Determining Concavity - Calculus 1

1. Let $h(x) = f(g(x))$, where $f$ and $g$ are twice-differentiable functions. Given the following values: $g(2)=3$, $g'(2)=0$, $g''(2)=-1$, $f'(3)=5$, and $f''(3)=2$. Which statement describes the graph of $h(x)$ at $x=2$?
   Answer: The graph is concave down at $x=2$. To determine the concavity of $h(x)$ at $x=2$, we need to find the sign of $h''(2)$. First, we find the first and second derivatives of $h(x)$ using the chain rule and product rule. $h'(x) = f'(g(x)) \cdot g'(x)$ $h''(x) = f''(g(x)) \cdot [g'(x)]^2 + f'(g(x)) \cdot g''(x)$ Now, substitute $x=2$ and the given values: $h''(2) = f''(g(2)) \cdot [g'(2)]^2 + f'(g(2)) \cdot g''(2)$ $h''(2) = f''(3) \cdot [0]^2 + f'(3) \cdot (-1)$ $h''(2) = (2) \cdot 0 + (5) \cdot (-1) = -5$. Since $h''(2) = -5 < 0$, the graph of $h(x)$ is concave down at $x=2$.
2. Let $f(x) = x e^{-kx}$ for some constant $k > 0$. The graph of $f(x)$ has a point of inflection at $x=1$. On what interval is $f(x)$ concave up?
   Answer: $(1, \infty)$ First, we find the second derivative of $f(x)$. $f'(x) = 1 \cdot e^{-kx} + x \cdot e^{-kx}(-k) = e^{-kx}(1-kx)$. $f''(x) = e^{-kx}(-k)(1-kx) + e^{-kx}(-k) = -k e^{-kx}(1-kx+1) = -k e^{-kx}(2-kx)$. An inflection point occurs where $f''(x)$ changes sign. We set $f''(x)=0$. Since $-k e^{-kx} \neq 0$, we must have $2-kx=0$, which implies $x=2/k$. We are given that the inflection point is at $x=1$. So, $1 = 2/k$, which gives $k=2$. Now we substitute $k=2$ back into the second derivative: $f''(x) = -2e^{-2x}(2-2x) = 4e^{-2x}(x-1)$. The graph of $f(x)$ is concave up when $f''(x) > 0$. We need to solve $4e^{-2x}(x-1) > 0$. Since $4e^{-2x}$ is always positive, the sign of $f''(x)$ is determined by the sign of $(x-1)$. $x-1 > 0 \implies x > 1$. Therefore, $f(x)$ is concave up on the interval $(1, \infty)$.
3. Let $f(x)=(x^2-2x-1)e^x$. The graph of $f$ has points of inflection at which of the following $x$-values?
   Answer: $x = -3$ and $x=1$ Points of inflection occur where the second derivative changes sign, which means you need to find where $f''(x) = 0$ and verify the sign change. Starting with $f(x) = (x^2 - 2x - 1)e^x$, you'll need the product rule twice. First, find $f'(x)$: $f'(x) = (2x - 2)e^x + (x^2 - 2x - 1)e^x = (x^2 - 1)e^x$ Now find $f''(x)$ using the product rule again: $f''(x) = (2x)e^x + (x^2 - 1)e^x = (x^2 + 2x - 1)e^x$ Since $e^x > 0$ for all $x$, the sign of $f''(x)$ depends only on $(x^2 + 2x - 1)$. Setting this equal to zero: $x^2 + 2x - 1 = 0$ Using the quadratic formula: $x = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$ This gives $x = -1 - \sqrt{2} \approx -3.41$ and $x = -1 + \sqrt{2} \approx 0.41$. Since $-1 - \sqrt{2} \approx -2.414$, this is closest to $-3$, and $-1 + \sqrt{2} \approx 0.414$ is closest to $1$. Choice A gives the wrong values entirely. Choice B only identifies one inflection point when there are two. Choice C shows $1 \pm \sqrt{2}$, which comes from incorrectly solving the quadratic or making an algebraic error with the discriminant. Key strategy: When finding inflection points for products involving $e^x$, remember that $e^x$ never equals zero, so focus entirely on where the polynomial factor in $f''(x)$ equals zero.
4. The graph of a twice-differentiable function $g$ is concave up on $(-\infty, 3)$ and concave down on $(3, \infty)$. Which of the following could be the equation for $g''(x)$?
