Free Math Worksheets

- What must the slope be when a tangent line is horizontal? - Which function gives the slope of the tangent line? - After finding the \(x\)-coordinates, how will you find the corresponding \(y\)-coordinates?

1. A horizontal tangent has slope \(0\), so solve \(f'(x) = 0\). 2. Differentiate: \(f'(x) = 6x^2 + 6x - 12\). 3. Solve \(6x^2 + 6x - 12 = 0\), or \(x^2 + x - 2 = 0\). Factoring gives \((x + 2)(x - 1) = 0\), so \(x = -2\) or \(x = 1\). 4. Evaluate the function: \(f(-2) = 24\) and \(f(1) = -3\). 5. The points are \((-2, 24)\) and \((1, -3)\).

\((-2, 24)\) and \((1, -3)\)

- What condition does the derivative satisfy at a horizontal tangent? - First factor out the common factor \(x\). - How many real zeros can the cubic derivative have at most?

1. A horizontal tangent occurs where \(g'(x) = 0\). 2. Differentiate: \(g'(x) = x^3 - 3x^2 + 2x\). 3. Factor: \(x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x - 1)(x - 2)\). 4. Set each factor equal to zero. The solutions are \(x = 0\), \(x = 1\), and \(x = 2\).

\(x = 0\), \(x = 1\), and \(x = 2\)

- Use the product rule on \(x\sin x\). - Combine like trigonometric terms after differentiating. - A horizontal tangent has derivative zero. - Use the zero-product property.

1. Differentiate the product and the cosine term: \(f'(x)=\sin x+x\cos x-\sin x=x\cos x\). 2. A horizontal tangent requires \(x\cos x=0\). Thus, either \(x=0\) or \(\cos x=0\). In \([0, 2\pi]\), the solutions are \(x=0\), \(x=\frac{\pi}{2}\), and \(x=\frac{3\pi}{2}\).

\(f'(x)=x\cos x\); horizontal tangents at \(x=0, \frac{\pi}{2}, \frac{3\pi}{2}\)

- The necessary condition at an interior local extremum is a zero derivative. - Differentiate using the power rule. - Rewrite a reciprocal as a negative power if helpful. - Check all solutions against the domain.

1. For part a), \(f'(x)=x^2-2x-3=(x-3)(x+1)\). Setting \(f'(x)=0\) gives \(x=-1\) and \(x=3\). 2. For part b), \(g'(x)=1-\frac{4}{x^2}\). Setting \(g'(x)=0\) gives \(x^2=4\), so \(x=-2\) and \(x=2\). Both are in the domain.

a) \(x=-1\) and \(x=3\) b) \(x=-2\) and \(x=2\)

- The derivative gives the slope of the graph. - Set the derivative equal to \(0\) and solve for \(x\). - Look for the point on the parabola where the graph changes from increasing to decreasing.

1. Differentiate: \(f'(x)=-2x+4\). 2. Set the slope equal to \(0\): \(-2x+4=0\), so \(x=2\). 3. The graph has a maximum at \(x=2\), where the tangent line is horizontal. This confirms that the slope is \(0\) there.

a) \(f'(x)=-2x+4\) b) \(x=2\) c) The graph has a horizontal tangent at its maximum when \(x=2\).

- A local maximum is at least as large as all nearby function values. - Check the actual filled point when a graph is discontinuous. - Compare the graph on both sides of \(x_0\).

1. In graph a), \(x_0\) is the vertex of an upward-opening parabola, so it is a local minimum. 2. In graph b), \(x_0\) is the vertex of a downward-opening parabola, so nearby values are less than or equal to \(f(x_0)\). Therefore, \(x_0\) is a local maximum. 3. In graph c), the function is increasing through \(x_0\), so there is no local extremum. 4. In graph d), the defined point at \(x_0\) lies below nearby values, so it is a local minimum.

b)

- A local minimum is no greater than all nearby function values. - At a discontinuity, use the filled point as the actual function value. - Compare values immediately to the left and right of \(x_0\).

1. In graph a), the function is decreasing through \(x_0\), so there is no local extremum. 2. In graph b), \(x_0\) is the vertex of a downward-opening parabola, so it is a local maximum. 3. In graph c), the filled point at \(x_0\) lies above the nearby curve values, so it is a local maximum, not a local minimum. 4. In graph d), the filled point at \(x_0\) lies below all nearby values. Therefore, \(x_0\) is a local minimum.

d)

- Look for peaks and valleys. - A local maximum occurs where the graph changes from increasing to decreasing. - A local minimum occurs where the graph changes from decreasing to increasing.

1. The graph changes from increasing to decreasing at \(x=-4\) and \(x=3\), so these are local-maximum locations. 2. The graph changes from decreasing to increasing at \(x=-1\) and \(x=5\), so these are local-minimum locations.

Local maxima at \(x=-4\) and \(x=3\); local minima at \(x=-1\) and \(x=5\)

- What conditions must the first and second derivatives satisfy at a stationary inflection point? - How is the number of local extrema related to the zeros of the first derivative? - Use the discriminant to determine when a quadratic equation has two distinct real solutions.

1. Differentiate: \(f_t'(x) = x^2 + 2tx + 9\) and \(f_t''(x) = 2x + 2t\). 2. At a stationary inflection point, \(f_t'(x) = 0\) and \(f_t''(x) = 0\). From \(2x + 2t = 0\), we get \(x = -t\). 3. Substitute into the first derivative: \(f_t'(-t) = t^2 - 2t^2 + 9 = 9 - t^2\). Setting this equal to zero gives \(t = \pm 3\). Since \(t > 0\), \(t = 3\). 4. The corresponding input is \(x = -3\), and \(f_3(-3) = \frac{1}{3}(-3)^3 + 3(-3)^2 + 9(-3) = -9\). The stationary inflection point is \((-3, -9)\). 5. Two local extrema occur when \(f_t'(x) = x^2 + 2tx + 9\) has two distinct real zeros. Its discriminant is \(D = (2t)^2 - 4 \cdot 1 \cdot 9 = 4t^2 - 36\). 6. Solve \(4t^2 - 36 > 0\): \(t^2 > 9\), so \(t < -3\) or \(t > 3\). Each simple zero of the derivative produces a sign change, so these values give two local extrema.

a) \(t = 3\); the stationary inflection point is \((-3, -9)\). b) \(t < -3\) or \(t > 3\)

- Find the first derivative and analyze how many real zeros it has. - Use the discriminant of the quadratic derivative. - A double zero of the derivative does not produce a local extremum. Check whether concavity changes there.

1. Differentiate: \(f_k'(x) = 3x^2 - 2kx + 3k\). 2. The number of critical points depends on the discriminant of \(3x^2 - 2kx + 3k = 0\): \(D = (-2k)^2 - 4 \cdot 3 \cdot 3k = 4k(k - 9)\). 3. If \(D > 0\), the derivative has two distinct real zeros. This occurs for \(k < 0\) or \(k > 9\). The derivative changes sign at each simple zero, so the graph has two local extrema. 4. If \(D < 0\), which occurs for \(0 < k < 9\), the derivative has no real zeros, so the graph has no local extrema. 5. If \(D = 0\), then \(k = 0\) or \(k = 9\). The derivative has a double zero and does not change sign, so there is no local extremum. Because \(f_k''(x) = 6x - 2k\) changes sign at the double zero, the graph has a stationary inflection point. 6. Therefore, the graph has no local extrema for every \(k\) in \([0, 9]\), with stationary inflection points occurring at the endpoint parameter values \(k = 0\) and \(k = 9\).

Two local extrema: \(k < 0\) or \(k > 9\) No local extrema: \(0 \le k \le 9\) Stationary inflection point: \(k = 0\) or \(k = 9\)

- What must the first derivative equal at a point with a horizontal tangent? - Substitute \(x = -3\) into the derivative. - Clear the denominator before solving the resulting quadratic equation.

1. Differentiate: \(g_a'(x) = \frac{3}{a}x^2 + 12x + 9a\). 2. A horizontal tangent at \(x = -3\) requires \(g_a'(-3) = 0\). Substitution gives \(\frac{27}{a} - 36 + 9a = 0\). 3. Because \(a \ne 0\), multiply by \(a\): \(9a^2 - 36a + 27 = 0\). 4. Divide by \(9\) and factor: \(a^2 - 4a + 3 = (a - 1)(a - 3) = 0\). 5. Therefore, \(a = 1\) or \(a = 3\). Both values satisfy the domain restriction.

\(a = 1\) or \(a = 3\)

- Find the critical numbers and identify the one that depends on \(k\). - Compute the corresponding function value in terms of \(k\). - Solve the \(x\)-coordinate equation for \(k\), then substitute into the \(y\)-coordinate.

1. Differentiate: \(f_k'(x) = x^2 - 2kx = x(x - 2k)\). The critical numbers are \(x = 0\) and \(x = 2k\). 2. The parameter-dependent critical number is \(x = 2k\). Its function value is \(f_k(2k) = \frac{8}{3}k^3 - 4k^3 + 4 = -\frac{4}{3}k^3 + 4\). 3. Solve \(x = 2k\) for the parameter: \(k = \frac{x}{2}\). 4. Substitute into the function value: \(y = -\frac{4}{3}\left(\frac{x}{2}\right)^3 + 4 = -\frac{1}{6}x^3 + 4\). 5. Because \(k \ne 0\), the parameter-dependent extrema have \(x \ne 0\). Therefore, the locus is \(y = -\frac{1}{6}x^3 + 4\), for \(x \ne 0\).

\(y = -\frac{1}{6}x^3 + 4\), for \(x \in \mathbb{R} \setminus \{0\}\)

- Factor the polynomial before finding its zeros. - Substitute the coordinates of the given point. - Set the derivative equal to zero and use \(k\ne0\).

