Second Derivative Test - Calculus 1

1. Let $f$ and $g$ be twice-differentiable functions. Suppose $g(1)=2$, $g'(1)=0$, and $g''(1)=3$. Also, $f'(2)=-4$. Let $h(x) = f(g(x))$. Use the Second Derivative Test to classify the point $x=1$ for the function $h(x)$.
   Answer: $h$ has a local maximum at $x=1$. First, determine if $x=1$ is a critical point for $h(x)$. Using the chain rule, $h'(x) = f'(g(x))g'(x)$. At $x=1$, $h'(1) = f'(g(1))g'(1) = f'(2) \cdot 0 = 0$. Since $h'(1)=0$, $x=1$ is a critical point. Next, find the second derivative of $h(x)$ using the product rule and chain rule: $h''(x) = f''(g(x))[g'(x)]^2 + f'(g(x))g''(x)$. Now evaluate at $x=1$: $h''(1) = f''(g(1))[g'(1)]^2 + f'(g(1))g''(1) = f''(2)[0]^2 + f'(2)(3) = 0 + (-4)(3) = -12$. Since $h''(1) = -12 < 0$, the Second Derivative Test indicates that $h(x)$ has a local maximum at $x=1$.
2. In which step does the student's first error appear?
   Answer: Step 5 Let's check each step. Step 1 correctly applies the product rule. Step 2 correctly solves for the critical point. Step 3 correctly finds the second derivative. Step 4 correctly evaluates the second derivative at the critical point. Step 5 correctly identifies that $f''(0) < 0$, but draws the wrong conclusion. The condition $f'(c)=0$ and $f''(c)<0$ implies that the function is concave down at the critical point, which corresponds to a local maximum, not a local minimum. The first error is in the final conclusion drawn from the results of the test.
3. Suppose $f'(c)=0$ for a twice-differentiable function $f$. Which of the following conditions is sufficient to conclude that $f$ has a local minimum at $x=c$?
   Answer: $f''(c) > 0$. The Second Derivative Test states that if $f'(c)=0$ and $f''(c)>0$, then $f$ has a local minimum at $x=c$. Choice D is the precise statement of this test. Choice A is a stronger condition than necessary; we only need to know the sign of $f''(c)$ at the point $c$ itself. Choice B is incorrect because if $f''(c)=0$, the test is inconclusive. Choice C describes a condition for an inflection point where the function is increasing, not a local minimum (this is related to the Higher-Order Derivative Test, but the conclusion is for an inflection point).
4. Let $f(x)$ be a function such that $f'(3)=0$ and $f''(3)=4$. If $g(x) = \ln(f(x))$ and $f(3)=e^2$, what does the Second Derivative Test reveal about the function $g$ at $x=3$?
   Answer: $g$ has a local minimum at $x=3$. First, we check if $x=3$ is a critical point for $g(x)$. Using the chain rule, $g'(x) = \frac{f'(x)}{f(x)}$. At $x=3$, $g'(3) = \frac{f'(3)}{f(3)} = \frac{0}{e^2} = 0$. So, $x=3$ is a critical point. Next, we find the second derivative of $g(x)$ using the quotient rule: $g''(x) = \frac{f''(x)f(x) - f'(x)f'(x)}{[f(x)]^2}$. Now we evaluate $g''(3)$: $g''(3) = \frac{f''(3)f(3) - [f'(3)]^2}{[f(3)]^2} = \frac{(4)(e^2) - (0)^2}{[e^2]^2} = \frac{4e^2}{e^4} = \frac{4}{e^2}$. Since $e>0$, $e^2>0$, so $g''(3) = 4/e^2 > 0$. Because $g'(3)=0$ and $g''(3)>0$, the Second Derivative Test shows that $g(x)$ has a local minimum at $x=3$.
5. Let $f(x) = -x^3 + kx^2 - 10$ for some constant $k$. If $f$ has a local maximum at $x=2$, what is the value of $k$?
   Answer: $k = 3$ For $f$ to have a local maximum at $x=2$, two conditions must be met: $f'(2)=0$ and $f''(2)<0$. First, find the derivative: $f'(x) = -3x^2 + 2kx$. Set $f'(2)=0$: $-3(2)^2 + 2k(2) = 0 \implies -12 + 4k = 0 \implies k=3$. Now, we must verify this value of $k$ satisfies the second condition. The second derivative is $f''(x) = -6x + 2k$. With $k=3$, this becomes $f''(x) = -6x+6$. Evaluating at $x=2$: $f''(2) = -6(2)+6 = -12+6 = -6$. Since $f''(2) = -6 < 0$, the function is concave down at the critical point, confirming a local maximum. Thus, $k=3$ is the correct value.
6. Consider the function $f(x) = (x-3)^4 + 5$. When using the Second Derivative Test to analyze the critical point at $x=3$, what is the conclusion?
   Answer: The test is inconclusive for the critical point at $x=3$. The Second Derivative Test requires evaluating the second derivative at a critical point. First, find the critical points by setting the first derivative to zero. $f'(x) = 4(x-3)^3$. Setting $f'(x)=0$ gives $x=3$. Next, find the second derivative: $f''(x) = 12(x-3)^2$. Now, evaluate the second derivative at the critical point: $f''(3) = 12(3-3)^2 = 0$. When the second derivative is zero at a critical point, the Second Derivative Test is inconclusive. It does not provide information about whether the point is a maximum, minimum, or neither. While further analysis with the First Derivative Test would show $x=3$ is a local minimum, the question specifically asks for the conclusion from the Second Derivative Test itself.
