Calculus 1 Quiz: Graphing Functions And Derivatives

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Graphing Functions And DerivativesQuestion 1 of 20

A function $f$ is defined such that its derivative is $f'(x) = x^2(x-3)$ . Which of the following statements accurately describes the graph of $f$ ?

f

has a local maximum at

x=0

and a local minimum at

x=3

f

has a local minimum at

x=3

and inflection points at

x=0

and

x=2

f

has a point of inflection at

x=0

and a local maximum at

x=3

f

has a local minimum at

x=3

and is concave down on the interval

(0, 2)

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Calculus 1 Quiz

Calculus 1 Quiz: Graphing Functions And Derivatives

Practice Graphing Functions And Derivatives 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 Graphing Functions And Derivatives, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

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.

All questions

Question 1

A function $f$ is defined such that its derivative is $f'(x) = x^2(x-3)$ . Which of the following statements accurately describes the graph of $f$ ?

$f$ has a local maximum at $x=0$ and a local minimum at $x=3$ .
$f$ has a local minimum at $x=3$ and inflection points at $x=0$ and $x=2$ . (correct answer)
$f$ has a point of inflection at $x=0$ and a local maximum at $x=3$ .
$f$ has a local minimum at $x=3$ and is concave down on the interval $(0, 2)$ .

Explanation: First, we find the critical points by setting

f'(x) = 0

, which gives

x=0

and

x=3

. Analyzing the sign of

f'(x)

: for

x<0

f'(x)<0

; for

0<x<3

f'(x)<0

; for

x>3

f'(x)>0

. Since

f'

does not change sign at

x=0

, there is no local extremum there. Since

f'

changes from negative to positive at

x=3

, there is a local minimum at

x=3

. Next, we find the second derivative:

f''(x) = \frac{d}{dx}(x^3-3x^2) = 3x^2-6x = 3x(x-2)

. Setting

f''(x)=0

gives

x=0

and

x=2

. Analyzing the sign of

f''

: for

x<0

f''>0

; for

0<x<2

f''<0

; for

x>2

f''>0

. Since the concavity changes at both

x=0

and

x=2

, these are the locations of inflection points. Thus,

f

has a local minimum at

x=3

and inflection points at

x=0

and

x=2

Question 2

The function $f(x)$ is continuous for all real numbers and its derivative is $f'(x) = (x-2)^{-1/3}$ . What feature does the graph of $y=f(x)$ have at $x=2$ ?

A vertical asymptote
A jump discontinuity
A corner
A vertical tangent (correct answer)

Explanation: The derivative

f'(x) = \frac{1}{\sqrt[3]{x-2}}

is undefined at

x=2

because the denominator is zero. As

x \to 2

, the magnitude of

f'(x)

approaches infinity. This indicates that the slope of the tangent line to the graph of

f(x)

becomes infinitely steep, which is a vertical tangent. The function is given as continuous, so it cannot have a vertical asymptote or a jump discontinuity. A corner has differing finite slopes from the left and right, which is not the case here.

Question 3

A function $f$ is twice differentiable and its second derivative is given by $f''(x) = (x-1)(x-3)^2(x-5)$ . How many points of inflection does the graph of $y=f(x)$ have?

One
Two (correct answer)
Three
Four

Explanation: Points of inflection can occur where

f''(x) = 0

or is undefined. Here,

f''(x) = 0

x=1, 3, 5

. A point of inflection exists only if

f''(x)

changes sign at the point. We test the intervals: For

x<1

f''(x) > 0

. For

1<x<3

f''(x) < 0

. For

3<x<5

f''(x) < 0

. For

x>5

f''(x) > 0

. The sign of

f''(x)

changes at

x=1

(from + to -) and at

x=5

(from - to +). The sign does not change at

x=3

due to the factor

(x-3)^2

having an even power. Thus, there are exactly two points of inflection.

Question 4

The graphs of $y=f(x)$ and $y=g(x)$ are both concave up. The graph of which of the following functions must also be concave up?

$h(x) = f(x) - g(x)$
$h(x) = f(x)g(x)$
$h(x) = f(g(x))$
$h(x) = f(x) + g(x)$ (correct answer)

Explanation: Concave up means the second derivative is positive. We are given

f''(x) > 0

and

g''(x) > 0

. We test the second derivative of each choice. For A,

h''(x) = f''(x) - g''(x)

, which could be positive or negative. For B,

h''(x) = f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x)

, which may not be positive. For C,

h''(x) = f''(g(x))(g'(x))^2 + f'(g(x))g''(x)

, which may not be positive. For D,

h(x) = f(x) + g(x)

, the second derivative is

h''(x) = f''(x) + g''(x)

. Since

f''(x) > 0

and

g''(x) > 0

, their sum must be positive. Therefore, the graph of

h(x) = f(x) + g(x)

must be concave up.

