2-k^2\) and \(f_k''(x)=2x\). The critical numbers are \(x=\pm k\). 2. Since \(f_k''(-k)=-2k<0\), the point at \(x=-k\) is the only local maximum. Its height is \(f_k(-k)=\frac{2}{3}k^3+k\), so the maximum is \(\left(-k,\frac{2}{3}k^3+k\right)\). The other critical number \(x=k\) has \(f_k''(k)=2k>0\) and is a local minimum. 3. At the maximum, \(x=-k<0\), so \(k=-x\). Substitution gives \(y=-\frac{2}{3}x^3-x\), for \(x<0\).

a) \(\left(-k,\frac{2}{3}k^3+k\right)\). b) \(y=-\frac{2}{3}x^3-x\) for \(x<0\).

- Use the derivative of \(\ln(x)\). - Apply the second derivative test. - Solve the minimum x-coordinate for \(t\). - Use logarithm properties to simplify the locus equation.

1. Differentiate: \(f_t'(x)=tx-\frac{1}{x}\). The critical-point equation gives \(tx^2=1\), so \(x=\frac{1}{\sqrt{t}}\). 2. Since \(f_t''(x)=t+\frac{1}{x^2}\), we have \(f_t''\left(\frac{1}{\sqrt{t}}\right)=2t>0\). The point is a local minimum by the second derivative test. 3. Its y-coordinate is \(\frac{1}{2}-\ln(t^{-1/2})=\frac{1}{2}+\frac{1}{2}\ln(t)\), so the minimum is \(\left(\frac{1}{\sqrt{t}},\frac{1}{2}+\frac{1}{2}\ln(t)\right)\). 4. At the minimum, \(t=\frac{1}{x^2}\). Thus the locus is \(y=\frac{1}{2}-\ln(x)\) for \(x>0\).

a) \(\left(\frac{1}{\sqrt{t}},\frac{1}{2}+\frac{1}{2}\ln(t)\right)\). b) \(y=\frac{1}{2}-\ln(x)\) for \(x>0\).

- Find the inflection points using the second derivative. - Use the symmetry of the graph about the y-axis. - Two vectors are perpendicular when their dot product is zero.

1. Differentiate twice: \(f_k'(x)=-2kxe^{-kx^2}\) and \(f_k''(x)=(4k^2x^2-2k)e^{-kx^2}\). 2. Since the exponential factor is positive, inflection-point candidates satisfy \(4k^2x^2-2k=0\), giving \(x=\pm\frac{1}{\sqrt{2k}}\). The second derivative changes sign at both values, so they are inflection points. 3. Their y-coordinate is \(e^{-1/2}=\frac{1}{\sqrt{e}}\). Thus the inflection points are \(\left(\frac{1}{\sqrt{2k}}, \frac{1}{\sqrt{e}}\right)\) and \(\left(-\frac{1}{\sqrt{2k}}, \frac{1}{\sqrt{e}}\right)\). 4. The angle at the origin is right when the position vectors to the two inflection points have dot product zero. Therefore, \(-\frac{1}{2k}+\frac{1}{e}=0\). 5. Solving gives \(2k=e\), so \(k=\frac{e}{2}\).

\(k=\frac{e}{2}\)

- Find the inflection points in terms of \(a\). - Use the horizontal segment between them as the base. - The height is the absolute value of their y-coordinate. - Express both sides of the final equation as powers with the same exponent.

1. Differentiate: \(f_a'(x)=4x^3-2ax\) and \(f_a''(x)=12x^2-2a\). The inflection-point equation gives \(x=\pm\sqrt{\frac{a}{6}}\). Since the third derivative is nonzero at both values, they are inflection points. 2. Their y-coordinate is \(\frac{a^2}{36}-\frac{a^2}{6}=-\frac{5a^2}{36}\). Thus the horizontal base of the triangle has length \(2\sqrt{\frac{a}{6}}\), and its height is \(\frac{5a^2}{36}\). 3. The area is \(A=\frac{1}{2}\cdot2\sqrt{\frac{a}{6}}\cdot\frac{5a^2}{36}=\frac{5a^{5/2}}{36\sqrt{6}}\). 4. Setting \(A=5\) gives \(a^{5/2}=36\sqrt{6}=6^{5/2}\). Because \(a>0\), \(a=6\).

\(a=6\)

- Find the inflection points in terms of \(k\). - Use their equal y-coordinates to choose a horizontal base. - Write the triangle’s area before solving for \(k\).

1. Differentiate twice: \(f_k'(x)=-2kxe^{-kx^2}\) and \(f_k''(x)=2k(2kx^2-1)e^{-kx^2}\). 2. Since the exponential factor is positive, the candidates have \(x=\pm\frac{1}{\sqrt{2k}}\). The factor \(2kx^2-1\) changes sign at both values, so they are inflection points. Their common y-coordinate is \(e^{-1/2}=\frac{1}{\sqrt{e}}\). 3. The triangle has base \(\frac{2}{\sqrt{2k}}\) and height \(\frac{1}{\sqrt{e}}\). Thus \(A=\frac{1}{2}\cdot\frac{2}{\sqrt{2k}}\cdot\frac{1}{\sqrt{e}}=\frac{1}{\sqrt{2ke}}\). 4. Setting \(\frac{1}{\sqrt{2ke}}=\frac{1}{\sqrt{2e}}\) gives \(k=1\).

\(k=1\)

- Set the second derivative equal to zero. - Verify the inflection point using the third derivative. - Eliminate \(a\) from the point coordinates. - Include the restriction implied by \(a\neq0\).

1. Differentiate: \(f_a''(x)=6x-12a\). The inflection-point equation gives \(x=2a\). 2. Since \(f_a'''(x)=6\neq0\), this point is an inflection point. 3. Its y-coordinate is \(f_a(2a)=8a^3-24a^3+2=-16a^3+2\). 4. From \(x=2a\), obtain \(a=\frac{x}{2}\). Substitution gives \(y=-16\left(\frac{x}{2}\right)^3+2=-2x^3+2\). 5. Because \(a\neq0\), the attainable x-values satisfy \(x\neq0\).

\(y=-2x^3+2\) for \(x\neq0\)

- Compare \(f_a(-x)\) with \(f_a(x)\). - Factor the first derivative and separate the cases \(a\leq0\) and \(a>0\). - Use the second derivative to classify the extrema. - Eliminate \(a\) from the minimum coordinates.

1. Since \(f_a(-x)=f_a(x)\), every graph is symmetric about the y-axis. 2. Differentiate: \(f_a'(x)=4x(ax^2-1)\) and \(f_a''(x)=12ax^2-4\). 3. If \(a\leq0\), the only critical number is \(x=0\). Since \(f_a''(0)=-4<0\), the origin is the only local extremum and is a local maximum. 4. If \(a>0\), the critical numbers are \(x=0\) and \(x=\pm\frac{1}{\sqrt{a}}\). The origin is a local maximum because \(f_a''(0)=-4<0\), while \(f_a''\left(\pm\frac{1}{\sqrt{a}}\right)=8>0\), so the other two points are local minima. 5. At a local minimum, \(x^2=\frac{1}{a}\), so \(a=\frac{1}{x^2}\). Substituting gives \(y=\frac{x^4}{x^2}-2x^2=-x^2\). Because these minima have \(x\neq0\), the attained locus is \(y=-x^2\) for \(x\neq0\).

a) Every graph is symmetric about the y-axis. b) For \(a\leq0\), there is one local maximum at \((0, 0)\). For \(a>0\), there is a local maximum at \((0, 0)\) and two local minima at \(\left(\pm\frac{1}{\sqrt{a}}, -\frac{1}{a}\right)\). c) \(y=-x^2\) for \(x\neq0\)

- Compare logarithmic growth with quadratic growth. - Use the derivative of \(\ln x\) and the domain restriction. - Apply the second derivative test. - How can you describe the set of maximum points without referring to the parameter?