1. Factor the function: \(f_k(x)=kx(x-3)^2\). Since \(k\ne0\), the zeros are \(x=0\) and \(x=3\), independent of \(k\). 2. Substituting \(P(1, 8)\) gives \(8=k(1-6+9)=4k\), so \(k=2\). 3. The derivative is \(f_k'(x)=k(3x^2-12x+9)=3k(x-1)(x-3)\). Since \(k\ne0\), the critical points occur at \(x=1\) and \(x=3\), independent of \(k\).

a) \(x=0\) and \(x=3\) b) \(k=2\) c) Critical-point x-coordinates: \(1\) and \(3\)

- A horizontal tangent has slope zero. - Differentiate while treating \(k\) as a constant. - Substitute the given input into the derivative. - Use the exact cosine value at \(\frac{\pi}{3}\).

1. Differentiate: \(f_k'(x)=4\cos x-k\). 2. A horizontal tangent requires \(f_k'\left(\frac{\pi}{3}\right)=0\). Therefore, \(4\cos\left(\frac{\pi}{3}\right)-k=0\). Since \(\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\), \(2-k=0\), so \(k=2\).

\(k=2\)

- What do the function value and derivative equal at a local extremum on the x-axis? - What is the derivative of \(e^x\)? - Apply the product rule to the entire expression for \(g\). - Why does multiplying by the always-positive factor \(e^x\) preserve the sign of \(f\)?

1. Since \(f\) has a local extremum at \((0, 0)\), \(f(0) = 0\) and \(f'(0) = 0\). 2. Evaluate \(g\) at the origin: \(g(0) = e^0f(0) = 0\). 3. Apply the product rule: \(g'(x) = e^xf(x) + e^xf'(x)\). Therefore, \(g'(0) = e^0f(0) + e^0f'(0) = 0\). 4. Also, \(e^x > 0\) for every real \(x\), so \(g(x)\) has the same sign as \(f(x)\) near \(x = 0\). Because \(f\) has a local extremum with value \(0\), it does not change sign there. Thus \(g\) does not cross the x-axis at the origin and has a horizontal tangent there, so its graph touches the x-axis at \((0, 0)\).

Since \(g(0) = 0\), \(g'(0) = 0\), and \(g\) has the same sign as \(f\) near \(0\), the graph of \(g\) touches the x-axis at \((0, 0)\).

- Use the known zeros of cosine. - Include the derivative of the inner quadratic expression when applying the chain rule. - Check systematically which integer values produce solutions inside the interval. - The range of cosine can help classify the critical points.

1. For the zeros, solve \(\cos\left(\frac{\pi}{4}x^2\right)=0\). Thus \(\frac{\pi}{4}x^2=\frac{\pi}{2}+k\pi\), so \(x^2=2+4k\). Within \(0\le x^2\le 9\), this gives \(x^2=2\) and \(x^2=6\). Therefore, the zeros are \(x=\pm\sqrt{2}\) and \(x=\pm\sqrt{6}\). 2. Differentiate: \(g'(x)=-\frac{\pi}{2}x\sin\left(\frac{\pi}{4}x^2\right)\). Critical numbers occur when \(x=0\) or \(\sin\left(\frac{\pi}{4}x^2\right)=0\). Thus \(x^2=4k\), which gives \(x=\pm2\) and \(x=\pm\sqrt{8}\), in addition to \(x=0\). 3. Evaluate the function at the critical numbers. Since \(g(0)=1\), \(g(\pm2)=-1\), and \(g(\pm\sqrt{8})=1\), the points with value \(1\) are local maxima and the points with value \(-1\) are local minima.

a) \(x=\pm\sqrt{2}\approx\pm1.414\) and \(x=\pm\sqrt{6}\approx\pm2.449\). b) Local maxima: \((0,1)\), \(\left(-\sqrt{8},1\right)\), and \(\left(\sqrt{8},1\right)\), where \(\sqrt{8}\approx2.828\). Local minima: \((-2,-1)\) and \((2,-1)\).

- Use the standard zeros of cosine. - Apply the chain rule when differentiating \(\cos(x^2)\). - Set each factor of the first derivative equal to zero. - Use the maximum and minimum values of cosine to classify the critical points. - Keep only values inside the stated interval.

1. Solve \(\cos(x^2)=0\). This requires \(x^2=\frac{\pi}{2}+k\pi\). Because \(0\le x^2\le4\), only \(x^2=\frac{\pi}{2}\) is possible. Thus the zeros are \(x=\pm\sqrt{\frac{\pi}{2}}\). 2. Differentiate using the chain rule: \(f'(x)=-2x\sin(x^2)\). Critical numbers satisfy \(x=0\) or \(\sin(x^2)=0\). Within the interval, this gives \(x=0\) and \(x=\pm\sqrt{\pi}\). 3. Since \(f(0)=1\), the point \((0,1)\) is a local maximum. Since \(f(\pm\sqrt{\pi})=-1\), the points \(\left(-\sqrt{\pi},-1\right)\) and \(\left(\sqrt{\pi},-1\right)\) are local minima.

a) \(x=\pm\sqrt{\frac{\pi}{2}}\approx\pm1.253\). b) Local maximum: \((0,1)\). Local minima: \(\left(-\sqrt{\pi},-1\right)\) and \(\left(\sqrt{\pi},-1\right)\), with x-coordinates approximately \(\pm1.772\).

- Recall when sine equals zero. - Apply the chain rule to the quadratic argument. - Set each factor of the first derivative equal to zero. - Use the maximum and minimum values of sine to classify the critical points. - Check that \(\sqrt{6}\) lies inside the interval.

1. Solve \(\sin\left(\frac{\pi}{4}x^2\right)=0\). Then \(\frac{\pi}{4}x^2=k\pi\), so \(x^2=4k\). Since \(0\le x^2\le6.25\), the possible values are \(x^2=0\) and \(x^2=4\). Thus the zeros are \(x=-2,0,2\). 2. Differentiate: \(h'(x)=\frac{\pi}{2}x\cos\left(\frac{\pi}{4}x^2\right)\). Critical numbers occur when \(x=0\) or \(\cos\left(\frac{\pi}{4}x^2\right)=0\). This gives \(x^2=2+4k\), so the interior critical numbers are \(x=0\), \(x=\pm\sqrt{2}\), and \(x=\pm\sqrt{6}\). 3. At \(x=0\), the derivative changes from negative to positive, so \((0,0)\) is a local minimum. At \(x=\pm\sqrt{2}\), the function value is \(1\), so these are local maxima. At \(x=\pm\sqrt{6}\), the function value is \(-1\), so these are local minima.

a) \(x=-2,0,2\). b) Local maxima: \(\left(-\sqrt{2},1\right)\) and \(\left(\sqrt{2},1\right)\). Local minima: \((0,0)\), \(\left(-\sqrt{6},-1\right)\), and \(\left(\sqrt{6},-1\right)\).

- The exponential factor is never zero. - Use the derivative at the x-intercept to find the crossing angle. - Apply the product and chain rules. - Eliminate \(k\) from the extremum coordinates to find the locus.

1. Since the exponential factor is positive, the x-intercept satisfies \(x-k=0\), so \(x=k\). 2. Differentiate: \(f_k'(x)=\left(2-\frac{x}{k}\right)e^{1-x/k}\). At the intercept, \(f_k'(k)=1\). Thus, the tangent slope is always \(1\), and the angle with the positive x-axis is \(\arctan(1)=45^\circ\). 3. The critical point satisfies \(2-\frac{x}{k}=0\), so \(x=2k\). The second derivative is \(f_k''(x)=\frac1k e^{1-x/k}\left(\frac{x}{k}-3\right)\). Therefore, \(f_k''(2k)=-\frac1{ke}<0\), so the extremum is a local maximum. Its coordinates are \(E_k=\left(2k, \frac{k}{e}\right)\). 4. From \(x=2k\), \(k=x/2\). Substituting into \(y=k/e\) gives \(y=\frac{x}{2e}\). Thus, all extrema lie on this line through the origin.

a) The x-intercept is \(x=k\), and the crossing angle is \(45^\circ\). b) \(E_k=\left(2k, \frac{k}{e}\right)\), a local maximum. c) \(y=\frac{x}{2e}\)

- At an interior maximum, the first derivative is zero. - Use the product rule and chain rule. - The exponential factor is never zero.

1. Differentiate using the product and chain rules: \(g'(t)=10e^{kt}(-1+60k-kt)\). 2. A maximum at \(t=20\) requires \(g'(20)=0\). Since \(10e^{20k}\neq0\), \(-1+40k=0\), so \(k=\frac{1}{40}=0.025\,\text{min}^{-1}\). 3. The second derivative at \(t=20\) is negative when \(k=0.025\), confirming that the critical point is a maximum.

\(k=0.025\,\text{min}^{-1}\)

- Find the critical time in terms of \(k\). - Substitute that time into the original function. - Evaluate the first derivative at \(t=0\). - Analyze how \(\frac{1}{k}\) changes as \(k\) increases.

1. Differentiate: \(f_k'(t)=ke^{-kt}(1-kt)\). The critical time is \(t=\frac{1}{k}\). Substitution gives \(f_k\left(\frac{1}{k}\right)=e^{-1}=\frac{1}{e}\), which is independent of \(k\). 2. \(f_k'(0)=k\). 3. The maximum occurs at \(t_{\max}=\frac{1}{k}\), so a larger \(k\) makes the maximum occur sooner. Since \(f_k'(0)=k\), a larger \(k\) also gives a steeper initial increase.