   Answer: $g''(x) = (3-x)$ When you see a question about concavity and the second derivative, remember that the sign of $g''(x)$ determines concavity: positive means concave up, negative means concave down. Since the function is concave up on $(-\infty, 3)$, we need $g''(x) > 0$ for all $x < 3$. Since it's concave down on $(3, \infty)$, we need $g''(x) < 0$ for all $x > 3$. The concavity changes at $x = 3$, so this is an inflection point where $g''(3) = 0$. Let's check each option systematically. For choice A, $g''(x) = (x-3)^2$ is always non-negative since it's a perfect square. It equals zero at $x = 3$ but is positive everywhere else, so the function would be concave up on both sides of $x = 3$. Choice B, $g''(x) = -(x-3)^2$, is always non-positive, making the function concave down everywhere except at $x = 3$. Choice C gives us $g''(x) = (x-3)$. This is negative when $x < 3$ (concave down) and positive when $x > 3$ (concave up) — exactly opposite to what we need. Choice D, $g''(x) = (3-x) = -(x-3)$, is positive when $x < 3$ (concave up) and negative when $x > 3$ (concave down). This perfectly matches our requirements. Study tip: When analyzing concavity changes, first identify where $g''(x)$ must be positive or negative, then test each option by picking specific values on either side of the inflection point.
5. Consider the function $f(x) = \frac{1}{20}x^5 - \frac{1}{3}x^4 + x^3 - 2x^2 + 5$. On which interval is the graph of $f$ concave down?
   Answer: $(-\infty, 2)$ First, find the second derivative of $f(x)$. $f'(x) = \frac{1}{4}x^4 - \frac{4}{3}x^3 + 3x^2 - 4x$ $f''(x) = x^3 - 4x^2 + 6x - 4$ To find where the graph is concave down, we need to find where $f''(x) < 0$. We can test for rational roots of the cubic. We find $f''(2) = 2^3 - 4(2^2) + 6(2) - 4 = 8 - 16 + 12 - 4 = 0$. So $(x-2)$ is a factor. By polynomial division, $f''(x) = (x-2)(x^2 - 2x + 2)$. The quadratic factor $x^2-2x+2$ has a discriminant of $b^2-4ac = (-2)^2 - 4(1)(2) = -4$, which is negative. Since the leading coefficient is positive, $x^2-2x+2$ is always positive. Therefore, the sign of $f''(x)$ is determined solely by the sign of $(x-2)$. The graph is concave down when $x-2 < 0$, which is $x < 2$.
6. The graph of the function $f(x) = \sin(x) + ax^2 + bx$ has a point of inflection at $(\pi/2, 5)$. What is the value of $b$?
   Answer: $8/\pi - \pi/4$ Points of inflection occur where the second derivative equals zero and changes sign. When you encounter inflection point problems, you need to work with the second derivative and use the given coordinates to find unknown parameters. Given $f(x) = \sin(x) + ax^2 + bx$, let's find the derivatives: - $f'(x) = \cos(x) + 2ax + b$ - $f''(x) = -\sin(x) + 2a$ At the inflection point $(\pi/2, 5)$, the second derivative must equal zero: $f''(\pi/2) = -\sin(\pi/2) + 2a = -1 + 2a = 0$ This gives us $a = 1/2$. Since the point $(\pi/2, 5)$ lies on the curve: $f(\pi/2) = \sin(\pi/2) + a(\pi/2)^2 + b(\pi/2) = 5$ $1 + \frac{1}{2} \cdot \frac{\pi^2}{4} + b \cdot \frac{\pi}{2} = 5$ $1 + \frac{\pi^2}{8} + \frac{b\pi}{2} = 5$ Solving for $b$: $\frac{b\pi}{2} = 4 - \frac{\pi^2}{8}$ $b = \frac{8}{\pi} - \frac{\pi}{4}$ Choice A gives $8/\pi + \pi/4$, incorrectly adding instead of subtracting the $\pi/4$ term. Choice B gives $1/2$, which is the value of $a$, not $b$ — a common mix-up. Choice C gives $4/\pi - \pi/8$, which uses incorrect coefficients in the algebraic manipulation. The correct answer is D. Study tip: For inflection point problems, always find both derivatives first, use the inflection condition ($f'' = 0$) to find one parameter, then use the point coordinates in the original function to find remaining parameters.
7. Let $f(x)$ be a polynomial function such that $f''(x) = c(x-a)^2$ for some non-zero constants $a$ and $c$. Which statement accurately describes the concavity of the graph of $f(x)$?
   Answer: The graph has a point of zero concavity at $x=a$ but no change in concavity. The second derivative is $f''(x) = c(x-a)^2$. An inflection point exists at $x=a$ only if $f''(a)=0$ and the sign of $f''(x)$ changes as $x$ passes through $a$. Here, $f''(a)=0$. However, the term $(x-a)^2$ is always non-negative. Thus, the sign of $f''(x)$ is determined by the sign of the constant $c$ and does not change. For $x \neq a$, $f''(x)$ will have the same sign as $c$. Since the concavity does not change at $x=a$, there is no inflection point. There is, however, a point where the concavity is momentarily zero.
