Calculus 1 Quiz: Differentiability And Continuity

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Differentiability And ContinuityQuestion 1 of 20

Consider the function $f(x) = |x^2 - 5x + 4|$ . Which of the following describes the set of points where $f(x)$ is not differentiable?

x = 1

only

x = 4

only

x = 1

and

x = 4

x = 2.5

only

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Calculus 1 Quiz

Calculus 1 Quiz: Differentiability And Continuity

Practice Differentiability And Continuity 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 Differentiability And Continuity, 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

Consider the function $f(x) = |x^2 - 5x + 4|$ . Which of the following describes the set of points where $f(x)$ is not differentiable?

$x = 1$ only
$x = 4$ only
$x = 1$ and $x = 4$ (correct answer)
$x = 2.5$ only

Explanation: A function involving an absolute value,

|g(x)|

, is potentially not differentiable at the points where

g(x)=0

. Here,

g(x) = x^2 - 5x + 4

. Setting this to zero gives

(x-1)(x-4)=0

, so the points of interest are

x=1

and

x=4

. At these points, the graph of

f(x)

will have sharp corners because the function inside the absolute value changes sign. The derivative of

x^2 - 5x + 4

2x-5

. At

x=1

, the slope changes from

-(2(1)-5)=3

2(1)-5=-3

. At

x=4

, the slope changes from

-(2(4)-5)=-3

2(4)-5=3

. Since the left and right derivatives are not equal at these points, the function is not differentiable at

x=1

and

x=4

Question 2

If $\lim_{x \to c} f(x) = L$ exists, but $f'(c)$ does not exist, what is a possible classification of the point $x=c$ on the graph of $y=f(x)$ ?

A removable discontinuity.
A jump discontinuity.
A sharp corner. (correct answer)
An inflection point.

Explanation: The condition

\lim_{x \to c} f(x) = L

means the function approaches the same value from both sides. For the function to be continuous, we also need

f(c) = L

. The problem doesn't state this, but options A and C assume continuity while B does not. If

f(c) eq L

or is undefined, it's a removable discontinuity. At a removable discontinuity, the function is not differentiable. If

f(c)=L

, the function is continuous. If

f'(c)

does not exist at a point of continuity, it could be a corner, cusp, or vertical tangent. A sharp corner fits this description perfectly. A jump discontinuity (B) would mean the limit does not exist. An inflection point (D) is a point where the function is typically differentiable.

Question 3

Find the sum $a+b$ if the function $f(x) = \begin{cases} e^{3x} & x \le 0 \\ ax+b & x > 0 \end{cases}$ is differentiable for all real numbers.

$1$
$3$
$4$ (correct answer)
$e+3$

Explanation: For the function to be differentiable at

x=0

, it must be continuous and have matching slopes. Continuity requires

e^{3(0)} = a(0)+b

, which simplifies to

1=b

. For the slopes to match, the derivatives of the pieces must be equal at

x=0

. The derivative of

e^{3x}

3e^{3x}

. At

x=0

, this is

3e^0 = 3

. The derivative of

ax+b

a

. Therefore, we must have

a=3

. The sum is

a+b = 3+1=4

Question 4

The function $f(x) = |\cos(x)|$ is not differentiable at which of the following x-values?

$x = \pi/4$
$x = \pi/2$ (correct answer)
$x = 2\pi/3$
$x = \pi$

Explanation: The function

f(x) = |g(x)|

is typically not differentiable where

g(x)=0

. In this case,

g(x)=\cos(x)

. We need to find where

\cos(x)=0

. This occurs at

x = \frac{\pi}{2} + n\pi

for any integer

n

. At these points, the graph of

|\cos(x)|

has sharp corners. Let's check

x=\pi/2

. The derivative of

\cos(x)

-\sin(x)

. For

x

slightly less than

\pi/2

\cos(x)>0

, so

f(x)=\cos(x)

and the slope is

-\sin(\pi/2)=-1

. For

x

slightly greater than

\pi/2

\cos(x)<0

, so

f(x)=-\cos(x)

and the slope is

\sin(\pi/2)=1

. Since

-1 \neq 1

, the function is not differentiable at

x=\pi/2

Question 5

For what value of $k$ is the function $f(x) = \begin{cases} x^2 - 1 & x < 2 \\ k(x-2) + 3 & x \ge 2 \end{cases}$ differentiable at $x=2$ ?