1. As \(x\to0^+\), \(\ln x\to-\infty\) while \(-\frac{1}{2}x^2\to0\), so \(f_a(x)\to-\infty\). As \(x\to\infty\), the negative quadratic term dominates the logarithmic term, so \(f_a(x)\to-\infty\). 2. Differentiate: \(f_a'(x)=\frac{a}{x}-x\). The critical-point equation gives \(x^2=a\), so \(x=\sqrt{a}\). 3. Since \(f_a''(x)=-\frac{a}{x^2}-1<0\), this point is a local maximum. Its y-coordinate is \(\frac{a}{2}\ln a-\frac{a}{2}\), giving the local maximum \(\left(\sqrt{a}, \frac{a}{2}(\ln a-1)\right)\). 4. At the maximum, \(a=x^2\). Therefore, \(y=\frac{x^2}{2}(\ln(x^2)-1)=x^2\ln x-\frac{1}{2}x^2\), with \(x>0\).

a) \(\lim_{x\to0^+}f_a(x)=-\infty\) and \(\lim_{x\to\infty}f_a(x)=-\infty\) b) \(\left(\sqrt{a}, \frac{a}{2}(\ln a-1)\right)\) c) \(y=x^2\ln x-\frac{1}{2}x^2\) for \(x>0\)

- Factor the function to find its zeros. - Compare \(f_a(-x)\) with \(f_a(x)\). - Use the first and second derivatives to locate and classify extrema. - Determine when \(\pm\sqrt{a}\) are distinct from zero. - For the final blueprint, distinguish the degenerate case \(a=0\) from \(a>0\); the derivative has a different zero structure in those two cases.

1. Factor \(f_a(x)=x(x^2-a)\). Graph p has only the zero \(x=0\), so \(a=0\). Graph q has zeros \(-1, 0, 1\), so \(a=1\). Graph r has zeros \(-2, 0, 2\), so \(a=4\). 2. Since \(f_a(-x)=-f_a(x)\), every graph is symmetric about the origin. 3. The first derivative is \(f_a'(x)=3x^2-a\), so the critical numbers are \(x=\pm\sqrt{\frac{a}{3}}\). The second derivative is \(f_a''(x)=6x\). 4. The local maximum is \(\left(-\sqrt{\frac{a}{3}}, \frac{2a}{3}\sqrt{\frac{a}{3}}\right)\), and the local minimum is \(\left(\sqrt{\frac{a}{3}}, -\frac{2a}{3}\sqrt{\frac{a}{3}}\right)\). 5. The zeros are \(0\) and \(\pm\sqrt{a}\). They are three distinct real numbers exactly when \(a>0\). 6. For \(a=0\), \(f_0(x)=x^3\): it is increasing on \(\mathbb{R}\), has no local extremum, has a stationary inflection point at \((0,0)\), and runs from \(-\infty\) to \(\infty\). 7. For \(a>0\), the graph increases on \(( -\infty,-\sqrt{a/3})\), decreases on \((-\sqrt{a/3},\sqrt{a/3})\), and increases on \((\sqrt{a/3},\infty)\). It has the local maximum and minimum from part c, a nonstationary inflection point at \((0,0)\), and the same cubic end behavior.

a) p: \(a=0\); q: \(a=1\); r: \(a=4\) b) Symmetric about the origin c) Local maximum: \(\left(-\sqrt{\frac{a}{3}}, \frac{2a}{3}\sqrt{\frac{a}{3}}\right)\); local minimum: \(\left(\sqrt{\frac{a}{3}}, -\frac{2a}{3}\sqrt{\frac{a}{3}}\right)\) d) \(a>0\) e) \(a=0\): increasing on \(\mathbb{R}\), no extrema, stationary inflection at \((0,0)\), cubic end behavior. For \(a>0\): increase → local maximum at \(x=-\sqrt{a/3}\) → decrease through the inflection at \(x=0\) → local minimum at \(x=\sqrt{a/3}\) → increase, with left end down and right end up.

- Compare where the two displayed curves reach their peaks. - Exponential decay eventually dominates linear growth. - Set the first derivative equal to zero for the maximum. - Check whether the parameter cancels from the critical-point equation. - The most rapid decrease occurs where the slope is smallest.

1. The two displayed curves appear to reach their maximum at the same time, near day \(10\). 2. Since \(te^{-0.1t}\to0\) as \(t\to\infty\), \(f_k(t)\to100\) signups per day. 3. The first derivative is \(f_k'(t)=ke^{-0.1t}(1-0.1t)\). Because \(k>0\) and the exponential factor is positive, the derivative is zero at \(t=10\), independent of \(k\). The second derivative there is negative, so this is the maximum. 4. For \(k=150\), \(f_{150}(10)=1500e^{-1}+100\approx651.82\), or about \(652\) signups per day. 5. The rate decreases most rapidly where the first derivative is minimal. The second derivative is \(f_k''(t)=ke^{-0.1t}(0.01t-0.2)\), which is zero at \(t=20\). The third derivative is positive there, so the first derivative has a minimum at \(t=20\).

a) The maximum occurs at the same time for both displayed curves, near day \(10\). b) \(100\) signups per day c) Day \(10\) d) \(652\) signups per day e) Day \(20\)

- What prevents the extrema's input coordinates from changing with the parameter? - What checks are needed before a stationary point can be classified? - How would you find both the slope and a point needed for the requested tangent?

1. The first derivative is \(f_a'(x)=ax(2-x)e^{-x}\). Since \(a>0\) and the exponential is positive, the critical numbers \(x=0\) and \(x=2\) do not depend on \(a\). 2. The second derivative is \(f_a''(x)=a(x^2-4x+2)e^{-x}\). Since \(f_a''(0)=2a>0\) and \(f_a''(2)=-2ae^{-2}<0\), \(x=0\) is a local minimum and \(x=2\) a local maximum. 3. The maximum value is \(f_a(2)=4ae^{-2}\). Setting this equal to \(4\) gives \(a=e^2\). 4. For \(a=10\), \(f_{10}(1)=\frac{10}{e}\) and \(f_{10}'(1)=\frac{10}{e}\). Thus the tangent is \(y-\frac{10}{e}=\frac{10}{e}(x-1)\), or \(y=\frac{10}{e}x\); it passes through the origin.

a) \(x=0\): local minimum; \(x=2\): local maximum. b) \(a=e^2\) c) \(y=\frac{10}{e}x\); it passes through the origin.

- Use the first two derivatives to locate and classify the points. - Treat \(k\) as a constant when differentiating. - Eliminate \(k\) separately for each locus. - Use \(k>0\) to determine the attainable x-values.