Question 5

The derivative of a function $f$ is given by $f'(x) = \cos(x) - x$ . On the interval $[0, 2\pi]$ , the function $f(x)$ is concave down on the set of all $x$ for which...

$\cos(x) < x$
$\sin(x) > -1$
$\sin(x) < -1$
$-\sin(x) < 1$ (correct answer)

Explanation: The graph of

f(x)

is concave down where

f''(x) < 0

. We find the second derivative:

f''(x) = \frac{d}{dx}(\cos(x) - x) = -\sin(x) - 1

. For concave down behavior, we need

-\sin(x) - 1 < 0

, which simplifies to

-\sin(x) < 1

. Since

-1 \leq \sin(x) \leq 1

, we have

-1 \leq -\sin(x) \leq 1

. Therefore

-\sin(x) < 1

is satisfied for all

x

[0, 2\pi]

except at

x = 3\pi/2

where

\sin(x) = -1

and

-\sin(x) = 1

Question 6

Let $h(x) = \int_{0}^{x} (t^2 - 3t + 2) \, dt$ . On which open interval is the graph of $h(x)$ both decreasing and concave up?

( $-\infty, 1$ )
( $1, 3/2$ )
( $3/2, 2$ ) (correct answer)
( $2, \infty$ )

Explanation: By the Fundamental Theorem of Calculus, Part 1,

h'(x) = x^2 - 3x + 2 = (x-1)(x-2)

. The function

h(x)

is decreasing when

h'(x) < 0

, which occurs on the interval

(1, 2)

. To determine concavity, we find the second derivative:

h''(x) = 2x - 3

. The graph of

h(x)

is concave up when

h''(x) > 0

, which means

2x - 3 > 0

, or

x > 3/2

. We need the interval where both conditions are met, which is the intersection of

(1, 2)

and

(3/2, \infty)

. This intersection is the interval

(3/2, 2)

Question 7

Suppose $f$ is a function such that for all $x$ in the interval $(-1, 1)$ , $f'(x) < 0$ and $f''(x) > 0$ . Which of the following statements must be true about the graph of its derivative, $y=f'(x)$ , on $(-1, 1)$ ?

The graph of $f'(x)$ is above the x-axis and is increasing.
The graph of $f'(x)$ is below the x-axis and is decreasing.
The graph of $f'(x)$ is above the x-axis and is decreasing.
The graph of $f'(x)$ is below the x-axis and is increasing. (correct answer)

Explanation: The condition

f'(x) < 0

means that the value of the function

f'(x)

is negative. On a graph, this means the graph of

y=f'(x)

lies below the x-axis. The condition

f''(x) > 0

means that the derivative of

f'(x)

is positive. When the derivative of a function is positive, that function is increasing. Therefore, the function

f'(x)

is increasing. Combining these two facts, the graph of

y=f'(x)

is below the x-axis and is increasing.

Question 8

Let $f$ be a differentiable function. The graph of $f$ passes through the points $(0,1)$ and $(2,5)$ . It is also known that $f'(x)$ is an increasing function. Which of the following statements must be true?

$f(1) < 3$ (correct answer)
$f'(1) > 2$
$f'(0) > 2$
$f(1) > 3$

Explanation: The condition that

f'(x)

is increasing means that

f(x)

is concave up. The secant line connecting the points

(0,1)

and

(2,5)

has the equation

y - 1 = \frac{5-1}{2-0}(x-0)

, which simplifies to

y = 2x + 1

. For a concave up function, the graph of the function on the interval

(0,2)

must lie below its secant line on that interval. At

x=1

, the value on the secant line is

y = 2(1) + 1 = 3

. Therefore, it must be true that

f(1) < 3

Question 9

The function $f$ is three times differentiable. A necessary condition for $f$ to have a point of inflection at $x=c$ is $f''(c) = 0$ . Which of the following is a sufficient condition?

$f'(c)=0$ and $f''(c)=0$
$f''(c)=0$ and $f'''(c) ≠ 0$ (correct answer)
$f''(c)=0$ and $f'''(c) = 0$
$f''(x)$ is continuous at $x=c$

Explanation: A point of inflection requires the concavity to change, which means the sign of

f''(x)

must change as

x

passes through

c

. The condition

f''(c) = 0

alone is not sufficient (e.g.,

f(x) = x^4

x=0

). The Third Derivative Test for inflection points states that if

f''(c) = 0

and

f'''(c) ≠ 0

, then

f''(x)

must change sign at

x=c

, guaranteeing a point of inflection. If

f'''(c) = 0

, the test is inconclusive.