8. The second derivative of a twice-differentiable function $f$ is given by $f''(x) = x^2(x-4)(x+2)$. On which of the following open intervals is the graph of $f$ concave down?
   Answer: $(-2, 4)$ The graph of $f$ is concave down where $f''(x) < 0$. We analyze the sign of $f''(x) = x^2(x-4)(x+2)$. The term $x^2$ is non-negative for all $x$ and does not affect the sign of $f''(x)$ except at $x=0$ where it makes $f''(x)=0$. The sign is determined by the product $(x-4)(x+2)$. This is a parabola opening upward with roots at $x=4$ and $x=-2$. The product is negative between the roots. Therefore, $f''(x) < 0$ on the interval $(-2, 4)$, excluding $x=0$. However, since the sign of $f''$ is negative on both sides of $x=0$, the function remains concave down across $x=0$. The entire interval is $(-2, 4)$.
9. Find all open intervals on which the graph of $f(x) = x + 2\cos(x)$ is concave up for $x \in [0, 2\pi]$.
   Answer: $(\pi/2, 3\pi/2)$ To determine concavity, we find the second derivative. $f'(x) = 1 - 2\sin(x)$ $f''(x) = -2\cos(x)$ The graph is concave up when $f''(x) > 0$, which means $-2\cos(x) > 0$, or $\cos(x) < 0$. In the interval $[0, 2\pi]$, $\cos(x)$ is negative when $x$ is in the second or third quadrants. This corresponds to the open interval $(\pi/2, 3\pi/2)$. The other choices correspond to intervals where the function is concave down (B) or where it is decreasing (C) and increasing (D).
10. A twice-differentiable function $f$ has the property that $f'(x) = x^2 - \sin(x) + 1$. On which interval is the graph of $f$ concave up?
    Answer: $(1/2, \infty)$ To determine the concavity of $f$, we need to analyze the sign of $f''(x)$. We are given $f'(x)$, so we differentiate it to find $f''(x)$. $f''(x) = \frac{d}{dx}(x^2 - \sin(x) + 1) = 2x - \cos(x)$. The graph of $f$ is concave up when $f''(x) = 2x - \cos(x) > 0$, or $2x > \cos(x)$. Let's analyze the functions $y=2x$ and $y=\cos(x)$. For $x \le -1/2$, $2x \le -1$ while $\cos(x) \ge -1$. At $x=0$, $2x=0$ and $\cos(x)=1$, so $2x < \cos(x)$. For $x \ge 1/2$, $2x \ge 1$ and $\cos(x) \le 1$. Since $2x$ is strictly increasing and $\cos(x)$ oscillates, the inequality $2x > \cos(x)$ will hold for all $x$ greater than some value in $(0, 1/2)$. Specifically, for $x \ge 1/2$, $2x \ge 1$ and $\cos(x) \le 1$, so $2x \ge \cos(x)$. Equality only holds if $x=1/2$ and $\cos(x)=1$, which is impossible. So for $x > 1/2$, $2x > 1 \ge \cos(x)$. Therefore, the graph is concave up on $(1/2, \infty)$.
11. Let $f$ be a twice-differentiable function. If the graph of $f$ is concave down on the interval $(a, b)$, which of the following statements must be true?
    Answer: $f'$ is a decreasing function on $(a,b)$. The concavity of the graph of a function $f$ is determined by the sign of its second derivative, $f''$. If the graph of $f$ is concave down on an interval, then $f''(x) < 0$ on that interval. The sign of the derivative of a function determines whether that function is increasing or decreasing. Since $f''$ is the derivative of $f'$, $f''(x) < 0$ implies that $f'$ is a decreasing function. A function can be increasing and concave down (e.g., $y=\sqrt{x}$), so A is incorrect. C and D are not necessarily true.
12. For the function $f(x) = (x-1)^{2/3} + 4$, determine the interval(s) on which its graph is concave down.
    Answer: $(-\infty, 1) \cup (1, \infty)$ First, find the second derivative of $f(x)$. $f'(x) = \frac{2}{3}(x-1)^{-1/3}$ $f''(x) = \frac{2}{3} \cdot (-\frac{1}{3})(x-1)^{-4/3} = -\frac{2}{9}(x-1)^{-4/3} = \frac{-2}{9(x-1)^{4/3}}$ The graph is concave down when $f''(x) < 0$. The numerator is a constant -2. The denominator is $9(x-1)^{4/3} = 9(\sqrt[3]{x-1})^4$. Since the power is even, the denominator is positive for all $x \neq 1$. Therefore, $f''(x)$ is a negative constant divided by a positive value, which means $f''(x)$ is always negative for all $x$ in its domain, $x \neq 1$. The function is concave down on $(-\infty, 1) \cup (1, \infty)$.