1. Differentiate: \(f_k'(x)=3x(x-2k)\) and \(f_k''(x)=6x-6k\). 2. Because \(k>0\), the first derivative changes from positive to negative at \(x=0\) and from negative to positive at \(x=2k\). Thus the local maximum is \((0, 0)\), and the local minimum is \((2k, -4k^3)\). 3. Since \(f_k''(x)=6(x-k)\) changes sign at \(x=k\), the inflection point is \((k, -2k^3)\). 4. All local maxima form the single point \((0, 0)\). For the local minima, \(x=2k>0\), so \(k=\frac{x}{2}\) and \(y=-\frac{1}{2}x^3\). For the inflection points, \(x=k>0\), so \(y=-2x^3\).

a) Local maximum: \((0, 0)\); local minimum: \((2k, -4k^3)\); inflection point: \((k, -2k^3)\) b) Maximum locus: \(\{(0, 0)\}\); minimum locus: \(y=-\frac{1}{2}x^3\) for \(x>0\); inflection-point locus: \(y=-2x^3\) for \(x>0\) c) The displayed curves and loci agree with these equations.

- What conditions must hold at a differentiable local minimum? - How could you distinguish a minimum from a different kind of stationary point? - What relationship between the two coordinates would describe the entire set of minima?

1. The derivative is \(g_a'(x)=e^x-a\). It is zero at \(x=\ln a\). 2. Since \(g_a''(\ln a)=e^{\ln a}=a>0\), this critical number gives a local minimum. 3. Its y-coordinate is \(g_a(\ln a)=a-a\ln a\), so the point is \((\ln a,a-a\ln a)\). 4. Since \(a=e^x\) at the minimum, the curve is \(y=e^x(1-x)\).

a) \((\ln a,a-a\ln a)\) b) \(y=e^x(1-x)\)

- What derivative conditions distinguish the two extrema? - How could one equation describe the extrema over all allowed parameter values? - What can the original algebraic form tell you about the zeros? - How could you test whether a proposed common point belongs to every graph?

1. The derivative is \(g_k'(x)=x^2-k^2\), with critical numbers \(x=\pm k\). 2. Since \(g_k''(x)=2x\), we have \(g_k''(-k)=-2k<0\) and \(g_k''(k)=2k>0\). Thus the local maximum is \((-k,\frac{2}{3}k^3)\) and the local minimum is \((k,-\frac{2}{3}k^3)\). 3. At the minimum, \(x=k\), giving \(y=-\frac{2}{3}x^3\). At the maximum, \(x=-k\), giving the same equation. Because \(k>0\), the attained locus has \(x\ne0\). 4. Factoring \(g_k(x)=x(\frac{1}{3}x^2-k^2)\) gives three zeros: \(x=0\) and \(x=\pm\sqrt{3} k\). 5. Since \(g_k(0)=0\) for every \(k\), all graphs pass through \((0,0)\). For \(x\ne0\), values change with \(k\), so no other point is common.

a) Local maximum \((-k,\frac{2}{3}k^3)\); local minimum \((k,-\frac{2}{3}k^3)\). b) \(y=-\frac{2}{3}x^3\), for \(x\ne0\). c) Three zeros: \(x=0\), \(x=\pm\sqrt{3} k\). d) \((0,0)\)

- Find the maximum of the given function. - Compare its exact maximum height with the marked graph maxima. - A multiplier outside the function creates a vertical scale.

1. The derivative of \(h\) is \(h'(x)=(2-x)e^{-0.5x}\). It changes from positive to negative at \(x=2\), so the maximum occurs there. Its value is \(h(2)=\frac{4}{e}\), which matches graph q. 2. All three graphs have the same zero and maximum x-coordinate, so they differ by a vertical scale factor. Their marked maximum heights are \(\frac{2}{e}\), \(\frac{4}{e}\), and \(\frac{6}{e}\). 3. Graph p has half the height of q, so its equation is \(xe^{-0.5x}\). Graph r has \(1.5\) times the height of q, so its equation is \(3xe^{-0.5x}\). 4. The family is \(f_k(x)=kxe^{-0.5x}\), where \(k>0\).

a) Graph q b) p: \(xe^{-0.5x}\); r: \(3xe^{-0.5x}\) c) \(f_k(x)=kxe^{-0.5x}\), \(k>0\)

- Compare the horizontal positions of the two displayed peaks. - Use the product rule and chain rule. - Find the zero of the first derivative and notice which factors can affect it. - Substitute the maximum time into the model. - The most rapid decrease occurs where the slope is minimal.

1. The two displayed curves reach their peaks at the same horizontal position, about \(t=4\) hours. 2. The first derivative is \(c_k'(t)=k(1-0.25t)e^{-0.25t}\). The positive factors do not affect its zeros, so the maximum occurs at \(t=4\) hours, independent of \(k\). 3. The maximum value is \(c_k(4)=\frac{4k}{e}\). Setting this equal to \(4\) gives \(k=e\). 4. Since exponential decay dominates linear growth, \(\lim_{t\to\infty}c_k(t)=0\). 5. The concentration decreases most rapidly where the first derivative is minimal. The second derivative is \(c_k''(t)=k(0.0625t-0.5)e^{-0.25t}\), which is zero at \(t=8\). It changes from negative to positive there, so the slope has a minimum at \(t=8\).

a) The peak time appears unchanged, at about \(4\) hours. b) \(4\) hours c) \(k=e\) d) \(0\,\text{mg/L}\) e) \(8\) hours

- Substitute the given point into the function to determine \(k\). - Factor the function to find its zeros and their multiplicities. - Find the critical numbers from the first derivative. - Use the second derivative and the sign of \(k\) to classify each critical point.

1. From \(f_k(2)=4\), \(4k-8=4\), giving \(k=3\). 2. Factor: \(f_k(x)=x^2(k-x)\). Thus, \(x=0\) is a double zero and \(x=k\) is a simple zero because \(k\ne0\). 3. The derivative is \(f_k'(x)=x(2k-3x)\), with critical numbers \(0\) and \(\frac{2k}{3}\). The second derivative is \(f_k''(x)=2k-6x\), so \(f_k''(0)=2k\) and \(f_k''(\frac{2k}{3})=-2k\). 4. The corresponding points are \((0,0)\) and \((\frac{2k}{3},\frac{4k^3}{27})\). If \(k>0\), the second derivative test gives a minimum at the first and a maximum at the second. If \(k<0\), the types reverse.

a) \(k=3\) b) \(x=0\), multiplicity \(2\); \(x=k\), multiplicity \(1\) c) If \(k>0\): minimum at \((0,0)\) and maximum at \((\frac{2k}{3},\frac{4k^3}{27})\). If \(k<0\): maximum at \((0,0)\) and minimum at \((\frac{2k}{3},\frac{4k^3}{27})\).

- Factor the function to find its zeros. - Use the first derivative to find critical numbers and the second derivative to classify them. - Eliminate \(t\) from the coordinates of the local minimum. - Write and solve an inequality using the local maximum value.

1. Factor: \(f_t(x)=x(\frac{1}{3}x^2-t^2)\), giving zeros \(0\) and \(\pm\sqrt{3}t\). 2. Since \(f_t'(x)=x^2-t^2\), the critical numbers are \(x=\pm t\). The second derivative is \(f_t''(x)=2x\). At \(-t\), \(f_t''(-t)=-2t<0\), so there is a local maximum at \((-t,\frac{2}{3}t^3)\). At \(t\), \(f_t''(t)=2t>0\), so there is a local minimum at \((t,-\frac{2}{3}t^3)\). 3. For the minima, \(x=t>0\); eliminating \(t\) gives the locus \(y=-\frac{2}{3}x^3\), with \(x>0\). 4. The maximum height exceeds \(18\) when \(\frac{2}{3}t^3>18\), or \(t>3\).

a) \(x=-\sqrt{3}t,0,\sqrt{3}t\) b) Local maximum: \((-t,\frac{2}{3}t^3)\); local minimum: \((t,-\frac{2}{3}t^3)\) c) \(y=-\frac{2}{3}x^3\), with \(x>0\) d) \(t>3\)

- Factor both functions and compare their zeros. - Find the critical numbers of \(q_a\) and use the second derivative test. - Eliminate \(a\) from the coordinates of the local minimum. - Remember to reverse the inequality when dividing by a negative number.