1. Maximum value: \(\frac{1}{e}\approx0.368\) 2. \(f_k'(0)=k\) 3. Larger \(k\) means a faster initial increase and an earlier maximum

- Solve \(f_k(x)=0\) and use the squared form. - Differentiate to count the critical points. - Compare the two ends of the graph when testing y-axis symmetry. - Evaluate the derivative at \(x=0\).

1. The function equals zero when \(e^x=k\), so \(x=\ln k\). Because the function is a square and is never negative, this zero is a tangency point. The statement is true for every \(k>0\). 2. Differentiate: \(f_k'(x)=2e^x(e^x-k)\). Since \(e^x>0\), the only critical point is \(x=\ln k\). The statement is true for every \(k>0\). 3. As \(x\to\infty\), \(f_k(x)\to\infty\), while as \(x\to-\infty\), \(f_k(x)\to k^2\). Therefore, the graph cannot be symmetric about the y-axis for any \(k>0\). The statement is false for every \(k\). 4. At \(x=0\), \(f_k'(0)=2(1-k)\). This value is positive when \(0<k<1\), zero when \(k=1\), and negative when \(k>1\). Therefore, the statement depends on \(k\).

1) True for every \(k>0\). 2) True for every \(k>0\). 3) False for every \(k>0\). 4) Depends on \(k\); the slope is positive exactly when \(0<k<1\).

- Identify the domain before finding critical numbers. - Use the quotient rule to differentiate. - Analyze the discriminant of the numerator of the derivative. - Check that any derivative zeros lie in the function's domain and produce a sign change.

1. The domain is \(\mathbb{R} \setminus \{1\}\). 2. Apply the quotient rule: \(f_a'(x) = \frac{2x(x - 1) - (x^2 + a)}{(x - 1)^2} = \frac{x^2 - 2x - a}{(x - 1)^2}\). 3. Critical numbers must satisfy \(x^2 - 2x - a = 0\). The discriminant is \(D = 4 + 4a = 4(1 + a)\). 4. If \(a < -1\), then \(D < 0\), so the derivative has no real zeros and the function has no local extrema. 5. If \(a = -1\), the numerator of the derivative is \((x - 1)^2\), whose only zero is \(x = 1\). This value is not in the domain, so there are no local extrema. 6. If \(a > -1\), the derivative has two distinct zeros \(x = 1 \pm \sqrt{1 + a}\). Both are in the domain, and the numerator changes sign at each simple zero while the denominator remains positive. Therefore, the function has exactly two local extrema.

No local extrema for \(a \le -1\) Exactly two local extrema for \(a > -1\)

- Use the product and chain rules. - The exponential function has no zeros. - Identify which factor controls the zeros of the derivative. - A quadratic equation has at most two real solutions.

1. Apply the product and chain rules: \(f'(x)=(2ax+b)e^{-x}-(ax^2+bx+c)e^{-x}\). 2. Factor the exponential term: \(f'(x)=e^{-x}\bigl(-ax^2+(2a-b)x+b-c\bigr)\). 3. Since \(e^{-x}>0\), zeros of the derivative can occur only when the quadratic factor is zero. 4. Because \(a\neq0\), this factor is a quadratic polynomial and therefore has at most two real zeros. 5. Local extrema can occur only at zeros of the derivative, so \(f\) has at most two local extrema.

The derivative is \(f'(x)=e^{-x}Q(x)\), where \(Q\) is quadratic. Since \(e^{-x}\neq0\) and a quadratic has at most two real zeros, \(f\) can have at most two local extrema.

- Require the expression under the square root to be nonnegative. - Use the symmetry of a parabola or complete the square to find its vertex. - A strictly increasing outer function preserves the ordering of inner function values. - Apply the chain rule to verify the critical number.

1. a) The zeros of \(f\) are \(-1\) and \(7\). Since the parabola opens downward, \(f(x)\ge0\) on \([-1,7]\). Thus, \(D_g=[-1,7]\). 2. b) The vertex lies midway between the zeros at \(x=3\), and \(f(3)=16\). Thus, the vertex is \((3,16)\). 3. The square-root function is strictly increasing on \([0,\infty)\), so maximizing \(f(x)\) also maximizes \(\sqrt{f(x)}\). Therefore, \(g\) has its maximum at \(x=3\). 4. c) By the chain rule, \(g'(x)=\frac{-2x+6}{2\sqrt{-x^2+6x+7}}=\frac{3-x}{\sqrt{-x^2+6x+7}}\), for \(-1<x<7\). 5. The derivative is zero only at \(x=3\), confirming the maximum location. Also, \(g(3)=4\), so the maximum point of \(g\) is \((3,4)\).

a) \(D_g=[-1,7]\) b) The vertex of \(f\) is \((3,16)\); because the square-root function is strictly increasing, \(g\) is maximized at \(x=3\). c) \(g'(x)=\frac{3-x}{\sqrt{-x^2+6x+7}}\); the maximum point is \((3,4)\).

- Ground-level width is the distance between the x-intercepts. - Because the square root is increasing, maximize the radicand. - Compare critical points with domain endpoints. - The slope at a point is the derivative value there.

1. a) The graph meets the ground at \(x=-3\) and \(x=6\), so the width is \(6-(-3)=9\,\text{m}\). 2. Let \(u(x)=(x+3)^2(6-x)=-x^3+27x+54\). Because the square root is increasing, \(f\) is maximized where \(u\) is maximized. 3. Since \(u'(x)=-3x^2+27\), the interior critical number is \(x=3\). Comparing \(u(3)\) with the endpoint values at \(x=-3\) and \(x=6\) shows the maximum occurs at \(x=3\). 4. b) \(f(3)=\frac{1}{4}\sqrt{108}=\frac{3\sqrt{3}}{2}\approx2.60\,\text{m}\). 5. c) For interior points, \(f'(x)=\frac{-3x^2+27}{8\sqrt{-x^3+27x+54}}\). Thus, \(f'(0)=\frac{9}{8\sqrt{6}}\approx0.46\).

a) \(9\,\text{m}\) b) \(2.60\,\text{m}\) c) \(f'(0)=\frac{9}{8\sqrt{6}}\approx0.46\)

- What is the slope of a line parallel to the x-axis? - Use the derivative to represent the tangent slope. - Set the derivative equal to \(0\). - After finding each x-coordinate, substitute it into \(f\) to find the corresponding y-coordinate.

1. Differentiate: \(f'(x)=x^2-4x+3\). 2. A horizontal tangent has slope \(0\), so solve \(f'(x)=0\): \(x^2-4x+3=(x-1)(x-3)=0\). Thus \(x=1\) or \(x=3\). 3. Evaluate the function: \(f(1)=\frac{19}{3}\) and \(f(3)=5\). 4. The points are \(\left(1,\frac{19}{3}\right)\) and \((3,5)\).

\(\left(1,\frac{19}{3}\right)\) and \((3,5)\)

- Differentiate using the chain rule. - Set the first derivative equal to zero. - Use the zeros of cosine in the relevant angle interval. - Solve each resulting equation for \(x\).

1. Differentiate using the chain rule: \(g'(x)=6\cos(2x-\pi)\). 2. Set the derivative equal to zero: \(\cos(2x-\pi)=0\). 3. Since \(x\in[0, \pi]\), the angle \(2x-\pi\) ranges from \(-\pi\) to \(\pi\). In this interval, cosine is zero at \(2x-\pi=-\frac{\pi}{2}\) and \(2x-\pi=\frac{\pi}{2}\). 4. Solving gives \(x=\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\). Both values lie in the interior of the stated interval.

\(x=\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\)

- Relate local extrema to the real zeros of the first derivative. - Use the discriminant of the quadratic derivative. - Determine what happens when the derivative has a double zero.

1. Differentiate: \(g_a'(x) = 3x^2 - 6ax + 6a - 3\). 2. The derivative has discriminant \(D = (-6a)^2 - 4 \cdot 3(6a - 3) = 36a^2 - 72a + 36 = 36(a - 1)^2\). 3. Since \(D \ge 0\) for every \(a\), the only way to avoid two distinct critical numbers is to have \(D = 0\). 4. Solve \(36(a - 1)^2 = 0\): \(a = 1\). 5. For \(a = 1\), \(g_1'(x) = 3(x - 1)^2\), so the derivative does not change sign at \(x = 1\). Therefore, the graph has no local extremum there. Also, \(g_1''(x) = 6x - 6\) changes sign at \(x = 1\), so the point is a stationary inflection point.

Yes. For \(a = 1\), the function has no local extrema.

- Compare \(f(0)\) with values immediately to the left and right. - A local maximum must be at least as large as all nearby values. - A local minimum must be at most as large as all nearby values. - Analyze the two formulas separately.

1. The function value is \(f(0)=1\). 2. For every \(x>0\), \(f(x)=x^2+1>1=f(0)\). 3. For every \(x<0\), \(f(x)=x<0<1=f(0)\). 4. Every neighborhood of \(0\) therefore contains function values both greater than and less than \(f(0)\). Thus \(f(0)\) is neither a local maximum nor a local minimum.

There is no local extremum at \(x=0\) because every neighborhood of \(0\) contains values above and below \(f(0)=1\).

- Test statements with lines, parabolas, and cubic functions. - Distinguish necessary conditions from sufficient conditions. - An open interval may still contain an interior point where a maximum is attained. - Fermat's theorem applies to differentiable functions at interior local extrema.

1. Statement (1) is true. Strict increase prevents any interior point from being locally greatest or least, and on all of \(\mathbb{R}\) there is no greatest or least value. 2. Statement (2) is false. For \(f(x)=x^3\), \(f'(0)=0\), but \(x=0\) is not a local extremum. 3. Statement (3) is false. The function \(f(x)=-x^2\) has an absolute maximum at \(x=0\) on \((-1,1)\). 4. Statement (4) is true by Fermat's theorem.