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

Explanation: First, the function must be continuous at

x=2

. This means the limits from both sides must be equal to the function value.

\lim_{x \to 2^-} (x^2 - 1) = 2^2 - 1 = 3

. The value from the right is

f(2) = k(2-2) + 3 = 3

. Since both equal 3, the function is continuous for any value of

k

. Next, the derivatives must match. The derivative of

x^2 - 1

2x

. At

x=2

, the left-hand derivative is

2(2) = 4

. The derivative of

k(x-2)+3 = kx - 2k + 3

k

. For the function to be differentiable, we must set the derivatives equal:

k=4

Question 6

The one-sided limits of the difference quotient for a function $f(x)$ at $x=c$ are $\lim_{h \to 0^+} \frac{f(c+h)-f(c)}{h} = L$ and $\lim_{h \to 0^-} \frac{f(c+h)-f(c)}{h} = M$ . If $L$ and $M$ are finite numbers and $L \neq M$ , which conclusion is guaranteed?

$f(x)$ must have a jump discontinuity at $x=c$ .
$f(x)$ must have a vertical tangent at $x=c$ .
$f(x)$ is continuous but not differentiable at $x=c$ . (correct answer)
The overall limit, $\lim_{x \to c} f(x)$ , does not exist.

Explanation: The existence of finite one-sided derivatives

L

and

M

implies that the function must be continuous at

x=c

. However, for the function to be differentiable, the derivative must exist, which requires the left-hand and right-hand derivatives to be equal (i.e.,

L=M

). Since the problem states

L \neq M

, the function is not differentiable at

x=c

. This describes a 'corner' point, where the function is continuous but has different slopes from the left and right.

Question 7

Consider the function $f(x) = |x^2-6x+8|$ . The derivative $f'(a)$ can be found by evaluating the limit $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$ . For which of the following values of $a$ does this limit fail to exist?

$a=1$
$a=4$ (correct answer)
$a=3$
$a=5$

Explanation: When you encounter absolute value functions and questions about derivatives, you need to identify where the function inside the absolute value equals zero—these are potential points where the derivative might not exist. First, let's find where the expression inside the absolute value equals zero:

x^2 - 6x + 8 = 0

. Factoring gives us

(x-2)(x-4) = 0

, so

x = 2

and

x = 4

. At these critical points, the absolute value function creates "corners" or "cusps" where the function changes from decreasing to increasing (or vice versa). Let's check

x = 4

: For values slightly less than 4,

x^2 - 6x + 8 > 0

, so

f(x) = x^2 - 6x + 8

. For values slightly greater than 4,

x^2 - 6x + 8 < 0

, so

f(x) = -(x^2 - 6x + 8)

. This means the left-hand and right-hand derivatives at

x = 4

are different, making the limit

\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}

fail to exist. Now for the wrong answers: A) At

x = 1

, the expression

1 - 6 + 8 = 3 > 0

, so we're not at a corner point and the derivative exists. C) At

x = 3

, we get

9 - 18 + 8 = -1 < 0

, so again no corner point. D) At

x = 5

, we have

25 - 30 + 8 = 3 > 0

, so the derivative exists. Strategy tip: For absolute value functions, always find where the inside expression equals zero—these are your candidate points where derivatives might fail to exist due to sharp corners in the graph.

Question 8

Let $f$ be a function such that its derivative $f'(1)$ exists. Which of the following statements must be true?

I. $f$ is continuous at $x=1$ .

II. $\lim_{x \to 1} f(x) = f(1)$ .

III. $\lim_{h \to 0} \frac{f(1+h) - f(1)}{h}$ exists and is finite.

III only
I and II only
I and III only
I, II, and III (correct answer)

Explanation: The premise is that

f'(1)

exists, which means the function is differentiable at

x=1

. Statement (III) is the limit definition of the derivative

f'(1)

. Since we are told

f'(1)

exists, this limit must exist and be a finite number. This is true by definition. A fundamental theorem in calculus states that if a function is differentiable at a point, it must be continuous at that point. Therefore, statement (I) must be true. Statement (II) is the definition of continuity at

x=1

. Since we concluded that

f

must be continuous at

x=1

, statement (II) must also be true. Thus, all three statements are necessary consequences of

f

being differentiable at

x=1

Question 9

Let $g(x) = \begin{cases} f(x) & x \le c \\ m(x-c) + k & x > c \end{cases}$ . If $f(x)$ is a differentiable function for all real numbers, what must be the values of $k$ and $m$ so that $g(x)$ is also differentiable at $x=c$ ?