13. On which of the following open intervals is the graph of the function $f(x) = x e^{-2x}$ concave up?
    Answer: $(1, \infty)$ To determine concavity, we must find the second derivative, $f''(x)$. Using the product and chain rules: $f'(x) = e^{-2x} - 2xe^{-2x} = e^{-2x}(1-2x)$ $f''(x) = -2e^{-2x}(1-2x) + e^{-2x}(-2) = -2e^{-2x}(1-2x+1) = -2e^{-2x}(2-2x) = 4e^{-2x}(x-1)$ The term $4e^{-2x}$ is always positive. Thus, the sign of $f''(x)$ is determined by the sign of $(x-1)$. The graph is concave up when $f''(x) > 0$, which occurs when $x-1 > 0$, or $x > 1$. This corresponds to the interval $(1, \infty)$.
14. Let $f(x) = e^x(x^2 + kx)$, where $k$ is a constant. The graph of $f$ has an inflection point at $x=1$. What is the value of $k$?
    Answer: $-7/3$ An inflection point can occur where the second derivative is zero. We must find $f''(x)$. $f'(x) = e^x(x^2+kx) + e^x(2x+k) = e^x(x^2 + (k+2)x + k)$ $f''(x) = e^x(x^2 + (k+2)x + k) + e^x(2x + k+2) = e^x(x^2 + (k+4)x + 2k+2)$ For an inflection point at $x=1$, we must have $f''(1)=0$. $f''(1) = e^1(1^2 + (k+4)(1) + 2k+2) = e(1 + k + 4 + 2k + 2) = e(3k+7)$ Setting $e(3k+7) = 0$ gives $3k+7=0$, so $k=-7/3$. We must also ensure concavity changes, which it does as $f''(x) = e^x(x-1)(x+8/3)$, so the sign of $f''$ changes at $x=1$.
15. Let $f(x)$ be a twice-differentiable function with $f(x) > 0$ and $f''(x) < 0$ for all real numbers $x$. Let $g(x) = \ln(f(x))$. What can be concluded about the concavity of the graph of $g(x)$?
    Answer: The graph of $g(x)$ is always concave down. When you encounter a function composition like $g(x) = \ln(f(x))$ and need to determine concavity, you must find the second derivative of the composite function and analyze its sign. To find $g''(x)$, start with the chain rule. Since $g(x) = \ln(f(x))$, we have: $g'(x) = \frac{f'(x)}{f(x)}$ Taking the derivative again using the quotient rule: $g''(x) = \frac{f''(x) \cdot f(x) - [f'(x)]^2}{[f(x)]^2}$ Now analyze the sign of $g''(x)$. Since $f(x) > 0$ for all $x$, the denominator $[f(x)]^2$ is always positive. The sign of $g''(x)$ depends entirely on the numerator: $f''(x) \cdot f(x) - [f'(x)]^2$. Given that $f''(x) < 0$ and $f(x) > 0$, the first term $f''(x) \cdot f(x)$ is negative. The second term $[f'(x)]^2$ is always non-negative (it's a square). Therefore, the entire numerator is negative, making $g''(x) < 0$ for all $x$. This means $g(x)$ is concave down everywhere. Answer D is correct. Answer A is wrong because we have sufficient information to determine concavity definitively. Answer B incorrectly suggests concave up behavior. Answer C is incorrect because while $f'(x)$ appears in the formula, the overall sign of $g''(x)$ is always negative regardless of whether $f'(x)$ is positive or negative. Study tip: When analyzing concavity of composite functions, always compute the second derivative completely before making conclusions about sign—don't assume the concavity transfers directly from the inner function.
16. Let $g$ be a twice-differentiable function such that its derivative, $g'$, is an increasing function on the interval $(-3, 5)$. Which statement about $g$ must be true on this interval?
    Answer: The graph of $g$ is concave up on $(-3, 5)$. The statement that the derivative, $g'$, is an increasing function on $(-3, 5)$ means that the derivative of $g'$, which is $g''$, must be positive on that interval. If $g''(x) > 0$ on an interval, then the graph of the original function, $g$, is concave up on that interval. Choice A is incorrect because $g'$ could be increasing but have negative values (e.g., increasing from -5 to -1), which would mean $g$ is decreasing. Choices B and D are not guaranteed by the given information.