Question 10

For a function $f$ , $f(3) = 5$ and $f'(3) = 0$ . Also, $f'(x) > 0$ for $x > 3$ and $f'(x) < 0$ for $x < 3$ . Which of the following statements about the graph of $f$ is consistent with this information?

The graph has a local maximum at $(3,5)$ and is concave up at $x=3$ .
The graph has a local minimum at $(3,5)$ and is concave up at $x=3$ . (correct answer)
The graph has a local minimum at $(3,5)$ and is concave down at $x=3$ .
The graph has an inflection point at $(3,5)$ .

Explanation: The information about

f'(x)

indicates that it changes from negative to positive at

x=3

. By the First Derivative Test, this means

f

has a local minimum at

x=3

. The value of the minimum is

f(3)=5

. The information

f'(x) < 0

for

x<3

and

f'(x) > 0

for

x>3

means that the function

f'(x)

is increasing as it passes through

x=3

. A function is increasing when its derivative is positive. The derivative of

f'(x)

f''(x)

. Therefore, at

x=3

, we can infer that

f''(3) \ge 0

. A positive second derivative indicates that the graph of

f

is concave up. Thus, the graph has a local minimum at

(3,5)

and is concave up at

x=3

Question 11

Let $f(x) = x^k e^{-x}$ , where $k$ is a positive integer. For what values of $k$ does the graph of $f(x)$ have a point of inflection at $x=2+\sqrt{2}$ ?

$k=1$ only
$k=2$ only (correct answer)
$k=3$ only
$k=4$ only

Explanation: We need to find the second derivative and set it to zero.

f'(x) = kx^{k-1}e^{-x} - x^k e^{-x} = e^{-x}(kx^{k-1} - x^k)

. Then

f''(x) = -e^{-x}(kx^{k-1} - x^k) + e^{-x}(k(k-1)x^{k-2} - kx^{k-1}) = e^{-x}[-kx^{k-1} + x^k + k(k-1)x^{k-2} - kx^{k-1}] = x^{k-2}e^{-x}[-kx + x^2 + k(k-1) - kx] = x^{k-2}e^{-x}[x^2 - 2kx + k(k-1)]

. For an inflection point, we set the bracketed quadratic to zero:

x^2 - 2kx + k(k-1) = 0

. We can solve for

x

using the quadratic formula:

x = \frac{2k \pm \sqrt{4k^2 - 4(k^2-k)}}{2} = k \pm \sqrt{k^2 - (k^2-k)} = k \pm \sqrt{k}

. We are given that an inflection point is at

x=2+\sqrt{2}

. Comparing this with

k \pm \sqrt{k}

, we can see by inspection that

k=2

. For

k=2

, the inflection points are at

x = 2 \pm \sqrt{2}

. Thus,

k=2

is the correct value.

Question 12

The graphs of a function $f(x)$ and its derivative $f'(x)$ are related in many ways. If the graph of $f(x)$ has an inflection point at $x=c$ , which of the following graphical features corresponds to this on the graph of $f'(x)$ ?

An x-intercept at $x=c$ .
A local maximum or a local minimum at $x=c$ . (correct answer)
A vertical asymptote at $x=c$ .
The y-coordinate of the graph is zero at $x=c$ .

Explanation: A point of inflection on the graph of

f(x)

occurs at

x=c

when the concavity changes. This means that the sign of the second derivative,

f''(x)

, changes at

x=c

. Since

f''(x)

is the derivative of

f'(x)

, a sign change in

f''(x)

means that the function

f'(x)

changes from increasing to decreasing, or vice-versa. A point where a function changes from increasing to decreasing (or vice versa) is a local maximum or a local minimum. Therefore, an inflection point of

f(x)

corresponds to a local extremum of

f'(x)

Question 13

Let $f$ be a twice-differentiable function on $(-\infty, \infty)$ . The graph of $f$ is concave up on $(-\infty, 3)$ and concave down on $(3, \infty)$ . Which of the following statements must be true about the derivative function $f'$ ?

$f'$ has a local minimum at $x=3$ .
$f'$ is positive on $(-\infty, 3)$ and negative on $(3, \infty)$ .
$f'$ is increasing on $(-\infty, 3)$ and decreasing on $(3, \infty)$ . (correct answer)
$f'(3)=0$ .