$k = f(c)$ and $m = f'(c)$ (correct answer)
$k = f'(c)$ and $m = f(c)$
$k = f(c)$ and $m = 0$
$k=0$ and $m=f'(c)$

Explanation: For

g(x)

to be differentiable at

x=c

, it must first be continuous. This requires the limit from the left to equal the limit from the right:

\lim_{x \to c^-} g(x) = f(c)

and

\lim_{x \to c^+} g(x) = m(c-c)+k = k

. Thus, we must have

k = f(c)

. Next, the derivatives from the left and right must match. The derivative from the left is

f'(c)

. The derivative of the linear function

m(x-c)+k

m

. So, the derivative from the right is

m

. For differentiability, we must have

m = f'(c)

. The line

y=m(x-c)+k

must be the tangent line to

f(x)

x=c

Question 10

Let a function be defined by a piecewise formula: $f(x) = \begin{cases} ax^2 + b & x \le 1 \\ 4x - 2 & x > 1 \end{cases}$ For which values of the constants $a$ and $b$ is the function $f(x)$ differentiable at $x=1$ ?

$a = 1, b = 1$
$a = 2, b = 0$ (correct answer)
$a = 2, b = -1$
There are no such values of $a$ and $b$ .

Explanation: For

f(x)

to be differentiable at

x=1

, it must first be continuous at

x=1

. This means

\lim_{x \to 1^-} f(x) = f(1)

must equal

\lim_{x \to 1^+} f(x)

. Evaluating this gives

a(1)^2 + b = 4(1) - 2

, which simplifies to

a + b = 2

. Second, the derivatives from the left and right must be equal. The derivative is

f'(x) = \begin{cases} 2ax & x < 1 \\ 4 & x > 1 \end{cases}

x=1

, we must have

2a(1) = 4

, which implies

a=2

. Substituting

a=2

into the continuity equation gives

2 + b = 2

, so

b=0

Question 11

A function $f(x)$ is defined by $f(x) = \begin{cases} g(x) & x < a \\ h(x) & x \ge a \end{cases}$ , where $g(x)$ and $h(x)$ are polynomials. Which condition guarantees that $f$ is continuous at $x=a$ but NOT differentiable at $x=a$ ?

$g(a) = h(a)$ and $g'(a) = h'(a)$
$g(a) \neq h(a)$ and $g'(a) = h'(a)$
$g(a) = h(a)$ and $g'(a) \neq h'(a)$ (correct answer)
$g(a) \neq h(a)$ and $g'(a) \neq h'(a)$

Explanation: For

f

to be continuous at

x=a

, the values of the pieces must match, so

\lim_{x \to a^-} g(x) = h(a)

, which means

g(a) = h(a)

. For

f

to be NOT differentiable at

x=a

, the slopes of the pieces must not match, which means

g'(a) \neq h'(a)

. This situation describes a 'corner' point. Therefore, the condition that guarantees continuity but not differentiability is

g(a) = h(a)

and

g'(a) \neq h'(a)

Question 12

Which of the following functions is continuous for all real numbers but not differentiable at $x=0$ ?

$f(x) = \frac{1}{x}$
$f(x) = x|x|$
$f(x) = x^{4/3}$
$f(x) = |x| + 1$ (correct answer)

Explanation: A)

f(x) = 1/x

is not continuous at

x=0

. B)

f(x) = x|x|

can be written as

x^2

for

x \ge 0

and

-x^2

for

x<0

. Its derivative is

2x

and

-2x

respectively. At

x=0

, both derivatives are 0, so it IS differentiable at

x=0

. C)

f(x) = x^{4/3}

. Its derivative is

\frac{4}{3}x^{1/3}

. At

x=0

, the derivative is 0. So it IS differentiable at

x=0

. D)

f(x) = |x|+1

is the graph of

|x|

shifted up by 1. It is continuous everywhere, but has a sharp corner at

x=0

and is therefore not differentiable at

x=0

Question 13

Let $f(x)$ be a function such that $f(2)=5$ and $f'(2)=3$ . What is the value of $\lim_{x \to 2} f(x)$ ?