(1) True (2) False; for example, \(f(x)=x^3\) (3) False; for example, \(f(x)=-x^2\) on \((-1,1)\) (4) True

- Use the definition of strict decrease at the left endpoint. - Local extrema compare only nearby values. - Recall the hypotheses of the extreme value theorem.

1. Statement (1) is true. Strict decrease gives \(f(a)>f(x)\) for every \(x>a\) in the interval. 2. Statement (2) is true. “Local” compares values only in a neighborhood, so a local maximum in one region can lie below a distant local minimum. 3. Statement (3) is true by the extreme value theorem.

(1) True (2) True (3) True

- Use continuity on the interval between the two minima. - Consider the end behavior of an even-degree polynomial. - An absolute extremum in the interior is also a local extremum.

1. Statement a) is true. On the closed interval between two local minima, continuity guarantees a maximum. It occurs between the minima; if the maximum is attained on a flat interval, interior points of that interval are still local maxima. 2. Statement b) is true. A quartic has the same end behavior on both sides, approaching either \(+\infty\) at both ends or \(-\infty\) at both ends. By continuity, it attains an absolute minimum or maximum, which is also a local extremum.

a) True b) True

- Collect the terms containing \(a\) to find parameter-independent points. - Use the first derivative condition at \(x = -1\). - Analyze the discriminant of the quadratic derivative. - Show that the discriminant is positive for every real \(a\).

1. Group the parameter terms: \(f_a(x) = x^3 - x + a(x^2 - x)\). A point is shared by every graph when \(x^2 - x = 0\), so \(x = 0\) or \(x = 1\). In both cases, \(f_a(x) = 0\). The common points are \((0, 0)\) and \((1, 0)\). 2. Differentiate: \(f_a'(x) = 3x^2 + 2ax - a - 1\). Require \(f_a'(-1) = 0\): \(3 - 2a - a - 1 = 2 - 3a = 0\), so \(a = \frac{2}{3}\). 3. The second derivative is \(f_a''(x) = 6x + 2a\). At \(a = \frac{2}{3}\) and \(x = -1\), it is \(-\frac{14}{3} \ne 0\), so the critical point is a local extremum. 4. For any \(a\), the discriminant of the quadratic derivative is \(D = (2a)^2 - 4 \cdot 3(-a - 1) = 4(a^2 + 3a + 3)\). 5. Complete the square: \(a^2 + 3a + 3 = \left(a + \frac{3}{2}\right)^2 + \frac{3}{4} > 0\). Thus, \(D > 0\) for every real \(a\). The derivative always has two distinct real zeros and changes sign at each, so every graph has exactly two local extrema.

a) \((0, 0)\) and \((1, 0)\) b) \(a = \frac{2}{3}\) c) The derivative has two distinct real zeros for every \(a\), so every graph has exactly two local extrema.

- Apply the product rule and replace \(\cos^2x\) using the Pythagorean identity. - Horizontal tangents occur where the first derivative is zero. - In part c, substitute \(u=\sin x\). - Reject any solution outside the range \([-1, 1]\) of sine. - A positive sine value occurs twice in one full cycle.

1. Apply the product rule: \(f_k'(x)=-\sin x(k+\sin x)+\cos^2x\). Using \(\cos^2x=1-\sin^2x\), \(f_k'(x)=1-k\sin x-2\sin^2x\). 2. For \(k=0\), horizontal tangents satisfy \(1-2\sin^2x=0\). Since \(1-2\sin^2x=\cos(2x)\), solve \(\cos(2x)=0\). In \([0, 2\pi)\), this gives \(x\in\left\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right\}\). 3. For \(k=\frac{7}{3}\), set the derivative equal to zero: \(1-\frac{7}{3}\sin x-2\sin^2x=0\). Let \(u=\sin x\). Then \(6u^2+7u-3=0\), so \(u=\frac{1}{3}\) or \(u=-\frac{3}{2}\). The second value is outside the range of sine. Thus, \(x=\arcsin\left(\frac{1}{3}\right)\approx0.34\) or \(x=\pi-\arcsin\left(\frac{1}{3}\right)\approx2.80\). The second derivative is nonzero at both inputs, so both are local extrema.

a) \(f_k'(x)=1-k\sin x-2\sin^2x\) b) \(x\in\left\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right\}\) c) \(x\approx0.34\) and \(x\approx2.80\)

- Since \(a\ne0\), determine when the trigonometric factor equals zero. - Rewrite each trigonometric equation using tangent. - Local extrema can occur where the first derivative equals zero. - Examine what happens to the factor \(a\) after setting the derivative equal to zero.

1. Since \(a\ne0\), the zeros satisfy \(\sin x-\sqrt{3}\cos x=0\). Thus, \(\tan x=\sqrt{3}\), which gives \(x=\frac{\pi}{3}\) and \(x=\frac{4\pi}{3}\) in \([0, 2\pi]\). 2. Differentiate: \(f_a'(x)=a(\cos x+\sqrt{3}\sin x)\). Critical numbers satisfy \(\cos x+\sqrt{3}\sin x=0\), so \(\tan x=-\frac{1}{\sqrt{3}}\). In \([0, 2\pi]\), the solutions are \(x=\frac{5\pi}{6}\) and \(x=\frac{11\pi}{6}\). Both are local extrema. Their classifications reverse when the sign of \(a\) changes. 3. The factor \(a\) can be divided out of the equation \(f_a'(x)=0\) because \(a\ne0\). The remaining equation contains no \(a\), so the critical numbers do not depend on the parameter.

a) \(x=\frac{\pi}{3}\) and \(x=\frac{4\pi}{3}\) b) \(x=\frac{5\pi}{6}\) and \(x=\frac{11\pi}{6}\) c) The nonzero factor \(a\) cancels from the equation \(f_a'(x)=0\), leaving an equation independent of \(a\).

- State the necessary derivative condition for an interior local extremum. - Use \(f'(x)=\cos x\). - Solve \(\cos x=0\) on the given interval. - Check how the sign of cosine changes at each zero.

1. If a differentiable function has an interior local extremum at \(x_0\), then the necessary condition is \(f'(x_0)=0\). Since \(f'(x)=\cos x\), every interior local extremum of sine must occur at a zero of cosine. 2. Solve \(\cos x=0\) on \([0, 2\pi]\). The solutions are \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). 3. Cosine changes from positive to negative at \(\frac{\pi}{2}\), so sine has a local maximum there. Cosine changes from negative to positive at \(\frac{3\pi}{2}\), so sine has a local minimum there.

The necessary condition shows that an interior local extremum of sine must be a zero of cosine. In \([0, 2\pi]\), the zeros are \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). The first is a local maximum of sine, and the second is a local minimum.

- Read the \(x\)-coordinates of the local maximum and local minimum from the graph. - The first derivative equals zero at a local extremum. - Substitute one critical number into the derivative to solve for \(a\). - Use the discriminant to determine when the quadratic derivative does not have two distinct real zeros.

1. Differentiate: \(f_a'(x) = 0.75x^2 - 1.5ax + 2a\). 2. From the graph, the local maximum occurs at \(x = 2\) and the local minimum occurs at \(x = 4\). Use either critical number. Setting \(f_a'(2) = 0\) gives \(0.75(2)^2 - 1.5a(2) + 2a = 3 - a = 0\), so \(a = 3\). 3. This agrees with the other critical number: \(f_a'(4) = 12 - 4a = 0\), which also gives \(a = 3\). 4. The function has no local extrema when the quadratic derivative has at most one distinct real zero. Using the standard discriminant, \(D = (-1.5a)^2 - 4(0.75)(2a) = 2.25a^2 - 6a = 2.25a\left(a - \frac{8}{3}\right)\). 5. Solve \(D \le 0\): \(0 \le a \le \frac{8}{3}\). At the endpoint values, the derivative has a double zero and does not change sign.

a) \(a = 3\) b) \(0 \le a \le \frac{8}{3}\)

- Use the sign of the leading coefficient to determine the end behavior. - Find the derivative and evaluate the slope at \(x = 0\). - Set the derivative equal to zero at \(x = 2\) and solve for \(k\).

1. Since \(k > 0\), the leading coefficient is positive. Therefore, \(f_k(x) \to \infty\) as \(x \to \infty\), and \(f_k(x) \to -\infty\) as \(x \to -\infty\). Only graph 1 has this end behavior. 2. Also, \(f_k'(x) = 3kx^2 - 3\), so \(f_k'(0) = -3\). The graph must be decreasing as it passes through the origin. Again, only graph 1 has this behavior. 3. For a local extremum at \(x = 2\), require \(f_k'(2) = 0\): \(3k(2)^2 - 3 = 12k - 3 = 0\). Thus, \(k = \frac{1}{4}\). 4. For this value, \(f_k''(2) = 6k(2) = 3 > 0\), so the extremum at \(x = 2\) is a local minimum.

a) Graph 1. It has the required end behavior for a positive leading coefficient and is decreasing at the origin, where the slope is \(-3\). b) \(k = \frac{1}{4}\)

- Apply the definition using function values in a small neighborhood. - Differentiability is not required for a local extremum. - Compare the values on both sides of \(x_1\). - On a constant interval, nearby values equal the value at the marked point.

1. A function has a local maximum at \(x_0\) when \(f(x)\le f(x_0)\) for all \(x\) sufficiently close to \(x_0\). It has a local minimum when \(f(x)\ge f(x_0)\) nearby. 2. At \(x_1=-2\), \(f(-2)=4\), and every nearby function value is less than \(4\). Therefore, \(x=-2\) is a local-maximum location. The corner does not prevent a local extremum. 3. At \(x_2=1.5\), the function is constant with value \(1\) throughout a neighborhood of \(1.5\). Thus, every nearby value is both less than or equal to and greater than or equal to \(f(1.5)\). Under the stated non-strict definition, \(x=1.5\) is both a local-maximum and a local-minimum location.