7. Consider the function $f(x) = \frac{x^2 - 8}{x+3}$. The function has a local maximum at which value of $x$?
   Answer: $x=-4$ First, find the critical points using the quotient rule for $f'(x)$. $f'(x) = \frac{2x(x+3) - (x^2-8)(1)}{(x+3)^2} = \frac{2x^2+6x-x^2+8}{(x+3)^2} = \frac{x^2+6x+8}{(x+3)^2} = \frac{(x+4)(x+2)}{(x+3)^2}$. Setting $f'(x)=0$ gives critical points at $x=-4$ and $x=-2$. (Note $x=-3$ is a vertical asymptote). Next, find $f''(x)$. The numerator of $f'(x)$ is $N(x)=x^2+6x+8$ and the denominator is $D(x)=(x+3)^2$. Instead of a full quotient rule, it's easier to find the derivative of the simplified $f'(x)$'s numerator for the sign of $f''(x)$ at critical points, but the formal second derivative is $f''(x) = \frac{(2x+6)(x+3)^2 - (x^2+6x+8) \cdot 2(x+3)}{(x+3)^4} = \frac{(2x+6)(x+3) - 2(x^2+6x+8)}{(x+3)^3} = \frac{2x^2+12x+18 - 2x^2-12x-16}{(x+3)^3} = \frac{2}{(x+3)^3}$. Test the critical points: $f''(-4) = \frac{2}{(-4+3)^3} = -2 < 0$. This indicates a local maximum. $f''(-2) = \frac{2}{(-2+3)^3} = 2 > 0$. This indicates a local minimum. Therefore, the local maximum is at $x=-4$.
8. Let $f(x) = \frac{1}{4}x^4 - x^3 - 2x^2 + 5$. At which of its critical points does $f$ have a local maximum?
   Answer: $x = 0$ First, find the critical points by setting the first derivative equal to zero. $f'(x) = x^3 - 3x^2 - 4x = x(x^2 - 3x - 4) = x(x-4)(x+1)$. The critical points are $x=-1$, $x=0$, and $x=4$. Next, find the second derivative to classify these points: $f''(x) = 3x^2 - 6x - 4$. Now, evaluate $f''(x)$ at each critical point: At $x=-1$: $f''(-1) = 3(-1)^2 - 6(-1) - 4 = 3+6-4 = 5 > 0$. This is a local minimum. At $x=0$: $f''(0) = 3(0)^2 - 6(0) - 4 = -4 < 0$. This is a local maximum. At $x=4$: $f''(4) = 3(4)^2 - 6(4) - 4 = 48-24-4 = 20 > 0$. This is a local minimum. Therefore, the function has a local maximum only at $x=0$.
9. The function $f(x) = x - 2\cos(x)$ has a local maximum on the interval $(0, 2\pi)$ at which of the following $x$-values?
   Answer: $x = 7\pi/6$ First, find the critical points by setting $f'(x)=0$. The derivative is $f'(x) = 1 + 2\sin(x)$. Setting this to zero gives $1+2\sin(x)=0$, or $\sin(x) = -1/2$. In the interval $(0, 2\pi)$, the solutions are $x=7\pi/6$ and $x=11\pi/6$. Next, use the Second Derivative Test to classify these points. The second derivative is $f''(x) = 2\cos(x)$. Evaluate $f''(x)$ at each critical point. At $x=7\pi/6$, $f''(7\pi/6) = 2\cos(7\pi/6) = 2(-\sqrt{3}/2) = -\sqrt{3} < 0$, which indicates a local maximum. At $x=11\pi/6$, $f''(11\pi/6) = 2\cos(11\pi/6) = 2(\sqrt{3}/2) = \sqrt{3} > 0$, which indicates a local minimum. Therefore, the local maximum occurs at $x=7\pi/6$.
10. A function $f(x)$ is twice differentiable and has a critical point at $x=2$. If the second derivative is given by $f''(x) = \ln(x^2 - 2x + 2)$, what does the Second Derivative Test imply about the point $x=2$?
    Answer: $f$ has a local minimum at $x=2$. We are given that $x=2$ is a critical point, so we know $f'(2)=0$. To use the Second Derivative Test, we must evaluate the sign of $f''(2)$. Plugging $x=2$ into the expression for the second derivative gives: $f''(2) = \ln(2^2 - 2(2) + 2) = \ln(4 - 4 + 2) = \ln(2)$. Since $e^0=1$ and $e^1\approx2.718$, we know that $0 < \ln(2) < 1$. Because $f''(2) = \ln(2)$ is a positive number, the function is concave up at the critical point $x=2$. Therefore, the Second Derivative Test implies that $f$ has a local minimum at $x=2$.
11. Let $f(x)$ be a twice-differentiable function with critical points at $x=2$ and $x=6$. The graph of its second derivative, $f''(x)$, is a parabola opening downward with roots at $x=1$ and $x=7$. Which of the following statements must be true?
    Answer: $f$ has local minima at both $x=2$ and $x=6$. The graph of $f''(x)$ is a downward-opening parabola with roots at $x=1$ and $x=7$. This means $f''(x)$ is positive for $x$ between 1 and 7, and negative otherwise. We need to classify the critical points $x=2$ and $x=6$. Both of these points lie in the interval $(1, 7)$. For any $x$ in $(1, 7)$, the value of $f''(x)$ is positive. Therefore, $f''(2) > 0$ and $f''(6) > 0$. According to the Second Derivative Test, if $f'(c)=0$ and $f''(c)>0$, then $f$ has a local minimum at $x=c$. Since this condition holds for both $x=2$ and $x=6$, the function $f$ must have local minima at both points.

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