Calculus 1 Quiz: Determining Concavity

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Determining ConcavityQuestion 1 of 20

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 $b$ ?

8/\pi + \pi/4

1/2

4/\pi - \pi/8

8/\pi - \pi/4

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Calculus 1 Quiz

Calculus 1 Quiz: Determining Concavity

Practice Determining Concavity in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Determining Concavity, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

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 $b$ ?

$8/\pi + \pi/4$
$1/2$
$4/\pi - \pi/8$
$8/\pi - \pi/4$ (correct answer)

Explanation: 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.

Question 2

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 $g''(x)$ ?

$g''(x) = (x-3)^2$
$g''(x) = -(x-3)^2$
$g''(x) = (x-3)$
$g''(x) = (3-x)$ (correct answer)

Explanation: 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.

Question 3

A particle moves along the x-axis with position given by $s(t) = t^3 - 6t^2 + 9t + 1$ for $t \ge 0$ . The velocity of the particle is increasing when the graph of the position function is...

concave up. (correct answer)
concave down.
increasing.
decreasing.

Explanation: Velocity is the derivative of position,

v(t) = s'(t)

. The velocity is increasing when its derivative,

v'(t)

, is positive. The derivative of velocity is acceleration,

a(t) = v'(t)

. Since

v(t) = s'(t)

, we have

a(t) = s''(t)

. Therefore, the velocity is increasing when

s''(t) > 0

. The condition

s''(t) > 0

means that the graph of the position function,

s(t)

, is concave up. This is a conceptual question linking the physical meaning of derivatives to the geometric property of concavity.

Question 4

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 $k$ ?

$-3/2$
$-7/3$ (correct answer)
$-2$
$5$

Explanation: 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

Question 5

The value of a stock, $V(t)$ , in dollars, is modeled by a twice-differentiable function of time $t$ in months. For $t \in [0, 6]$ , the rate of change of the stock's value, $V'(t)$ , is increasing for the first two months, and then decreasing for the next four months. Which of the following statements about the graph of $V(t)$ must be true?

The value of the stock, $V(t)$ , increased over the entire interval $[0, 6]$ .
The value of the stock, $V(t)$ , was at its maximum at $t=2$ .
The graph of $V(t)$ changes from concave up to concave down at $t=2$ . (correct answer)
The graph of $V(t)$ is concave up over the entire interval $[0, 6]$ .

Explanation: The concavity of the graph of

V(t)

is determined by the sign of its second derivative,

V''(t)

. The rate of change of the stock's value is

V'(t)

. The statement that

V'(t)

is increasing means that its derivative,

(V'(t))' = V''(t)

, is positive. The statement that

V'(t)

is decreasing means that its derivative,

V''(t)

, is negative.

For $t \in (0, 2)$ , $V'(t)$ is increasing, so $V''(t) > 0$ . This means the graph of $V(t)$ is concave up.
For $t \in (2, 6)$ , $V'(t)$ is decreasing, so $V''(t) < 0$ . This means the graph of $V(t)$ is concave down. Since the concavity of the graph of $V(t)$ changes from concave up to concave down at $t=2$ , this is the correct description. A change in concavity signifies a point of inflection.

Question 6

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 $x=2$ ?

The graph is concave up at $x=2$ .
The graph is concave down at $x=2$ . (correct answer)
The graph has a point of inflection at $x=2$ .
The concavity cannot be determined from the information given.

Explanation: To determine the concavity of

h(x)

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

Question 7

Let $f$ be a twice-differentiable function such that $f'(x) > 0$ and $f''(x) > 0$ for all real numbers $x$ . Let $g(x) = f(x^2)$ . On which interval is the graph of $g$ guaranteed to be concave up?

$(0, \infty)$
$(-\infty, 0)$
$(-\infty, \infty)$ (correct answer)
The concavity of $g$ cannot be determined.