\(x_1=-2\): local maximum \(x_2=1.5\): both a local maximum and a local minimum under the non-strict definition

- Must a function have derivative \(0\) in order to have a local extremum? - Compare the function value at \(x_1=-1\) with nearby values. - Use the shape and opening direction of the parabola near \(x_2=3\).

1. Near \(x_1=-1\), the graph has a V-shaped point and \(f(x)\ge f(-1)=0\). Therefore, \(f\) has a local minimum at \(x_1=-1\). The function is not differentiable there, but a local extremum is defined by nearby function values and does not require differentiability. 2. At \(x_2=3\), the graph is the vertex of a downward-opening parabola. The function changes from increasing to decreasing, so \(f\) has a local maximum there.

a) Yes. The function has a local minimum at \(x_1=-1\), even though it is not differentiable there. b) The function has a local maximum at \(x_2=3\).

- What does differentiability look like on a graph? - Compare the slope immediately before and after \(x_e=2\). - Recall the derivative as a limit of difference quotients. - What assumption is required before setting a derivative equal to \(0\)?

1. At \(x_e=2\), the graph has a corner. The slope from the left is \(1\), while the slope from the right is \(-1\). Since the one-sided slopes are different, the derivative does not exist at \(x_e=2\). 2. The condition that the derivative equal \(0\) applies only at points where the derivative exists. Because \(f\) is not differentiable at \(x_e=2\), this local maximum cannot be found by solving an equation that sets the derivative equal to \(0\).

1. The function is not differentiable at \(x_e=2\) because the left-hand slope is \(1\) and the right-hand slope is \(-1\). 2. The derivative does not exist at the maximum, so the condition that the derivative equal \(0\) does not apply.

- The exponential factor is never zero. - The zero comes from the linear factor. - Use the product and chain rules. - At a local minimum, the first derivative is zero and the second derivative is positive.

1. Since \(e^{0.5x}>0\), the only zero of \(g_a\) is \(x=a\). Graph 1 has its zero at \(x=1\), so it represents \(a=1\). Graph 2 has its zero at \(x=-1\), so it represents \(a=-1\). 2. Differentiate: \(g_a'(x)=e^{0.5x}(1+0.5x-0.5a)\). Requiring a critical point at \(x=4\) gives \(1+2-0.5a=0\), so \(a=6\). 3. For \(a=6\), \(g_a''(4)=0.5e^2>0\), confirming a local minimum.

a) Graph 1: \(a=1\); Graph 2: \(a=-1\) b) \(a=6\)

- Start with the graph whose y-intercept is zero. - Use a visible extremum to create one function-value equation and one derivative equation. - Add a constant for a vertical translation.

1. Use the graph through the origin as the unshifted base graph. It has a local minimum at \((1, -1)\). 2. Since \(g(1)=-1\), \(a+b=-1\). Since the derivative \(g'(x)=3ax^2+b\) is zero at \(x=1\), \(3a+b=0\). 3. Solving gives \(a=0.5\) and \(b=-1.5\). Thus \(g(x)=0.5x^3-1.5x\). 4. A vertical shift by \(c\) gives \(f_c(x)=0.5x^3-1.5x+c\).

\(f_c(x)=0.5x^3-1.5x+c\)

- A quotient is zero only when its numerator is zero. - Track the sign of the denominator on each side of \(x=-2\). - Compare exponential and linear growth for large positive \(x\). - Use the quotient rule to find critical numbers.

1. A quotient is zero only when its numerator is zero. Since \(e^x>0\) for every real \(x\), \(h\) has no zeros. 2. Near \(x=-2\), the numerator approaches \(e^{-2}>0\). The denominator approaches \(0^+\) from the right and \(0^-\) from the left, so \(\lim_{x\to-2^+}h(x)=\infty\) and \(\lim_{x\to-2^-}h(x)=-\infty\). As \(x\to\infty\), exponential growth dominates the linear denominator, so \(h(x)\to\infty\). As \(x\to-\infty\), \(e^x\to0\), so \(h(x)\to0\). 3. By the quotient rule, \(h'(x)=\frac{e^x(x+1)}{(x+2)^2}\). The only critical number is \(x=-1\). The derivative changes from negative to positive, so the graph has a local minimum at \(\left(-1, \frac{1}{e}\right)\).

a) No zeros b) \(\lim_{x\to-2^+}h(x)=\infty\), \(\lim_{x\to-2^-}h(x)=-\infty\), \(\lim_{x\to\infty}h(x)=\infty\), and \(\lim_{x\to-\infty}h(x)=0\) c) Local minimum at \(\left(-1, \frac{1}{e}\right)\)

- Use the initial value at \(x=0\) to find \(b\). - At an interior maximum, the first derivative is zero. - Differentiate with the product and chain rules.

1. The initial condition gives \(c(0)=b=10\,\text{mg/L}\). 2. Differentiate: \(c'(x)=e^{-0.1x}(a-0.1ax-1)\). 3. A maximum at \(x=5\) requires \(c'(5)=0\). Since the exponential factor is nonzero, \(a-0.5a-1=0\), so \(a=2\,\text{mg}/(\text{L}\cdot\text{h})\). 4. With \(a=2\), \(c'(x)=e^{-0.1x}(1-0.2x)\), which changes from positive to negative at \(x=5\). Thus, the point is a maximum.

\(a=2\,\text{mg}/(\text{L}\cdot\text{h})\) and \(b=10\,\text{mg/L}\)

- Use the necessary condition for an interior maximum in part a). - For part b), identify where the graph lies on or above the horizontal threshold line. - For part c), examine what the formula predicts after a very long time. - Also consider what the model predicts at \(t=0\).

1. Differentiate: \(c'(t)=\frac{12(t^2+4)-24t^2}{(t^2+4)^2}=\frac{48-12t^2}{(t^2+4)^2}\). 2. Set the derivative equal to zero: \(48-12t^2=0\), so \(t=2\) for \(t\ge 0\). The derivative changes from positive to negative, so the concentration is maximized at \(t=2\). 3. The maximum concentration is \(c(2)=\frac{24}{8}=3\,\text{mg/L}\). 4. The effectiveness endpoints satisfy \(\frac{12t}{t^2+4}=1.5\). This simplifies to \(t^2-8t+4=0\), giving \(t=4\pm 2\sqrt{3}\). 5. Therefore, the graph is at or above the threshold on \([4-2\sqrt{3},4+2\sqrt{3}]\), approximately \([0.54,7.46]\). The duration is \(4\sqrt{3}\approx 6.93\) hours. 6. As \(t\to\infty\), \(c(t)\to 0\), but the model remains positive for every finite \(t>0\). It is an idealized curve that does not account for individual metabolism, body size, detection limits, or more complex absorption and elimination processes.

a) The maximum occurs at \(t=2\,\text{h}\), and the maximum concentration is \(3\,\text{mg/L}\). b) The medication is effective for \(t\in[4-2\sqrt{3},4+2\sqrt{3}]\), approximately from \(0.54\) to \(7.46\) hours, for a duration of about \(6.93\) hours. c) The model approaches zero but never reaches it at a finite positive time and does not capture individual or physiological complexity.

- A horizontal tangent has derivative \(0\). - Find the full coordinates of all three points. - Notice whether two points form a horizontal base. - Use \(\frac{1}{2}(\text{base})(\text{height})\).

1. Differentiate: \(f'(x)=x^3-4x=x(x^2-4)\). 2. Horizontal tangent lines occur where \(f'(x)=0\). Thus, \(x=-2\), \(x=0\), and \(x=2\). 3. Evaluate the function: \(f(-2)=-3\), \(f(0)=1\), and \(f(2)=-3\). The points are \((-2,-3)\), \((0,1)\), and \((2,-3)\). 4. The horizontal base from \((-2,-3)\) to \((2,-3)\) has length \(4\), and the vertical height to \((0,1)\) is \(4\). 5. The area is \(\frac{1}{2}\cdot 4\cdot 4=8\).

\(8\) square units

- What happens to the degree of the function when the leading coefficient is zero? - Differentiate and determine how many real zeros the derivative has. - Use the discriminant for the quadratic derivative when \(a \ne 0\). - Check whether the derivative changes sign at each critical number.

1. First consider \(a = 0\). Then \(g_0(x) = 3x^2 + 3x\), a quadratic function. Its derivative is \(g_0'(x) = 6x + 3\), which changes sign at \(x = -\frac{1}{2}\). Thus, \(g_0\) has exactly one local extremum. 2. For \(a \ne 0\), differentiate: \(g_a'(x) = 3ax^2 + 6x + 3\). 3. The discriminant of the derivative equation is \(D = 6^2 - 4 \cdot 3a \cdot 3 = 36(1 - a)\). 4. If \(a < 1\) and \(a \ne 0\), then \(D > 0\). The derivative has two distinct real zeros and changes sign at both, so the graph has two local extrema. 5. If \(a = 1\), then \(g_1'(x) = 3(x + 1)^2\). The derivative has a double zero without a sign change, so the graph has no local extremum. 6. If \(a > 1\), then \(D < 0\), so the derivative has no real zeros and the graph has no local extrema. 7. Combining the cases, there are two local extrema for \(a < 1\) with \(a \ne 0\), one local extremum for \(a = 0\), and no local extrema for \(a \ge 1\).

Two local extrema: \(a < 1\) with \(a \ne 0\) One local extremum: \(a = 0\) No local extrema: \(a \ge 1\)

- Factor the first derivative to find the critical numbers. - Determine when the two critical numbers are equal, then check whether the derivative changes sign. - Group the function into a parameter-free part plus \(a\) times another expression. - Common points occur where the coefficient of \(a\) is zero.