17. On which open interval is the graph of the function $h(x) = \frac{x+1}{x-3}$ concave down?
    Answer: $(-\infty, 3)$ We must find the second derivative of $h(x)$ using the quotient rule. $h'(x) = \frac{1(x-3) - (x+1)(1)}{(x-3)^2} = \frac{-4}{(x-3)^2} = -4(x-3)^{-2}$ $h''(x) = (-4)(-2)(x-3)^{-3}(1) = 8(x-3)^{-3} = \frac{8}{(x-3)^3}$ The graph is concave down when $h''(x) < 0$. Since the numerator is a positive constant (8), the sign of $h''(x)$ is determined by the denominator, $(x-3)^3$. The expression $(x-3)^3$ is negative when $x-3 < 0$, which means $x < 3$. Thus, the graph is concave down on the interval $(-\infty, 3)$.
18. A function $f$ is continuous and twice-differentiable for all real numbers. It is known that $f(2)=1$, $f'(2)=0$, and $f''(2)=-3$. What can be concluded about the function $f$ at $x=2$?
    Answer: $f$ has a local maximum at $x=2$. This question applies the Second Derivative Test. The condition $f'(2)=0$ indicates that $x=2$ is a critical point of the function (a horizontal tangent). The condition $f''(2)=-3$ means the second derivative is negative at this critical point. According to the Second Derivative Test, if $f'(c)=0$ and $f''(c)<0$, the function has a local maximum at $x=c$. The negative second derivative implies the graph is concave down at $x=2$, so the critical point must be a peak.
19. Let $f$ be a twice-differentiable function. If $f'(2) = 0$ and the graph of $f(x)$ is concave down on the interval $(0, 4)$, which of the following statements must be true?
    Answer: $f(2)$ is a local maximum value of $f$. The graph of $f(x)$ being concave down on $(0, 4)$ means that its first derivative, $f'(x)$, is a decreasing function on this interval. We are given that $f'(2) = 0$. Since $f'(x)$ is decreasing on $(0, 4)$: - For $x$ in $(0, 2)$, we have $x < 2$, so $f'(x) > f'(2) = 0$. This means $f(x)$ is increasing on $(0, 2)$. - For $x$ in $(2, 4)$, we have $x > 2$, so $f'(x) < f'(2) = 0$. This means $f(x)$ is decreasing on $(2, 4)$. A function that is increasing to the left of a point and decreasing to the right has a local maximum at that point. Therefore, $f(2)$ is a local maximum value of $f$. This is also consistent with the Second Derivative Test, as concave down implies $f''(x) \le 0$. Since $f'(2)=0$ and $f''(2) < 0$ (or at least $f''(x) \le 0$ in the neighborhood), it indicates a local maximum.
20. Consider the curve defined by the equation $y^2 - xy + x^2 = 3$. It can be shown that the curve is concave down at the point $(1, 2)$. What is the concavity of the curve at the point $(1, -1)$?
    Answer: The curve is concave up. First, verify that $(1, -1)$ is on the curve: $(-1)^2 - (1)(-1) + (1)^2 = 1 + 1 + 1 = 3$. To find the concavity, we need to find the sign of the second derivative, $y''$, at this point. We use implicit differentiation. Differentiating $y^2 - xy + x^2 = 3$ with respect to $x$: $2y \frac{dy}{dx} - (y + x \frac{dy}{dx}) + 2x = 0$ $\frac{dy}{dx}(2y - x) = y - 2x \implies \frac{dy}{dx} = \frac{y - 2x}{2y - x}$. At $(1, -1)$, $\frac{dy}{dx} = \frac{-1 - 2(1)}{2(-1) - 1} = \frac{-3}{-3} = 1$. Now, differentiate $\frac{dy}{dx}$ to find $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = \frac{(\frac{dy}{dx} - 2)(2y - x) - (y - 2x)(2\frac{dy}{dx} - 1)}{(2y - x)^2}$. Substitute $x=1, y=-1,$ and $\frac{dy}{dx}=1$: $\frac{d^2y}{dx^2} = \frac{(1 - 2)(2(-1) - 1) - (-1 - 2(1))(2(1) - 1)}{(2(-1) - 1)^2} = \frac{(-1)(-3) - (-3)(1)}{(-3)^2} = \frac{3+3}{9} = \frac{6}{9} = \frac{2}{3}$. Since $y'' = 2/3 > 0$, the curve is concave up at $(1, -1)$. The information about the point $(1, 2)$ is extraneous.

Start this quiz