Explanation: To determine the concavity of

g(x)

, we find its second derivative using the chain rule twice.

g'(x) = f'(x^2) \cdot 2x

g''(x) = (f''(x^2) \cdot 2x) \cdot 2x + f'(x^2) \cdot 2 = 4x^2 f''(x^2) + 2f'(x^2)

We are given that

f'(x) > 0

and

f''(x) > 0

for all inputs. The input for

f'

and

f''

x^2

, which is always non-negative. Therefore,

f'(x^2) > 0

and

f''(x^2) > 0

for all

x

. Also,

4x^2 \ge 0

. The term

4x^2 f''(x^2)

is non-negative, and the term

2f'(x^2)

is strictly positive. The sum of a non-negative term and a strictly positive term is always strictly positive. Thus,

g''(x) > 0

for all

x

, and the graph of

g

is concave up on

(-\infty, \infty)

Question 8

Let $f(x) = x \sqrt{4-x}$ . On which interval is the graph of $f$ concave down? Note the domain of $f$ is $x \le 4$ .

$(8/3, 4)$
The graph is never concave down.
$(0, 8/3)$
$(-\infty, 4)$ (correct answer)

Explanation: To determine where a function is concave down, you need to find where the second derivative is negative. This requires computing both the first and second derivatives of the given function. For

f(x) = x\sqrt{4-x}

, start by finding the first derivative using the product rule. Let

u = x

and

v = \sqrt{4-x} = (4-x)^{1/2}

. Then

u' = 1

and

v' = \frac{1}{2}(4-x)^{-1/2} \cdot (-1) = -\frac{1}{2\sqrt{4-x}}

. Using the product rule:

f'(x) = 1 \cdot \sqrt{4-x} + x \cdot \left(-\frac{1}{2\sqrt{4-x}}\right) = \sqrt{4-x} - \frac{x}{2\sqrt{4-x}}

Combining over a common denominator:

f'(x) = \frac{2(4-x) - x}{2\sqrt{4-x}} = \frac{8-3x}{2\sqrt{4-x}}

Now find the second derivative using the quotient rule. After careful calculation (involving both the quotient rule and chain rule), you get:

f''(x) = -\frac{3x+8}{4(4-x)^{3/2}}

Since

x \leq 4

(the domain), we have

3x + 8 > 0

for all values in the domain, and

(4-x)^{3/2} > 0

for

x < 4

. Therefore,

f''(x) < 0

throughout the entire domain where the function is differentiable. Choice A)

(8/3, 4)

and C)

(0, 8/3)

only give partial intervals. Choice B) incorrectly claims the function is never concave down. Choice D)

(-\infty, 4)

correctly identifies that the function is concave down everywhere in its domain. Study tip: When checking concavity, always compute the second derivative carefully and consider the function's entire domain, not just where it might change behavior.

Question 9

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?

The graph of $g$ is increasing on $(-3, 5)$ .
The graph of $g$ has a local minimum in $(-3, 5)$ .
The graph of $g$ is concave up on $(-3, 5)$ . (correct answer)
The graph of $g$ has an inflection point in $(-3, 5)$ .

Explanation: 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.

Question 10

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 $(1, -1)$ ?

The curve is concave up. (correct answer)
The curve is concave down.
The curve has a point of inflection.
The concavity is the same as at the point $(1, 2)$ .

Explanation: 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.

Question 11

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?

$f(2)$ is a local minimum value of $f$ .
$f(2)$ is a local maximum value of $f$ . (correct answer)
The graph of $f$ has a point of inflection at $x=2$ .
$f(x) < f(0)$ for all $x \in (0, 4)$ .

Explanation: 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.

Question 12

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 $g(x)$ ?

The concavity cannot be determined without more information about $f(x)$ .
The graph of $g(x)$ is always concave up.
The concavity of the graph of $g(x)$ depends on the sign of $f'(x)$ .
The graph of $g(x)$ is always concave down. (correct answer)

Explanation: 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.

Question 13

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?

$(1, \infty)$ (correct answer)
$(-\infty, 1)$
$(2, \infty)$
$(1/2, \infty)$

Explanation: 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)

Question 14

On which of the following open intervals is the graph of the function $f(x) = x e^{-2x}$ concave up?

$(-\infty, 1/2)$
$(1/2, \infty)$
$(-\infty, 1)$
$(1, \infty)$ (correct answer)

Explanation: 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)

Question 15

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?