1. Differentiate: \(f_a'(x) = 3x^2 - 6ax + 12a - 12 = 3(x - 2)(x - 2a + 2)\). 2. The critical numbers are \(x = 2\) and \(x = 2a - 2\). They are distinct when \(a \ne 2\), and each then produces a local extremum. 3. When \(a = 2\), the critical numbers coincide at \(x = 2\), and \(f_2'(x) = 3(x - 2)^2\). The derivative does not change sign, so there is no local extremum. Since \(f_2''(x) = 6(x - 2)\) changes sign at \(x = 2\), the graph has a stationary inflection point there. Also, \(f_2(2) = 18\), so the point is \((2, 18)\). 4. Separate the parameter term: \(f_a(x) = x^3 - 12x + 10 + a(-3x^2 + 12x)\). A point is common to every graph when the coefficient of \(a\) is zero. 5. Solve \(-3x^2 + 12x = -3x(x - 4) = 0\): \(x = 0\) or \(x = 4\). The corresponding values are \(f_a(0) = 10\) and \(f_a(4) = 26\). Thus, the common points are \((0, 10)\) and \((4, 26)\).

a) \(x = 2\) and \(x = 2a - 2\); for \(a = 2\), these coincide. b) \(a = 2\); the graph has a stationary inflection point at \((2, 18)\). c) \((0, 10)\) and \((4, 26)\)

- Apply both the product rule and the chain rule. - Factor the intersection equation so that one factor is \(x\). - Analyze the maximum value of \(w(x)=\frac{h(x)}{x}\) for \(x>0\). - Count distinct intersection points, not root multiplicity.

1. Differentiate using the product and chain rules: \(h'(x)=e^{-0.5x}(2x-0.5x^2)=xe^{-0.5x}(2-0.5x)\). 2. The interior critical number is \(x=4\). The derivative changes from positive to negative there, so the point is a local maximum. Its coordinates are \(\left(4,16e^{-2}\right)\). 3. Intersections satisfy \(x^2e^{-0.5x}=kx\), or \(x\left(xe^{-0.5x}-k\right)=0\). Thus \(x=0\) is always one intersection. 4. For positive intersections, analyze \(w(x)=xe^{-0.5x}\). Its derivative is \(w'(x)=(1-0.5x)e^{-0.5x}\), so \(w\) has a maximum at \(x=2\), where \(w(2)=\frac{2}{e}\). 5. If \(0<k<\frac{2}{e}\), the equation \(w(x)=k\) has two positive solutions, for a total of three intersections. If \(k=\frac{2}{e}\), it has one positive solution, for a total of two intersections. If \(k\le0\) or \(k>\frac{2}{e}\), it has no positive solution, so the origin is the only intersection.

a) \(\left(4,16e^{-2}\right)\). b) If \(0<k<\frac{2}{e}\), there are \(3\) intersections. If \(k=\frac{2}{e}\), there are \(2\) intersections. If \(k\le0\) or \(k>\frac{2}{e}\), there is \(1\) intersection.

- Compare the signs of both sides when \(x\leq0\). - For \(x>0\), take logarithms and isolate a constant. - Find the maximum of \(\frac{\ln(x)}{x}\). - Compare \(\frac{1}{n}\) with that maximum.

1. If \(x\leq0\), then \(x^n\leq0\) because \(n\) is odd, while \(e^x>0\). Thus, there are no nonpositive solutions. 2. For \(x>0\), take natural logarithms: \(x=n\ln(x)\), or \(\frac{\ln(x)}{x}=\frac{1}{n}\). 3. Let \(q(x)=\frac{\ln(x)}{x}\). Then \(q'(x)=\frac{1-\ln(x)}{x^2}\), so \(q\) increases on \((0, e)\), decreases on \((e, \infty)\), and has maximum \(q(e)=\frac{1}{e}\). 4. Since \(n\geq3\), \(0<\frac{1}{n}\leq\frac{1}{3}<\frac{1}{e}\). Also, \(q(x)\to-\infty\) as \(x\to0^+\) and \(q(x)\to0^+\) as \(x\to\infty\). 5. Therefore, the horizontal line \(y=\frac{1}{n}\) intersects the graph of \(q\) once in \((1, e)\) and once in \((e, \infty)\). Hence, the original equation has exactly two real solutions.

There are no solutions for \(x\leq0\), and exactly two solutions for \(x>0\). Therefore, \(e^x=x^n\) has exactly two real solutions.

- Substitute the given point into the function. - Compare two functions in the family to analyze common points. - A horizontal stretch by \(c\) replaces \(x\) with \(x/c\). - Use the critical-point condition to simplify the y-coordinate of each extremum.

1. Substitute \((2, 2)\): \(2=2e^{1-4k}\). Thus, \(e^{1-4k}=1\), so \(k=\frac14\). The exponential equation has only this solution. 2. For \(k=0\), \(f_0(x)=ex\), so the slope is \(e\). 3. Every graph passes through the origin. The second statement shows that no two distinct graphs share any other point. Therefore, the origin is the only common point of the family. 4. A horizontal stretch by \(c\) followed by a vertical stretch by \(c\) gives \(y=c f_k\left(\frac{x}{c}\right)\) \(=xe^{1-(k/c^2)x^2}\). This is \(f_{k^*}(x)\) with \(k^*=\frac{k}{c^2}\), so the transformed graph remains in the family. 5. Differentiate: \(f_k'(x)=(1-2kx^2)e^{1-kx^2}\). For \(k>0\), the two critical points satisfy \(kx^2=\frac12\). The derivative changes sign at both points, so they are local extrema. At either critical point, \(f_k(x)=xe^{1-1/2}=\sqrt e\, x\). Therefore, every local extremum lies on \(y=\sqrt e\, x\).

a) \(k=\frac14\) b) The slope is \(e\). c) All graphs intersect only at the origin. d) The stretched graph is \(f_{k/c^2}\). e) Every local extremum satisfies \(y=\sqrt e\, x\).

- A rational function is zero when its numerator is zero and its denominator is nonzero. - Analyze the behavior near the excluded x-value and as \(|x|\to\infty\). - Use the first and second derivatives to find and classify the extremum. - Set two functions with different parameters equal to test for intersections. - Split the integrand into powers of \(x\) before integrating.

1. The denominator cannot be zero, so \(D_k=\mathbb R\setminus\{0\}\). The numerator is zero when \(x=\frac{k}{2}\), which is the x-intercept. The vertical asymptote is \(x=0\), and because \(f_k(x)\to0\) as \(x\to\pm\infty\), the horizontal asymptote is \(y=0\). 2. Differentiate: \(f_k'(x)=\frac{2x-2k}{x^3}\). The only critical point is \(x=k\). The second derivative is \(f_k''(x)=\frac{6k-4x}{x^4}\), so \(f_k''(k)=\frac{2}{k^3}\). The extremum has coordinates \(\left(k, -\frac1k\right)\). It is a local minimum when \(k>0\) and a local maximum when \(k<0\). 3. If two graphs intersect at a point in their domain, then \(\frac{k_1-2x}{x^2}=\frac{k_2-2x}{x^2}\), which implies \(k_1=k_2\). Therefore, distinct graphs in the family do not intersect. 4. An antiderivative is \(F_k(x)=-\frac{k}{x}-2\ln|x|\). For \(k>0\), \(\int_{k/2}^{k}f_k(x)\, \mathrm{d}x\) \(=\left[-\frac{k}{x}-2\ln x\right]_{k/2}^{k}\) \(=1-2\ln2\), which does not depend on \(k\).

a) \(D_k=\mathbb R\setminus\{0\}\); x-intercept \(x=\frac{k}{2}\); vertical asymptote \(x=0\); horizontal asymptote \(y=0\). b) The extremum is \(\left(k, -\frac1k\right)\). It is a local minimum for \(k>0\) and a local maximum for \(k<0\). c) Distinct graphs do not intersect. d) \(\int_{k/2}^{k}f_k(x)\, \mathrm{d}x=1-2\ln2\)

- Determine when the first derivative can equal zero. - Eliminate the parameter from the coordinates of the extrema. - Write the tangent equation at a general point \(x_0\). - Require the origin to lie on that tangent.

1. Differentiate: \(f_a'(x)=3-ae^x\) and \(f_a''(x)=-ae^x\). The critical-point equation is \(e^x=\frac3a\). This equation has a solution only when \(a>0\), in which case \(x=\ln\left(\frac3a\right)\). At this point, the second derivative is \(-3<0\), so the extremum is a local maximum. Its coordinates are \(H=\left(\ln\left(\frac3a\right), 3\ln\left(\frac3a\right)-3\right)\). For \(a<0\), there are no extrema. 2. Writing the maximum point as \((x, y)\), its coordinates satisfy \(y=3x-3\). Thus, the line containing all extrema is \(y=3x-3\). 3. Let \(x_0\) be a point of tangency. The tangent at \(x_0\) passes through the origin exactly when \(f_a(x_0)-x_0f_a'(x_0)=0\). Substitution gives \(ae^{x_0}(x_0-1)=0\). Since \(a\ne0\) and \(e^{x_0}>0\), the unique solution is \(x_0=1\). Therefore, the unique point of tangency is \(P_a=(1, 3-ae)\).

a) For \(a>0\), the graph has a local maximum at \(\left(\ln\left(\frac3a\right), 3\ln\left(\frac3a\right)-3\right)\). For \(a<0\), it has no extrema. b) \(y=3x-3\) c) \(P_a=(1, 3-ae)\)

- A tangency point on the x-axis is a zero that is also a local minimum. - Factor the first derivative to find its zeros. - Compare the x- and y-coordinates of \(H_a\). - Substitute the specified x-value into the derivative.