1. Factor: \(p_a(x)=x(x-2a)\) and \(q_a(x)=x^2(\frac{x}{a}-2)\). Both have zeros \(0\) and \(2a\). 2. The derivatives are \(q_a'(x)=x(\frac{3x}{a}-4)\) and \(q_a''(x)=\frac{6x}{a}-4\). The critical numbers are \(0\) and \(\frac{4a}{3}\). Since \(q_a''(0)=-4<0\), the origin is a local maximum. Since \(q_a''(\frac{4a}{3})=4>0\), the local minimum is \((\frac{4a}{3},-\frac{32a^2}{27})\). 3. From \(x=\frac{4a}{3}\), \(a=\frac{3x}{4}\). Substituting gives \(y=-\frac{2}{3}x^2\), where \(x>0\). 4. The minimum is below \(-32\) when \(-\frac{32a^2}{27}<-32\), or \(a^2>27\). Since \(a>0\), this gives \(a>3\sqrt{3}\).

a) Both have zeros \(x=0\) and \(x=2a\). b) Local minimum: \((\frac{4a}{3},-\frac{32a^2}{27})\) c) \(y=-\frac{2}{3}x^2\), with \(x>0\) d) \(a>3\sqrt{3}\)

- How can you rewrite the expression to find its zeros? - Separate the cases \(a>0\), \(a=0\), and \(a<0\) when classifying the critical point at zero. - Evaluate the second derivative at every critical number before deciding whether the test works. - Check whether each possible inflection point has a change of concavity.

1. Factoring gives \(f_a(x)=x^2(x^2-2a)\). There is always a zero at \(x=0\); when \(a>0\), there are also zeros at \(x=\pm\sqrt{2a}\). 2. The derivatives are \(f_a'(x)=4x(x^2-a)\) and \(f_a''(x)=4(3x^2-a)\). 3. For \(a>0\), the critical numbers are \(0\) and \(\pm\sqrt{a}\). Because \(f_a''(0)=-4a<0\), there is a local maximum at \((0,0)\). Because \(f_a''(\pm\sqrt{a})=8a>0\), there are local minima at \((\pm\sqrt{a},-a^2)\). 4. For \(a<0\), the only critical number is \(0\), and \(f_a''(0)=-4a>0\). The second derivative test gives a local minimum at \((0,0)\). 5. For \(a=0\), the only critical number is \(0\), and \(f_0''(0)=0\), so the second derivative test is inconclusive. Since \(f_0'(x)=4x^3\) changes from negative to positive at \(0\), there is a local minimum at \((0,0)\). 6. Inflection points require a sign change in \(f_a''=4(3x^2-a)\). This occurs at \(x=\pm\sqrt{a/3}\) when \(a>0\); when \(a\le0\), there are none.

Zeros: \(x=0\); when \(a>0\), also \(x=\pm\sqrt{2a}\). For \(a>0\): local maximum \((0,0)\); local minima \((\pm\sqrt{a},-a^2)\). For \(a\le0\): local minimum \((0,0)\). Inflection-point x-values: \(\pm\sqrt{a/3}\) when \(a>0\); none when \(a\le0\).

- Factor out common factors after using the product rule. - The exponential factor is never zero. - Set each remaining factor of the first derivative equal to zero. - Solve the quadratic from the second derivative, then check its sign around each root.

1. Since \(e^x>0\), the only zero is \(x=k\), with multiplicity \(2\). 2. The first derivative is \(g_k'(x)=(x-k)(x-k+2)e^x\), so the critical numbers are \(k\) and \(k-2\). 3. The second derivative is \(g_k''(x)=[(x-k)^2+4(x-k)+2]e^x\). At \(x=k\), \(g_k''(k)=2e^k>0\), giving a local minimum at \((k,0)\). At \(x=k-2\), \(g_k''(k-2)=-2e^{k-2}<0\), giving a local maximum at \((k-2,4e^{k-2})\). 4. The second derivative is zero when \((x-k)^2+4(x-k)+2=0\), or \(x=k-2\pm\sqrt{2}\). Its quadratic factor changes sign at both distinct roots, so both are inflection-point x-values.

Zero: \(x=k\), multiplicity \(2\) Local minimum: \((k,0)\) Local maximum: \((k-2,4e^{k-2})\) Inflection-point x-values: \(x=k-2\pm\sqrt{2}\)

- A smooth join requires both position and slope to match. - Check the two endpoints separately. - Substitute \(x=1\) to determine the parameter. - Use the second derivative and verify a change in concavity.

1. A smooth connection requires matching function values and slopes. The derivative is \(h_k'(x)=4kx^3-12kx^2+(8k+1)x\). At \(x=0\), \(h_k(0)=0\) and \(h_k'(0)=0\), matching the x-axis. At \(x=2\), \(h_k(2)=2\) and \(h_k'(2)=2\), matching the value and slope of \(y=2x-2\). 2. At \(x=1\), \(h_k(1)=k-4k+4k+0.5=k+0.5\). Setting this equal to \(1\) gives \(k=0.5\). 3. For \(k=1\), \(h_1''(x)=12x^2-24x+9\). Solving \(h_1''(x)=0\) gives \(x=0.5\) and \(x=1.5\). The third derivative is \(h_1'''(x)=24x-24\), which is nonzero at both values, so both are inflection points. Their coordinates are \(W_1=(0.5, 0.6875)\) and \(W_2=(1.5, 1.6875)\).

a) \(h_k(0)=0\), \(h_k'(0)=0\), \(h_k(2)=2\), and \(h_k'(2)=2\), so the values and slopes match at both connections. b) \(k=0.5\) c) The inflection points are \((0.5, 0.6875)\) and \((1.5, 1.6875)\).

- Treat \(k\) as a positive constant. - Remember that an equation \(x^2=c\) may have two solutions, and verify a change in concavity at each candidate. - The derivative value is the tangent slope. - Set the slope expression equal to the required value and solve for \(k\).

1. Differentiate: \(f_k'(x)=2kxe^{1-x^2}\) and \(f_k''(x)=2ke^{1-x^2}(1-2x^2)\). 2. Since the exponential factor is positive, \(f_k'(x)=0\) only at \(x=0\). Because \(f_k''(0)=2ke>0\), the unique local extremum is the local minimum \((0, -ke)\). 3. Inflection-point candidates satisfy \(1-2x^2=0\), so \(x=\pm\frac{1}{\sqrt{2}}\). Since \(1-2x^2\) changes sign at both values, both are x-coordinates of inflection points. 4. At either x-value, \(f_k(x)=-ke^{1/2}=-k\sqrt{e}\). Thus, the inflection points are \(\left(\pm\frac{1}{\sqrt{2}}, -k\sqrt{e}\right)\). 5. At \(x=\frac{1}{\sqrt{2}}\), the slope is \(k\sqrt{2e}\). Setting this equal to \(1\) gives \(k=\frac{1}{\sqrt{2e}}\).

a) Local minimum: \((0, -ke)\) b) \(\left(\pm\frac{1}{\sqrt{2}}, -k\sqrt{e}\right)\) c) \(k=\frac{1}{\sqrt{2e}}\)

- Find each graph's local maximum in terms of \(k\), then eliminate the parameter. - For part b, hold \(x\) fixed and view the expression as a function of \(k\). - Find the greatest y-value the family can attain at that fixed x-coordinate.