Explanation: The concavity of

f

is determined by the sign of

f''

. Concave up means

f''(x) > 0

, and concave down means

f''(x) < 0

. The behavior of

f'

(whether it is increasing or decreasing) is also determined by the sign of its derivative,

f''

. Therefore, on

(-\infty, 3)

, since

f

is concave up,

f''(x) > 0

, which implies that

f'

is increasing. On

(3, \infty)

, since

f

is concave down,

f''(x) < 0

, which implies that

f'

is decreasing. This describes the behavior of

f'

correctly.

Question 14

A twice-differentiable function $f$ has a horizontal tangent at $x=1$ and is decreasing on the interval $(1, 3)$ . The graph of $f$ has a point of inflection at $x=2$ . Which of the following statements about the derivative, $f'$ , must be true?

$f'$ has a local minimum at $x=2$ . (correct answer)
$f'$ has a local maximum at $x=2$ .
$f'$ is positive on the interval $(1, 2)$ .
$f'(2) = 0$ and $f'(1)=0$ .

Explanation: A horizontal tangent at

x=1

means

f'(1) = 0

. The function

f

is decreasing on

(1, 3)

, which means

f'(x) \le 0

on this interval. A point of inflection for

f

x=2

means that

f''(x)

changes sign at

x=2

, which implies that

f'

has a local extremum at

x=2

. Since

f'(1)=0

and

f'(x)

is negative for

x

just to the right of 1,

f'

must be decreasing as it leaves

x=1

. For

f'

to have an extremum at

x=2

, it must change direction. Since it starts by decreasing from 0, the extremum at

x=2

must be a local minimum.

Question 15

Let $f$ be a twice-differentiable function. If $f'(2) = 0$ and the graph of $f''(x)$ is a line with a positive slope and an x-intercept at $x=2$ , what feature does the graph of $f(x)$ have at $x=2$ ?

A local maximum
A local minimum
A point of inflection (correct answer)
A cusp

Explanation: We are given

f'(2) = 0

, so

x=2

is a critical point. The information about

f''(x)

tells us its properties. An x-intercept at

x=2

means

f''(2) = 0

. Because

f''(2)=0

, the Second Derivative Test for extrema is inconclusive. A line with positive slope means that for

x<2

f''(x)

is negative, and for

x>2

f''(x)

is positive. Since

f''(x)

changes sign at

x=2

, the graph of

f(x)

has a point of inflection at

x=2

. Because the sign of

f'

does not change (it is decreasing to

f'(2)=0

and then continues decreasing since

f''

is negative then positive... wait, this is wrong. Let's re-analyze

f'

. We know

f'(2)=0

and

f''

changes from neg to pos. This means

f'

is decreasing for

x<2

and increasing for

x>2

. This implies

f'

has a local minimum at

x=2

. Since

f'(2)=0

, it must be that

f'(x) > 0

for

x

near 2. So

f

is always increasing. Thus no extremum. Let's re-read my analysis. At

x=2

f'(2)=0

and

f''

changes sign. The First Derivative Test still applies. Since

f'' < 0

for

x < 2

f'

is decreasing. Since

f'' > 0

for

x > 2

f'

is increasing. This means

f'

has a minimum at

x=2

. Since

f'(2)=0

f'(x)

must be positive on both sides of

x=2

(for

x

near 2). So

f'

does not change sign. Therefore, there is no local extremum. There is, however, an inflection point because

f''

changes sign.

Question 16

Let $f$ be a differentiable function such that $f(-x) = f(x)$ for all $x$ in its domain. Which of the following statements must be true about its derivative, $f'(x)$ ?

$f'(x)$ is an even function.
$f'(x)$ is an odd function. (correct answer)
The graph of $f'(x)$ is a horizontal line.
$f'(x)$ is a positive function for all $x$ .

Explanation: The condition

f(-x) = f(x)

defines an even function. To find the property of its derivative, we differentiate both sides of the equation with respect to

x

. Using the chain rule on the left side, we get

\frac{d}{dx}f(-x) = f'(-x) \cdot (-1) = -f'(-x)

. The derivative of the right side is

\frac{d}{dx}f(x) = f'(x)

. Setting them equal gives

-f'(-x) = f'(x)

, or

f'(-x) = -f'(x)

. This is the definition of an odd function.

Question 17

Which of the following graphical features does the function $f(x) = \frac{x^2-3}{x-2}$ possess?