1. Because \(x>0\), \(f_a(x)=0\) exactly when \(\ln x-a=0\). Thus, \(x=e^a\). Since \(f_a(x)\ge0\) for all \(x>0\), the graph is tangent to the x-axis at \(B_a=(e^a, 0)\). 2. Differentiate: \(f_a'(x)=(\ln x-a)(\ln x-a+2)\). Besides \(x=e^a\), the other critical point satisfies \(\ln x=a-2\), so \(x=e^{a-2}\). The derivative changes from positive to negative there, so this point is a local maximum. Its y-coordinate is \(f_a(e^{a-2})=4e^{a-2}\). Therefore, \(H_a=(e^{a-2}, 4e^{a-2})\). 3. The coordinates of \(H_a\) satisfy \(y=4x\), so the locus is the line \(y=4x\). 4. Substitute \(x=e^{a-1}\) into the derivative: \(f_a'(e^{a-1})=(-1)(1)=-1\). The tangent slope is always \(-1\).

a) \(B_a=(e^a, 0)\) b) \(H_a=(e^{a-2}, 4e^{a-2})\) c) \(y=4x\) d) The tangent slope is \(-1\).

- Use \(a>0\) to analyze the denominator. - Apply the quotient rule and analyze the sign of the derivative. - Eliminate \(a\) from the local-maximum coordinates. - Compare the degrees of the numerator and denominator for end behavior. - Look for a denominator whose derivative matches the numerator up to a constant factor.

1. Since \(a>0\), the denominator \(x^2+a\) is positive for every real \(x\). Thus, \(D_a=\mathbb R\). The only zero is \(x=0\). Also, \(f_a(-x)=-f_a(x)\), so the graph is symmetric about the origin. 2. Differentiate: \(f_a'(x)=\frac{a^2(a-x^2)}{(x^2+a)^2}\). The critical points occur at \(x=\pm\sqrt a\). The derivative changes from negative to positive at \(-\sqrt a\) and from positive to negative at \(\sqrt a\). Therefore, \(T=\left(-\sqrt a, -\frac12a\sqrt a\right)\) is a local minimum, and \(H=\left(\sqrt a, \frac12a\sqrt a\right)\) is a local maximum. At a local maximum, \(x=\sqrt a\), so \(a=x^2\). Hence, \(y=\frac12a\sqrt a=\frac12x^3\), proving that all local maxima lie on \(h(x)=0.5x^3\). 3. The denominator has higher degree than the numerator, so \(\lim_{x\to\pm\infty}f_a(x)=0\). Thus, the horizontal asymptote is \(y=0\). 4. An antiderivative is \(F_a(x)=\frac{a^2}{2}\ln(x^2+a)\). Therefore, \(\int_0^{\sqrt a}f_a(x)\, \mathrm{d}x\) \(=\frac{a^2}{2}[\ln(2a)-\ln a]\) \(=\frac{a^2}{2}\ln2\).

a) \(D_a=\mathbb R\); zero \(x=0\); origin symmetry. b) \(T=\left(-\sqrt a, -\frac12a\sqrt a\right)\) and \(H=\left(\sqrt a, \frac12a\sqrt a\right)\). The local maxima lie on \(y=0.5x^3\). c) \(\lim_{x\to\pm\infty}f_a(x)=0\), so the horizontal asymptote is \(y=0\). d) \(\frac{a^2}{2}\ln2\)

- Replace \(x\) with \(-x\) to test symmetry. - Differentiate using the quotient and chain rules. - Compare two functions with different parameters to find common points. - Recognize the reverse chain rule when finding an antiderivative. - Use an improper integral for the unbounded interval.

1. Since \(g_k(-x)=-g_k(x)\), the graph is symmetric about the origin. The only zero is \(x=0\). 2. Differentiate: \(g_k'(x)=\frac{2k-6x^2}{(x^2+k)^3}\). The critical points satisfy \(x=\pm\sqrt{\frac{k}{3}}\). The positive critical point is a local maximum, and the negative critical point is a local minimum. Their coordinates are \(H=\left(\sqrt{\frac{k}{3}}, \frac{3\sqrt3}{8k^{3/2}}\right)\) and \(T=\left(-\sqrt{\frac{k}{3}}, -\frac{3\sqrt3}{8k^{3/2}}\right)\). 3. If two graphs with parameters \(k_1\ne k_2\) intersect, then \(\frac{2x}{(x^2+k_1)^2}=\frac{2x}{(x^2+k_2)^2}\). The equation always holds at \(x=0\). For \(x\ne0\), positivity of the denominators would force \(k_1=k_2\), a contradiction. Therefore, the only common point is \((0, 0)\). Also, \(\lim_{x\to\pm\infty}g_k(x)=0\), so the horizontal asymptote is \(y=0\). 4. An antiderivative is \(G_k(x)=-\frac1{x^2+k}\). Since \(g_k(x)\ge0\) for \(x\ge0\), the area is \(\int_0^\infty g_k(x)\, \mathrm{d}x\) \(=\lim_{b\to\infty}\left[-\frac1{x^2+k}\right]_0^b\) \(=\frac1k\). The area is finite.

a) The graph has origin symmetry and the only zero is \(x=0\). b) \(H=\left(\sqrt{\frac{k}{3}}, \frac{3\sqrt3}{8k^{3/2}}\right)\) and \(T=\left(-\sqrt{\frac{k}{3}}, -\frac{3\sqrt3}{8k^{3/2}}\right)\) c) The common point is \((0, 0)\), and the horizontal asymptote is \(y=0\). d) One antiderivative is \(-\frac1{x^2+k}\), and the area is finite with value \(\frac1k\).

- Use the first and second derivatives to locate extrema and inflection points. - Compare exponential and polynomial growth for the end behavior. - Differentiate the proposed antiderivative. - Use an improper integral for the unbounded region.

1. Differentiate: \(f_k'(x)=k(1-kx)e^{-kx}\) and \(f_k''(x)=k^2(kx-2)e^{-kx}\). The critical point is \(x=1/k\), where \(f_k(1/k)=1/e\). The derivative changes from positive to negative, so \(H_k=\left(\frac1k, \frac1e\right)\). The possible inflection point is at \(x=2/k\), where \(f_k(2/k)=2/e^2\). Since the third derivative is nonzero there, the inflection point is \(W_k=\left(\frac2k, \frac2{e^2}\right)\). 2. Exponential decay dominates the linear factor as \(x\to\infty\), so \(\lim_{x\to\infty}f_k(x)=0\). As \(x\to-\infty\), \(kx\to-\infty\) while \(e^{-kx}\to\infty\), so \(\lim_{x\to-\infty}f_k(x)=-\infty\). 3. Differentiate \(F_k\): \(F_k'(x)=-e^{-kx}+k\left(x+\frac1k\right)e^{-kx}\) \(=kxe^{-kx}=f_k(x)\). 4. Since \(f_k(x)\ge0\) for \(x\ge0\), the area is \(\int_0^\infty f_k(x)\, \mathrm{d}x\) \(=\lim_{b\to\infty}[F_k(x)]_0^b\) \(=0-\left(-\frac1k\right)=\frac1k\).

a) \(H_k=\left(\frac1k, \frac1e\right)\), \(W_k=\left(\frac2k, \frac2{e^2}\right)\) b) \(\lim_{x\to\infty}f_k(x)=0\) and \(\lim_{x\to-\infty}f_k(x)=-\infty\) c) \(F_k'(x)=f_k(x)\) d) The area is \(\frac1k\) square units.

- Analyze the derivative separately for positive and nonpositive values of \(a\). - Determine the end behavior of the function. - For \(a>0\), find the minimum value. - Compare the minimum value with zero.

1. Differentiate: \(f_a'(x)=ae^x-1\). 2. If \(a<0\), then \(f_a'(x)<0\) for every \(x\), so the function is strictly decreasing. Also, \(\lim_{x\to-\infty}f_a(x)=\infty\) and \(\lim_{x\to\infty}f_a(x)=-\infty\). Therefore, there is exactly one zero. If \(a=0\), then \(f_0(x)=-x\), which also has exactly one zero. 3. If \(a>0\), the only critical point satisfies \(ae^x=1\), so \(x=-\ln a\). Since \(f_a(x)\to\infty\) as \(x\to\pm\infty\), this critical point is the absolute minimum. Its value is \(f_a(-\ln a)=1+\ln a\). 4. The minimum is positive when \(a>1/e\), zero when \(a=1/e\), and negative when \(0<a<1/e\). Therefore: - there are no zeros when \(a>1/e\); - there is exactly one zero when \(a=1/e\) or \(a\le0\); - there are exactly two zeros when \(0<a<1/e\).

There are no real zeros for \(a>\frac1e\). There is exactly one real zero for \(a=\frac1e\) or \(a\le0\). There are exactly two real zeros for \(0<a<\frac1e\).

- Keep the domain \(x>0\) in mind. - Separate the cases \(k<0\), \(k=0\), and \(k>0\). - Use end behavior and monotonicity when \(k<0\). - For \(k>0\), find the minimum and compare it with zero.