1. Differentiate with respect to \(x\): \(g_k'(x)=e^{k-x}(1-x+k)\). The only critical number is \(x=k+1\). Since \(g_k''(k+1)=-e^{-1}<0\), the graph has a local maximum there. Its coordinates are \(\left(k+1, k+\frac1e\right)\). Since \(x=k+1\), we have \(k=x-1\), so the maxima lie on \(y=x-1+\frac1e\). 2. Fix \(x\) and vary \(k\). Let \(t=k-x\). Then \(g_k(x)=x+t-te^t\). As a function of \(t\), its derivative is \(1-(t+1)e^t\). This derivative is positive for \(t<0\), zero at \(t=0\), and negative for \(t>0\). Also, the expression approaches \(-\infty\) as \(t\to\pm\infty\). Therefore, its absolute maximum occurs at \(t=0\), or \(k=x\), and the maximum value is \(x\). Thus, for each fixed \(x\), the values attained by the family are exactly \((-\infty, x]\). A point lies on no graph exactly when \(y>x\).

a) \(y=x-1+\frac1e\) b) All points \((x, y)\) such that \(y>x\).

- Examine the end behavior as \(x\to\pm\infty\). - Find the inflection-point x-values in terms of \(k\), then substitute them into \(g_k\). - Use the graph's symmetry to find the tangent intercepts efficiently. - Write the triangle's area in terms of \(k\), then set it equal to \(2\).

1. Since \(k>0\), \(-kx^2\to-\infty\) as \(x\to\pm\infty\). Therefore, \(\lim_{x\to\pm\infty}e^{-kx^2}=0\), so \(y=0\) is a horizontal asymptote. 2. Differentiate: \(g_k'(x)=-2kxe^{-kx^2}\) and \(g_k''(x)=(4k^2x^2-2k)e^{-kx^2}\). 3. Because the exponential factor is positive, \(g_k''(x)=0\) when \(4k^2x^2-2k=0\). Thus \(x=\pm\frac{1}{\sqrt{2k}}\). The factor changes sign at both values, so both are inflection points. 4. At either inflection point, \(g_k\left(\pm\frac{1}{\sqrt{2k}}\right)=e^{-1/2}=\frac{1}{\sqrt{e}}\). 5. Let \(a=\frac{1}{\sqrt{2k}}\). At \(x=a\), the tangent slope is \(-\frac{\sqrt{2k}}{\sqrt{e}}\). This tangent meets the x-axis at \(x=2a=\frac{\sqrt{2}}{\sqrt{k}}\). By symmetry, the other x-intercept is \(-\frac{\sqrt{2}}{\sqrt{k}}\), so the triangle's base is \(\frac{2\sqrt{2}}{\sqrt{k}}\). 6. The two tangent lines meet on the y-axis at height \(\frac{2}{\sqrt{e}}\). Therefore, \(A(k)=\frac{1}{2}\cdot\frac{2\sqrt{2}}{\sqrt{k}}\cdot\frac{2}{\sqrt{e}}=\frac{2\sqrt{2}}{\sqrt{ke}}\). 7. Set the area equal to \(2\): \(\frac{2\sqrt{2}}{\sqrt{ke}}=2\). Then \(ke=2\), so \(k=\frac{2}{e}\).

a) \(\lim_{x\to\pm\infty}g_k(x)=0\), so \(y=0\) is a horizontal asymptote. b) The inflection points have y-coordinate \(\frac{1}{\sqrt{e}}\). c) \(k=\frac{2}{e}\)

- The radicand must be nonnegative. - A product is zero when at least one factor is zero. - Before differentiating at \(x=-2\), determine when that value is in the domain and whether it is an endpoint. - Apply the product and chain rules. - Use the first- and second-derivative conditions for an interior local minimum. - Concavity is determined by the sign of the second derivative.

1. The square root requires \(x+k\geq0\), so the domain is \([-k,\infty)\). 2. The zeros are \(x=-k\) and \(x=\frac{k}{2}\). 3. Differentiate: \(f_k'(x)=\frac{6x+3k}{2\sqrt{x+k}}\) for \(x>-k\). 4. For the point with x-coordinate \(-2\) to be on the graph, \(k\geq2\). If \(k=2\), then \(x=-2\) is the left endpoint, and nearby function values are negative, so the endpoint is a local maximum, not a local minimum. 5. If \(k>2\), then \(x=-2\) is an interior point. A local minimum there requires \(f_k'(-2)=0\), so \(-12+3k=0\) and \(k=4\). 6. The second derivative is \(f_k''(x)=\frac{6x+9k}{4(x+k)^{3/2}}\). For \(k=4\) and \(x=-2\), this is positive, confirming a local minimum. 7. For any \(k>0\) and \(x>-k\), the denominator is positive and \(6x+9k>6(-k)+9k=3k>0\). Therefore, \(f_k''(x)>0\), so the graph is concave up.

a) Domain: \([-k,\infty)\); zeros: \(x=-k\) and \(x=\frac{k}{2}\) b) \(k=4\) c) \(f_k''(x)=\frac{6x+9k}{4(x+k)^{3/2}}>0\) for \(x>-k\).

- Use the second derivative to locate an inflection point and check its sign on both sides. - Analyze the discriminant of the first derivative to determine when it has fewer than two distinct real zeros. - Interpret the sign of the first derivative in the cost context.

1. Differentiate: \(K'(x) = 0.3x^2 - 3x + c\), \(K''(x) = 0.6x - 3\), and \(K'''(x) = 0.6\). 2. Solve \(K''(x) = 0\): \(0.6x - 3 = 0\), so \(x = 5\). Because \(K''(x) < 0\) for \(x < 5\) and \(K''(x) > 0\) for \(x > 5\), the graph changes from concave down to concave up at \(x = 5\), independently of \(c\). 3. Interior local extrema occur where \(K'(x)\) changes sign. If the discriminant of \(K'(x) = 0.3x^2 - 3x + c\) is negative, then \(K'(x) > 0\) for all \(x\). If the discriminant is zero, the derivative has one double zero and does not change sign. If the discriminant is positive, the two distinct real zeros have sum \(10\), so at least one is positive; each simple zero changes the sign of the derivative and gives an interior local extremum. Thus, there are no interior local extrema exactly when the discriminant is nonpositive. 4. Compute \(D = (-3)^2 - 4(0.3)c = 9 - 1.2c\). Solve \(9 - 1.2c \le 0\): \(c \ge 7.5\). 5. For \(c \ge 7.5\), \(K'(x) \ge 0\) for all \(x\), with at most one isolated zero. Thus, total cost never decreases as production increases; there is no interior production level at which total cost changes from increasing to decreasing or vice versa.

a) The inflection point occurs at \(x = 5\), where the graph changes from concave down to concave up. b) \(c \ge 7.5\) c) Total cost does not decrease as production rises; it has no interior local maximum or minimum.