$f$ is a decreasing function on $(a,b)$ .
$f'$ is a decreasing function on $(a,b)$ . (correct answer)
$f'(c) = 0$ for some $c \in (a,b)$ .
The graph of $f'$ is concave down on $(a,b)$ .

Explanation: 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.

Question 16

Determine the complete set of intervals where the graph of the function $f(x) = \ln(x^2 + 4)$ is concave down.

$(-2, 2)$
$(-\infty, -2) \cup (2, \infty)$ (correct answer)
$(0, \infty)$
The graph is never concave down.

Explanation: First, find the second derivative of

f(x)

f'(x) = \frac{2x}{x^2+4}

Using the quotient rule for the second derivative:

f''(x) = \frac{2(x^2+4) - 2x(2x)}{(x^2+4)^2} = \frac{2x^2+8-4x^2}{(x^2+4)^2} = \frac{8-2x^2}{(x^2+4)^2} = \frac{2(4-x^2)}{(x^2+4)^2}

The graph is concave down when

f''(x) < 0

. Since the denominator

(x^2+4)^2

is always positive, the sign depends on the numerator,

2(4-x^2)

. We need

4-x^2 < 0

, which means

x^2 > 4

. This inequality holds for

x < -2

x > 2

. Therefore, the function is concave down on

(-\infty, -2) \cup (2, \infty)

Question 17

On which open interval is the graph of the function $h(x) = \frac{x+1}{x-3}$ concave down?

$(3, \infty)$
$(-\infty, 3)$ (correct answer)
$(-\infty, -1)$
The graph is never concave down.

Explanation: 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)

Question 18

Let $f(x) = x^4 - 4x^3 + 6x^2$ . Which of the following statements is true about the concavity of the graph of $f$ ?

The graph is concave up on $(-\infty, 1)$ and concave down on $(1, \infty)$ .
The graph is concave down on $(-\infty, 1)$ and concave up on $(1, \infty)$ .
The graph is always concave up for all real numbers. (correct answer)
The graph is concave up on $(-\infty, 0)$ and $(1, \infty)$ only.

Explanation: To determine concavity, we must find the second derivative,

f''(x)

f'(x) = 4x^3 - 12x^2 + 12x

f''(x) = 12x^2 - 24x + 12 = 12(x^2 - 2x + 1) = 12(x-1)^2

The graph is concave up when

f''(x) > 0

. Since

(x-1)^2

is a squared term, it is non-negative for all

x

. Specifically,

(x-1)^2 > 0

for

x \neq 1

and

(x-1)^2 = 0

for

x=1

. Thus,

f''(x) \ge 0

for all real numbers

x

. A function is considered concave up on an interval if its second derivative is non-negative. Since concavity does not change sign at

x=1

, the graph is considered always concave up.

Question 19

Let $F(x) = \int_0^x (t^2-4)e^{-t^2} dt$ . Determine the interval(s) on which the graph of $F(x)$ is concave down.

$(-2, 2)$
$(-\infty, -\sqrt{5}) \cup (0, \sqrt{5})$
$(-\sqrt{5}, 0) \cup (\sqrt{5}, \infty)$ (correct answer)
$(-\infty, -2) \cup (2, \infty)$

Explanation: To determine the concavity of

F(x)

, we need to find

F''(x)

. By the Fundamental Theorem of Calculus Part 1,

F'(x) = (x^2-4)e^{-x^2}

. Now we differentiate

F'(x)

using the product rule to find

F''(x)

F''(x) = (2x)e^{-x^2} + (x^2-4)e^{-x^2}(-2x) = 2xe^{-x^2}(1 - (x^2-4)) = 2xe^{-x^2}(5 - x^2)

. The graph is concave down when

F''(x) < 0

. Since

e^{-x^2}

is always positive, we analyze the sign of

2x(5-x^2)

. The roots are

x=0, x=\pm\sqrt{5}

. A sign analysis shows

F''(x) < 0

on the intervals

(-\sqrt{5}, 0)

and

(\sqrt{5}, \infty)

Question 20

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 $x=2$ ?

$f$ has a local maximum at $x=2$ . (correct answer)
$f$ has a local minimum at $x=2$ .
$f$ has an inflection point at $x=2$ .
$f$ is decreasing at $x=2$ .

Explanation: 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.