1. Differentiate: \(h_k'(x)=2x-\frac{k}{x}\). 2. If \(k<0\), then \(h_k'(x)=2x+\frac{|k|}{x}>0\) for all \(x>0\). The function is strictly increasing, with \(h_k(x)\to-\infty\) as \(x\to0^+\) and \(h_k(x)\to\infty\) as \(x\to\infty\). Therefore, it has exactly one zero. 3. If \(k=0\), then \(h_0(x)=x^2\), which has no zero in the domain \(x>0\). 4. If \(k>0\), the only critical point is \(x=\sqrt{\frac{k}{2}}\). The function approaches \(\infty\) at both ends of its domain, so this point is the absolute minimum. Its value is \(\frac{k}{2}\left(1-\ln\left(\frac{k}{2}\right)\right)\). This value is zero when \(k=2e\), positive when \(0<k<2e\), and negative when \(k>2e\). 5. Therefore: - there are no zeros when \(0\le k<2e\); - there is exactly one zero when \(k<0\) or \(k=2e\); - there are exactly two zeros when \(k>2e\).

There are no real zeros for \(0\le k<2e\). There is exactly one real zero for \(k<0\) or \(k=2e\). There are exactly two real zeros for \(k>2e\).

- Factor the polynomial to find the intercepts. - Compare \(f_k(x)\) and \(f_m(x)\) to locate common points of two different graphs. - Use the first and second derivatives to identify and classify local extrema. - Subtract the two functions before setting up the area integral.

1. Factor: \(f_k(x)=kx^3(x-4)\). Because \(k\ne0\), every graph has intercepts \((0, 0)\) and \((4, 0)\). If \(k\ne m\) and \(f_k(x)=f_m(x)\), then \((k-m)(x^4-4x^3)=0\). Therefore, \(x=0\) or \(x=4\), so there are no other common points. 2. Differentiate: \(f_k'(x)=4kx^2(x-3)\). The critical numbers are \(0\) and \(3\). At \(x=0\), the derivative does not change sign, so there is no local extremum. At \(x=3\), \(f_k(3)=-27k\) and \(f_k''(3)=36k\). Thus, \((3, -27k)\) is a local minimum when \(k>0\) and a local maximum when \(k<0\). 3. Every local extremum has x-coordinate \(3\), so the locus is the vertical line \(x=3\). 4. The difference between the two functions is \(f_{k+1}(x)-f_k(x)=x^4-4x^3\). On \((0, 4)\), this expression is negative. Therefore, the enclosed area is \(\int_0^4(4x^3-x^4)\, \mathrm{d}x=\left[x^4-\frac15x^5\right]_0^4=\frac{256}{5}\). This area is independent of \(k\).

a) The only common points are \((0, 0)\) and \((4, 0)\). b) The local extremum is \((3, -27k)\). It is a local minimum for \(k>0\) and a local maximum for \(k<0\). c) \(x=3\) d) \(\frac{256}{5}\) square units, or \(51.2\) square units.

- Rewrite the square root as a power before differentiating. - A fraction with a nonzero denominator equals zero exactly when its numerator equals zero. - Maximize the quadratic exponent before considering the exponential and square root. - Use the definition of a strictly increasing function to compare nearby function values.

1. a) Write \(h(x)=(f(x))^{\frac{1}{2}}\). By the chain rule, \(h'(x)=\frac{1}{2}(f(x))^{-\frac{1}{2}}f'(x)=\frac{f'(x)}{2\sqrt{f(x)}}\). 2. b) Since \(f(x)>0\), the denominator is always positive. Therefore, \(h'(x)=0\) if and only if \(f'(x)=0\). 3. c) The exponential function is strictly increasing, so \(f\) is maximized where the exponent \(-x^2+4x\) is maximized. Its vertex is at \(x=2\). The square-root function is also strictly increasing, so \(h\) has its maximum at \(x=2\). 4. d) A strictly increasing function preserves inequalities. If \(f(x)\le f(x_0)\) near \(x_0\), then \(\varphi(f(x))\le\varphi(f(x_0))\), so a local maximum remains a local maximum. The same argument applies to local minima.

a) \(h'(x)=\frac{f'(x)}{2\sqrt{f(x)}}\) b) \(h'(x)=0\iff f'(x)=0\) c) The maximum occurs at \(x=2\). d) Strictly increasing composition preserves local order, so local maxima and minima occur at the same inputs and keep their types.

- Substitute \(z=x^2\) to solve the quartic intercept equation as a quadratic. - Apply the chain rule to the square-root function. - Compare all critical-point values and endpoint values. - A horizontal tangent has slope \(0\).

1. a) Set the radicand equal to \(0\). With \(z=x^2\), \(-z^2+8z+9=0\), so \(z=9\) or \(z=-1\). Thus, the real intercepts are \(x=\pm3\), and the base width is \(6\) units. 2. For interior points, \(f'(x)=\frac{-2x^3+8x}{\sqrt{-x^4+8x^2+9}}\). Critical numbers occur at \(x=0\) and \(x=\pm2\). 3. The function values are \(f(0)=3\) and \(f(\pm2)=5\), while the endpoint values are \(0\). Therefore, the maximum points are \((-2,5)\) and \((2,5)\). 4. c) Since \(f'(0)=0\), the tangent is horizontal. Because \(f(0)=3\), its equation is \(y=3\).

a) \(6\) units b) \((-2,5)\) and \((2,5)\) c) \(f'(0)=0\), so the tangent line is \(y=3\).

- Substitute the given x-coordinates to check the common points. - Factor the first derivative to find all horizontal tangents. - Use the second derivative or a sign chart to classify critical points. - Evaluate \(f_k(k)\), then eliminate the parameter.

1. Direct substitution gives \(f_k(0)=0\) and \(f_k(4)=\frac{64}{6}-4(k+2)+4k=\frac83\). Both values are independent of \(k\). 2. Differentiate and factor: \(f_k'(x)=\frac12x^2-\frac12(k+2)x+k=\frac12(x-2)(x-k)\). Therefore, \(f_k'(2)=0\) for every \(k\), so every graph has a horizontal tangent at \(x=2\). The second derivative is \(f_k''(x)=x-\frac12(k+2)\), so \(f_k''(2)=1-\frac k2\). Thus, \(x=2\) is a local maximum when \(k>2\), a local minimum when \(k<2\), and a stationary inflection point when \(k=2\). 3. The factorization of the first derivative shows that its other zero is \(x=k\). When \(k\ne2\), this zero is simple, so the derivative changes sign and the point is a local extremum. 4. Evaluate the function at \(x=k\): \(f_k(k)=-\frac1{12}k^3+\frac12k^2\). Since the x-coordinate is \(k\), replace \(k\) by \(x\): \(h(x)=-\frac1{12}x^3+\frac12x^2\), with \(x\ne2\).

a) The common points are \((0, 0)\) and \(\left(4, \frac83\right)\). b) \(f_k'(2)=0\). The point is a local extremum for \(k\ne2\); for \(k>2\), it is a local maximum. c) For \(k\ne2\), the other local extremum occurs at \(x=k\). d) \(h(x)=-\frac1{12}x^3+\frac12x^2\), for \(x\ne2\).

- For part a, identify the intercepts and integrate over the bounded interval. - Factor the first derivative to find the critical numbers. - Use a sign chart to classify the critical points for the different parameter cases. - Eliminate the parameter from the coordinates of the off-axis extremum.

1. For \(a=4\), \(f_4(x)=(x-2)^2e^x\). The bounded region extends from \(x=0\) to \(x=2\). Two applications of integration by parts give the antiderivative \(F(x)=(x^2-6x+10)e^x\). Therefore, the area is \(F(2)-F(0)=2e^2-10\approx4.778\) square units. 2. Differentiate and factor: \(f_a'(x)=x(x-a+2)e^x\). The critical numbers are \(x=0\) and \(x=a-2\). When \(a=2\), these coincide, and the derivative does not change sign; \((0, 2)\) is a stationary inflection point, so there are no local extrema. When \(a>2\), the local maximum is \(E_1=(0, a)\), and the local minimum is \(E_2=\left(a-2, (4-a)e^{a-2}\right)\). When \(a<2\), \(E_1\) is the local minimum and \(E_2\) is the local maximum. 3. For the extremum not on the y-axis, let \(x=a-2\), so \(a=x+2\). Then \(y=(4-a)e^{a-2}=(2-x)e^x\). Because \(a\ne2\), the locus is \(y=(2-x)e^x\), with \(x\ne0\).

a) \(2e^2-10\approx4.778\) square units b) If \(a=2\), there are no local extrema; \((0, 2)\) is a stationary inflection point. If \(a>2\), \((0, a)\) is a local maximum and \(\left(a-2, (4-a)e^{a-2}\right)\) is a local minimum. If \(a<2\), the classifications are reversed. c) \(y=(2-x)e^x\), with \(x\ne0\)

- Translate each point and extremum condition into an equation. - A local extremum at \(x = a\) requires \(f'(a) = 0\). - Solve enough independent conditions to determine the polynomial, then test the remaining condition. - A contradiction shows that the full set of requirements is impossible.

1. Let \(f(x) = ax^3 + bx^2 + cx + d\). Then \(f'(x) = 3ax^2 + 2bx + c\). 2. The conditions at \((0, 4)\) give \(f(0) = 4\), so \(d = 4\), and \(f'(0) = 0\), so \(c = 0\). 3. The conditions at \((2, 0)\) give \(f(2) = 0\), so \(8a + 4b = -4\), and \(f'(2) = 0\), so \(12a + 4b = 0\). 4. Solving gives \(a = 1\) and \(b = -3\), so the only possible polynomial is \(f(x) = x^3 - 3x^2 + 4\). 5. Its second derivative is \(f''(x) = 6x - 6\). Since \(f''(0) < 0\) and \(f''(2) > 0\), the required maximum and minimum classifications are satisfied. 6. However, \(f(1) = 1 - 3 + 4 = 2\), not \(3\). Therefore, no cubic polynomial satisfies all three conditions.

No such cubic polynomial exists. The extremum conditions force \(f(x) = x^3 - 3x^2 + 4\), but this polynomial passes through \((1, 2)\), not \((1, 3)\).