- Find the two critical numbers in terms of \(a\). - Use the second derivative and consider the sign of \(a\). - Determine when the two critical numbers coincide. - Use the second and third derivatives to locate the inflection point.

1. Differentiate: \(g_a'(x)=\frac{3}{2}x(x-2a)\), \(g_a''(x)=3x-3a\), and \(g_a'''(x)=3\). 2. The critical numbers are \(x=0\) and \(x=2a\), with values \(g_a(0)=2\) and \(g_a(2a)=2-2a^3\). 3. If \(a>0\), \(g_a''(0)=-3a<0\) and \(g_a''(2a)=3a>0\). Thus \((0,2)\) is a local maximum and \((2a,2-2a^3)\) is a local minimum. 4. If \(a<0\), the signs reverse, giving a local minimum at \((0,2)\) and local maximum at \((2a,2-2a^3)\). 5. If \(a=0\), the critical numbers coincide at \(x=0\), where \(g_0''(0)=0\); the second derivative test is inconclusive. Since \(g_0(x)=\frac{1}{2}x^3+2\) increases through \(0\) and \(g_0'''(0)=3\ne0\), \((0,2)\) is a stationary inflection point and not an extremum. 6. For all \(a\), \(g_a''(x)=0\) gives \(x=a\). Since \(g_a'''=3\ne0\), this is an inflection point, at \((a,2-a^3)\).

For \(a>0\): local maximum \((0,2)\); local minimum \((2a,2-2a^3)\). For \(a<0\): local minimum \((0,2)\); local maximum \((2a,2-2a^3)\). For \(a=0\): stationary inflection point \((0,2)\), with no local extrema. Inflection point for all \(a\): \((a,2-a^3)\). The point on the y-axis is a local maximum when \(a>0\).

- Use symmetry when finding the two local maxima. - Eliminate \(a\) between the coordinates of a maximum. - Use the second derivative to locate the inflection points. - Simplify the ratio and check whether \(a\) cancels.

1. Differentiate: \(g_a'(x)=x\left(a-\frac{x^2}{3}\right)\). The nonzero critical numbers are \(x=\pm\sqrt{3a}\). Since \(g_a''(x)=a-x^2\), \(g_a''(\pm\sqrt{3a})=-2a<0\), giving maxima \(\left(\pm\sqrt{3a},\frac{3a^2}{4}\right)\). 2. At a maximum, \(x^2=3a\), so \(a=\frac{x^2}{3}\). Substitution gives \(y=\frac{3a^2}{4}=\frac{x^4}{12}\). Since \(a>0\), the locus has \(x\ne0\). 3. The second derivative changes sign at \(x=\pm\sqrt{a}\); there \(y_{\mathrm{infl}}=\frac{5a^2}{12}\). Therefore, \(\frac{y_{\max}}{y_{\mathrm{infl}}}=\frac{3a^2/4}{5a^2/12}=\frac{9}{5}\).

1. \(\left(\pm\sqrt{3a},\frac{3a^2}{4}\right)\). 2. \(y=\frac{1}{12}x^4\) for \(x\ne0\). 3. \(y_{\mathrm{infl}}=\frac{5a^2}{12}\) and \(\frac{y_{\max}}{y_{\mathrm{infl}}}=\frac{9}{5}\).

- Rewrite \(\frac{k}{x}\) using a negative exponent before differentiating. - At a critical point, use the derivative equation to replace \(k\). - Use the distance formula and square both sides. - A substitution for \(x^2\) turns the resulting equation into a quadratic.

1. Differentiate: \(f_k'(x)=2x-\frac{k}{x^2}\) and \(f_k''(x)=2+\frac{2k}{x^3}>0\). The critical-number equation gives \(2x^3=k\), so \(x=\sqrt[3]{\frac{k}{2}}\). Thus the critical point is a local minimum. 2. At this point, \(k=2x^3\), so \(y=x^2+\frac{2x^3}{x}=3x^2\). Therefore, the local minimum is \(\left(\sqrt[3]{\frac{k}{2}}, 3\left(\frac{k}{2}\right)^{2/3}\right)\), and the locus is \(y=3x^2\) for \(x>0\). 3. On the locus, the squared distance from the origin is \(x^2+y^2=x^2+9x^4\). Setting this equal to \(10\) gives \(9x^4+x^2-10=0\). 4. Let \(u=x^2\). Then \(9u^2+u-10=(u-1)(9u+10)=0\). Since \(x>0\), \(u=1\) and \(x=1\). From \(k=2x^3\), \(k=2\).

a) \(\left(\sqrt[3]{\frac{k}{2}}, 3\left(\frac{k}{2}\right)^{2/3}\right)\) b) \(y=3x^2\) for \(x>0\) c) \(k=2\)

- Isolate the term containing the parameter to find common points. - Apply the critical-point condition at \(x=4\), then use the second derivative. - A stationary inflection point has zero first and second derivatives. - Examine the number of zeros of the second derivative.

1. Rewrite the family as \(f_a(x)=\frac{1}{3}x^3-x+a\left(x-\frac{x^2}{2}\right)\). A point is independent of \(a\) when \(x-\frac{x^2}{2}=0\), so \(x=0\) or \(x=2\). The common points are \((0, 0)\) and \(\left(2, \frac{2}{3}\right)\). 2. Differentiate: \(f_a'(x)=x^2-ax+a-1\). The condition \(f_a'(4)=0\) gives \(16-4a+a-1=0\), so \(a=5\). Since \(f_5''(x)=2x-5\) and \(f_5''(4)=3>0\), the point at \(x=4\) is a local minimum. 3. A stationary inflection point satisfies \(f_a'(x)=0\) and \(f_a''(x)=0\). From \(f_a''(x)=2x-a\), we get \(x=\frac{a}{2}\). Substitution into the first derivative gives \(-\frac{a^2}{4}+a-1=0\), or \((a-2)^2=0\). Thus \(a=2\). Because \(f_a'''(x)=2\neq0\), this critical point is an inflection point. 4. For every \(a\), the equation \(f_a''(x)=2x-a=0\) has exactly one solution, \(x=\frac{a}{2}\). Since the third derivative is always \(2\), the concavity changes there, so each graph has exactly one inflection point.

a) \((0, 0)\) and \(\left(2, \frac{2}{3}\right)\) b) \(a=5\); the extremum at \(x=4\) is a local minimum. c) \(a=2\) d) \(f_a''(x)=2x-a\) has exactly one zero, and \(f_a'''(x)=2\neq0\).

- The exponential factor is never zero. - A horizontal tangent requires the first derivative to equal zero. - Use the discriminant to count the real roots of each quadratic factor. - An inflection point requires a sign change in the second derivative.