A local minimum at $(3, 6)$ and at least one inflection point.
A local maximum at $(1, 2)$ and a slant asymptote given by $y=x$ .
A local maximum at $(1, 2)$ and is concave up on the interval $(2, \infty)$ . (correct answer)
A local minimum at $(3, 6)$ and is concave down on the interval $(-\infty, \infty)$ .

Explanation: First, find the derivative:

f'(x) = \frac{2x(x-2) - (x^2-3)(1)}{(x-2)^2} = \frac{x^2-4x+3}{(x-2)^2} = \frac{(x-1)(x-3)}{(x-2)^2}

. Critical points are

x=1

and

x=3

. By the first derivative test,

f

has a local maximum at

x=1

and a local minimum at

x=3

. The value at the maximum is

f(1) = (1-3)/(1-2) = 2

. So, a local maximum exists at

(1, 2)

. Next, find the second derivative:

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

. Since

f''(x)

is never zero, there are no inflection points. We check concavity: for

x>2

f''(x) > 0

, so the graph is concave up on

(2, \infty)

. For

x<2

f''(x) < 0

, so it is concave down on

(-\infty, 2)

. Choice C combines a correct statement about the local maximum with a correct statement about concavity.

Question 18

Consider the function $f(x) = x^{2/3}(x-5)$ . Which statement correctly describes a key feature of the graph of $f$ ?

The graph has a local maximum at $(0, 0)$ and a local minimum at $(2, -3\sqrt[3]{4})$ . (correct answer)
The graph has local extrema at $x=0$ and $x=2$ , both at points of horizontal tangency.
The graph has a single critical point at $x=2$ , which corresponds to a local minimum.
The graph has inflection points at $x=-1$ and $x=0$ .

Explanation: The derivative is

f'(x) = \frac{d}{dx}(x^{5/3} - 5x^{2/3}) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5(x-2)}{3x^{1/3}}

. The critical points occur where

f'(x)=0

(at

x=2

) and where

f'(x)

is undefined (at

x=0

). Analyzing the sign of

f'

: for

x<0

f'>0

; for

0<x<2

f'<0

; for

x>2

f'>0

. This indicates a local maximum at

x=0

and a local minimum at

x=2

. The function values are

f(0)=0

and

f(2)=2^{2/3}(2-5) = -3\sqrt[3]{4}

. As

x \to 0

f'(x) \to \pm\infty

, indicating a cusp (a non-differentiable point) at

(0,0)

, not a horizontal tangent. The second derivative is

f''(x) = \frac{10(x+1)}{9x^{4/3}}

, which shows an inflection point at

x=-1

but not at

x=0

Question 19

Let $f$ be a function such that $f'(x) = (x-a)^2 g(x)$ where $a$ is a constant and $g$ is a differentiable function with $g(a) ≠ 0$ . What does the graph of $f$ have at $x=a$ ?

A local maximum or minimum, depending on the sign of $g(a)$ .
A point of inflection.
A vertical asymptote.
A horizontal tangent but no local extremum. (correct answer)

Explanation: A critical point occurs where

f'(x)=0

, which is at

x=a

. Thus, there is a horizontal tangent at

x=a

. To determine if it is a local extremum, we use the First Derivative Test. The factor

(x-a)^2

is always non-negative. Since

g(x)

is continuous and

g(a) ≠ 0

, the sign of

g(x)

will be the same in a small neighborhood around

x=a

. Therefore, the sign of

f'(x)

does not change as

x

passes through

a

. Because the sign of the derivative does not change,

f

has no local extremum at

x=a

Question 20

A function $f$ is twice-differentiable and its first derivative is given by $f'(x) = x^2(x-4)$ . Which of the following correctly describes the graph of $y=f(x)$ ?

One local minimum and two points of inflection. (correct answer)
One local maximum and one point of inflection.
A local minimum at $x=4$ and a local maximum at $x=0$ .
One local minimum and one point of inflection.

Explanation: Critical points of

f(x)

occur where

f'(x) = 0

, which are

x=0

and

x=4

. Analyzing the sign of

f'(x)

, for

x<0

f'(x)<0

. For

0<x<4

f'(x)<0

. For

x>4

f'(x)>0

. Since the sign of

f'(x)

does not change at

x=0

, there is no extremum there. The sign changes from negative to positive at

x=4

, so there is a local minimum at

x=4

. To find inflection points, we analyze

f''(x)

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

, so

f''(x) = 3x^2 - 8x = x(3x-8)

. The potential inflection points are at

x=0

and

x=8/3

. The sign of

f''(x)

changes at both of these values (from positive to negative at

x=0

, and from negative to positive at

x=8/3

). Therefore, there are two points of inflection. The function has one local minimum and two points of inflection.