1. Since \(e^x>0\), zeros come from \(x^2-k=0\). If \(k<0\), there are no real zeros. If \(k=0\), \(x=0\) is a double zero. If \(k>0\), the zeros \(x=\pm\sqrt{k}\) are both simple. 2. By the product rule, \(g_k'(x)=(x^2+2x-k)e^x\). The condition \(g_k'(1)=0\) gives \((3-k)e=0\), so \(k=3\). 3. Critical points satisfy \(x^2+2x-k=0\). This quadratic has two distinct real roots when its discriminant \(4+4k\) is positive, so exactly two local extrema occur when \(k>-1\). 4. The second derivative is \(g_k''(x)=(x^2+4x+2-k)e^x\). The quadratic factor has no sign-changing zeros when its discriminant \(8+4k\) is nonpositive. For \(k<-2\), there are no real zeros; for \(k=-2\), there is a double zero without a sign change. Therefore, there are no inflection points when \(k\le-2\).

a) \(k<0\): no real zeros; \(k=0\): \(x=0\), multiplicity \(2\); \(k>0\): \(x=\pm\sqrt{k}\), each with multiplicity \(1\) b) \(k=3\) c) \(k>-1\) d) \(k\le-2\)

- First determine the domain of each square-root function. - Use the chain rule on the square root. - The sign of the derivative is controlled by its numerator. - Eliminate \(k\) from the maximum coordinates.

1. The radicand is nonnegative on \([0, 4k]\). Every graph contains \((0, 0)\). If a point with \(x\neq0\) were common to graphs with distinct parameters \(k\) and \(m\), then \(kx-\frac{x^2}{4}=mx-\frac{x^2}{4}\), which would imply \(k=m\). Thus the origin is the only common point. 2. For interior points of the domain, the chain rule gives \(f_k'(x)=\frac{k-x/2}{2\sqrt{kx-x^2/4}}\). The derivative is zero at \(x=2k\), positive before this point, and negative after it, so the graph has a local maximum there. 3. Its y-coordinate is \(f_k(2k)=\sqrt{2k^2-k^2}=k\). Therefore, the local maximum is \((2k, k)\). 4. Since \(x=2k\) and \(y=k\), all maxima lie on \(y=\frac{1}{2}x\).

a) \((0, 0)\) b) \((2k, k)\) c) \(y=\frac{1}{2}x\)

- Use the product rule and chain rule. - Set the first and second derivatives equal to zero. - Horizontal distance is the difference between x-coordinates. - An exponential equals \(1\) only when its exponent is zero.

1. The first two derivatives are \(g_a'(x)=(1-a^2-ax)e^{-ax}\) and \(g_a''(x)=(a^2x+a^3-2a)e^{-ax}\). 2. Since the exponential factor is positive, \(g_a'(x)=0\) gives the local-maximum x-coordinate \(x=\frac{1}{a}-a\). The derivative changes from positive to negative there. 3. Similarly, \(g_a''(x)=0\) gives the inflection-point x-coordinate \(x=\frac{2}{a}-a\). 4. The horizontal distance is \(\left(\frac{2}{a}-a\right)-\left(\frac{1}{a}-a\right)=\frac{1}{a}\). 5. The slope at the inflection point is \(-e^{a^2-2}\). Setting this equal to \(-1\) gives \(e^{a^2-2}=1\), so \(a^2=2\). Since \(a>0\), \(a=\sqrt{2}\).

a) Local-maximum x-coordinate: \(\frac{1}{a}-a\); inflection-point x-coordinate: \(\frac{2}{a}-a\) b) The horizontal distance is \(\frac{1}{a}\). c) \(a=\sqrt{2}\)

- Find the inflection points from the second derivative. - Use symmetry to identify the triangle’s base. - The common y-coordinate gives the height. - Solve the resulting equation in \(k^{3/2}\).

1. Differentiate twice: \(f_k'(x)=-\frac{2k^2x}{(x^2+k)^2}\) and \(f_k''(x)=\frac{2k^2(3x^2-k)}{(x^2+k)^3}\). 2. The inflection-point equation gives \(x=\pm\sqrt{\frac{k}{3}}\). The second derivative changes sign at both values, so they are inflection points. 3. Their y-coordinate is \(\frac{k^2}{k/3+k}=\frac{3k}{4}\). The triangle therefore has base \(2\sqrt{\frac{k}{3}}\) and height \(\frac{3k}{4}\). 4. Its area is \(A=\frac{1}{2}\cdot2\sqrt{\frac{k}{3}}\cdot\frac{3k}{4}=\frac{\sqrt{3}}{4}k^{3/2}\). 5. Setting this equal to \(2\sqrt{3}\) gives \(k^{3/2}=8\), so \(k=4\).

\(k=4\)

- Use the product rule twice. - Solve the second-derivative equation for \(k\). - Substitute that expression into the original function. - Translate \(k>-2\) into a restriction on \(x\). - Verify the candidates using the third derivative.

1. By the product rule, \(f_k'(x)=(x^2+2x-k)e^x\) and \(f_k''(x)=(x^2+4x+2-k)e^x\). 2. Since \(e^x>0\), an inflection point must satisfy \(k=x^2+4x+2\). 3. Substituting this relation into the original function gives \(y=\left(x^2-(x^2+4x+2)\right)e^x=(-4x-2)e^x\). 4. The parameter restriction requires \(x^2+4x+2>-2\), or \((x+2)^2>0\). Therefore, \(x\neq-2\). 5. At a candidate point, the third derivative is \(2(x+2)e^x\), which is nonzero precisely because \(x\neq-2\). Thus every allowed point is an inflection point.

\(y=(-4x-2)e^x\) for \(x\neq-2\)

- Find the critical numbers and classify them with the second derivative. - Which graph features depend on the parameter, and which remain fixed? - What different contact patterns can a graph have with the x-axis? - How might turning-point heights limit the possible numbers of distinct roots?

1. The first derivative is \(f_k'(x)=4x(x^2-5)\), so the critical numbers are \(x=0\) and \(x=\pm\sqrt{5}\). 2. The second derivative is \(f_k''(x)=12x^2-20\). Since \(f_k''(0)=-20<0\), \((0, k)\) is a local maximum. Since \(f_k''(\pm\sqrt{5})=40>0\), \((-\sqrt{5}, k-25)\) and \((\sqrt{5}, k-25)\) are local minima. 3. Tangency at both minima requires \(k-25=0\), so \(k=25\). 4. Exactly three distinct zeros occur when the central local maximum lies on the x-axis while both minima lie below it. This gives \(k=0\), with zeros \(x=0\) and \(x=\pm\sqrt{10}\). 5. There are no real zeros when the absolute minima are above the x-axis: \(k-25>0\), so \(k>25\).

a) Local maximum: \((0, k)\); local minima: \((-\sqrt{5}, k-25)\) and \((\sqrt{5}, k-25)\) b) \(k=25\) c) \(k=0\) d) \(k>25\)

- Factor the second derivative and determine when the quadratic equation has real solutions. - Use the third derivative to verify each candidate. - Eliminate \(k\) from the nonzero inflection-point condition. - Compare the sizes of \(x^3\) and \(x^5\) near \(x=0\).

1. The second and third derivatives are \(g_k''(x)=2x(10kx^2-3)\) and \(g_k'''(x)=6(10kx^2-1)\). 2. The second derivative is zero at \(x=0\) for every \(k\). If \(k\leq0\), the equation \(10kx^2=3\) has no real solution, so the origin is the only inflection point. If \(k>0\), there are two additional candidates \(x=\pm\sqrt{\frac{3}{10k}}\). 3. At the origin, \(g_k'''(0)=-6\neq0\). At either nonzero candidate, the third derivative equals \(12\neq0\). Thus there is one inflection point when \(k\leq0\) and three when \(k>0\). 4. For a nonzero inflection point, \(k=\frac{3}{10x^2}\). Substitution gives \(y=\frac{3}{10x^2}x^5-x^3=-\frac{7}{10}x^3\). 5. Derivative argument: \(g_k''(0)=0\) and \(g_k'''(0)=-6\neq0\) for every \(k\), so the origin is always an inflection point. 6. Qualitative argument: near the origin, the fifth-degree term \(kx^5\) is much smaller in magnitude than the cubic term \(-x^3\). Therefore, the graph locally has the same concavity change as \(y=-x^3\).

a) One inflection point for \(k\leq0\); three inflection points for \(k>0\) b) \(y=-\frac{7}{10}x^3\) for the nonzero inflection points c) The origin is always an inflection point because \(g_k''(0)=0\) and \(g_k'''(0)=-6\neq0\); near the origin, \(-x^3\) dominates \(kx^5\).

- Compare two different parameter values to find the common point. - Use the second derivative test when \(k\ne0\). - When the second derivative is zero, examine the sign of the first derivative and the concavity.

1. If two distinct parameter values \(k\) and \(m\) give the same function value, then \((k-m)x^2e^{-x}=0\). Since \(e^{-x}>0\) and \(k\ne m\), this requires \(x=0\). Also, \(g_k(0)=0\), so the only common point is \((0, 0)\). 2. The first derivative is \(g_k'(x)=\left(-2x^3+(6-k)x^2+2kx\right)e^{-x}\), and \(g_k''(0)=2k\). 3. If \(k>0\), the second derivative is positive at \(0\), so \((0, 0)\) is a local minimum. If \(k<0\), it is a local maximum. 4. If \(k=0\), the second derivative test is inconclusive. In that case, \(g_0'(x)=2x^2(3-x)e^{-x}\). This derivative is positive on both sides of \(0\) near the point, so the function has no local extremum there. The third derivative at \(0\) equals \(12\ne0\), so the point is a stationary inflection point.

1) \((0, 0)\) 2) Local minimum for \(k>0\); local maximum for \(k<0\); stationary inflection point for \(k=0\).

- Compare the y-intercepts first. - Use the derivative to locate the two critical numbers. - Check whether the second extremum in a panel has the right height. - What relationship between the inflection point's coordinates holds for every allowed parameter?

1. Every family member has \(f_k(0)=4\); panels (2) and (3) instead have y-intercept \(2\). 2. Since \(f_k'(x)=3x(x-2k)\), critical numbers are \(0\) and \(2k\). Panel (4) would give \(k=\frac{1}{2}\), but then the second extremum should be \((1,\frac{7}{2})\), not \((1,2)\). Panel (1) has its second extremum at \(x=2\), giving \(k=1\), and \(f_1(2)=0\) agrees with its graph. 3. Since \(f_k''(x)=6x-6k\), the values at the critical numbers are \(f_k''(0)=-6k\) and \(f_k''(2k)=6k\). For \(k>0\), \((0,4)\) is a local maximum and \((2k,4-4k^3)\) is a local minimum; for \(k<0\), their types reverse. 4. Inflection points satisfy \(f_k''(x)=0\), so \(x=k\). The third derivative is \(6\), confirming the change of concavity. At that point, \(f_k(k)=4-2k^3\). Hence \(y=-2x^3+4\), with attained inputs \(x\ne0\).

a) Only panel (1), with \(k=1\). b) For \(k>0\): local maximum \((0,4)\), local minimum \((2k,4-4k^3)\); for \(k<0\), the types reverse. c) \(y=-2x^3+4\) for attained x-values \(x\ne0\).

- Read the time-coordinate of the highest point on the graph. - At an interior maximum, the first derivative is zero. - The most rapid decrease occurs where the derivative reaches its minimum. - Use the second derivative to locate a possible inflection point.

1. The graph has a maximum at about \(t=4\) hours. Since \(f_k'(t)=4e^{-kt}(1-kt)\), the maximum satisfies \(1-4k=0\), so \(k=0.25\). 2. In general, \(f_k'(t)=0\) when \(1-kt=0\), so \(t_{\max}=\frac{1}{k}\). Also, \(f_k''(t)=4ke^{-kt}(kt-2)\), and \(f_k''\left(\frac{1}{k}\right)=-4ke^{-1}<0\), confirming a maximum. 3. The concentration decreases most rapidly where its slope is minimal, which occurs at an inflection point of \(f_k\). Setting \(f_k''(t)=0\) gives \(kt-2=0\), so \(t=\frac{2}{k}\). Since \(\frac{2}{k}=2\left(\frac{1}{k}\right)\), this time is \(2t_{\max}\).

a) Maximum at about \(4\) hours; \(k=0.25\) b) \(t_{\max}=\frac{1}{k}\) c) The concentration decreases most rapidly at \(t=\frac{2}{k}=2t_{\max}\)

- Find the limit of the exponential term as \(t\to\infty\). - Use the first derivative to locate the maximum. - The most rapid decrease occurs where the first derivative is smallest. - Compare the long-term values of the two displayed graphs.

1. Since \((t-20)e^{-0.04t}\to0\) as \(t\to\infty\), \(\lim_{t\to\infty}v_a(t)=20a\). 2. Differentiate: \(v_a'(t)=ae^{-0.04t}(1.8-0.04t)\). The derivative is zero at \(t=45\), positive before \(45\), and negative after \(45\). Therefore, the greatest daily attendance occurs on day \(45\). 3. Differentiate again: \(v_a''(t)=-0.04ae^{-0.04t}(2.8-0.04t)\). This expression changes from negative to positive at \(t=70\), so \(v_a'\) has its minimum there. Thus, attendance is decreasing most rapidly on day \(70\). 4. For \(a=50\), the long-term value is \(1000\). For \(a=100\), it is \(2000\). Graph 1 approaches \(1000\), and graph 2 approaches \(2000\). Therefore, graph 1 corresponds to \(a=50\), and graph 2 corresponds to \(a=100\).

a) \(20a\) visitors per day b) Day \(45\) c) Day \(70\) d) Graph 1: \(a=50\); graph 2: \(a=100\).

- Factor the function to find its zeros. - Use the first and second derivatives to classify the critical points. - What relation between the maximum's two coordinates holds across the whole family? - Compare the positive x-intercept of each graph.

1. Factor: \(g_a(x)=x^2(a-x)\). The zeros are \(x=0\), multiplicity two, and \(x=a\). 2. The derivatives are \(g_a'(x)=x(2a-3x)\) and \(g_a''(x)=2a-6x\). The critical numbers are \(x=0\) and \(x=\frac{2a}{3}\). Since \(g_a''(0)=2a>0\), the origin is a local minimum. Since \(g_a''(\frac{2a}{3})=-2a<0\), the maximum is \(M_a=(\frac{2a}{3},\frac{4a^3}{27})\). 3. At the maximum, \(a=\frac{3x}{2}\). Substitution gives \(y=\frac{4}{27}(\frac{3x}{2})^3=\frac{1}{2}x^3\). Because \(a>0\), the locus has \(x>0\). 4. Setting \(\frac{4a^3}{27}=4\) gives \(a^3=27\), so \(a=3\). 5. The positive zero is at \(x=a\). Graph \(p\) has its positive zero at \(x=1.5\), and graph \(q\) at \(x=3\).

a) Zeros: \(x=0\), multiplicity two, and \(x=a\); local maximum \(M_a=(\frac{2a}{3},\frac{4a^3}{27})\) b) \(y=\frac{1}{2}x^3\), \(x>0\) c) \(a=3\) d) Graph \(p\): \(a=1.5\); graph \(q\): \(a=3\)