$2$
$3$
$5$ (correct answer)
The limit cannot be determined from the given information.

Explanation: The fact that

f'(2)

exists and is equal to 3 implies that the function

f(x)

must be differentiable at

x=2

. A core theorem of calculus states that if a function is differentiable at a point, it must also be continuous at that point. For a function to be continuous at

x=2

, the limit as

x

approaches 2 must be equal to the function's value at 2. Therefore,

\lim_{x \to 2} f(x) = f(2)

. Since we are given

f(2)=5

, the limit must be 5.

Question 14

The function $f(x) = (x^2-1)^{2/3}$ is not differentiable at which of the following points?

$x = 0$ only
$x = 1$ only
$x = -1$ and $x = 1$ (correct answer)
The function is differentiable everywhere.

Explanation: A function of the form

(g(x))^{p/q}

where

p/q < 1

is often not differentiable where

g(x)=0

. Here,

g(x) = x^2-1

. Setting

g(x)=0

gives

x^2-1=0

, which has solutions

x=1

and

x=-1

. Let's check the derivative:

f'(x) = \frac{2}{3}(x^2-1)^{-1/3}(2x) = \frac{4x}{3\sqrt[3]{x^2-1}}

. The derivative is undefined at

x=1

and

x=-1

. At both points, the denominator is zero, leading to cusps in the graph. Therefore, the function is not differentiable at

x=-1

and

x=1

Question 15

For the function $f(x) = x^{1/5}$ , which statement correctly describes the behavior at $x=0$ ?

The function is discontinuous and has a vertical asymptote.
The function is continuous and has a corner.
The function is continuous and has a vertical tangent. (correct answer)
The function is differentiable and $f'(0)=0$ .

Explanation: The function

f(x) = x^{1/5}

(or

\sqrt[5]{x}

) is continuous for all real numbers, including

x=0

. The derivative is

f'(x) = \frac{1}{5}x^{-4/5} = \frac{1}{5x^{4/5}}

. At

x=0

, the derivative is undefined. To classify this point, we examine the limit of the derivative as

x \to 0

. Because of the

4/5

power, the denominator

5x^{4/5}

is always positive for

x \neq 0

. Therefore,

\lim_{x \to 0} f'(x) = +\infty

. Since the limit of the derivative is infinite, the function has a vertical tangent at

x=0

Question 16

Let $f$ be a function defined by $f(x) = \begin{cases} ax^3 + 1 & x \le 1 \\ b(x-2)^2 & x > 1 \end{cases}$ . For what values of the constants $a$ and $b$ is the function $f$ differentiable at $x=1$ ?

$a = 2, b = 3$
$a = -2/5, b = 3/5$ (correct answer)
$a = 2, b = -3$
$a = 1, b = 2$

Explanation: For

f

to be differentiable at

x=1

, it must first be continuous at

x=1

. This means

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)

. So,

a(1)^3 + 1 = b(1-2)^2

, which simplifies to

a+1=b

. Second, the derivatives from the left and right must be equal. The derivative is

f'(x) = \begin{cases} 3ax^2 & x < 1 \\ 2b(x-2) & x > 1 \end{cases}

. At

x=1

, we must have

3a(1)^2 = 2b(1-2)

, which simplifies to

3a = -2b

. Now we solve the system of two equations:

b=a+1

and

3a=-2b

. Substituting the first into the second gives

3a = -2(a+1)

, so

3a = -2a - 2

, which gives

5a = -2

, or

a = -2/5

. Then

b = a+1 = -2/5 + 1 = 3/5

Question 17

Consider the function $f(x) = \begin{cases} x^2+1 & x \le 2 \\ 4x-5 & x > 2 \end{cases}$ . Which statement accurately describes the differentiability of $f$ at $x=2$ ?

$f$ is differentiable at $x=2$ because the derivatives of both pieces are equal to 4 at $x=2$ .
$f$ is not differentiable at $x=2$ because the graph of $f$ has a corner at $x=2$ .
$f$ is not differentiable at $x=2$ because $f$ is not continuous at $x=2$ . (correct answer)
$f$ is not differentiable at $x=2$ because the derivative from the left is 2 and the derivative from the right is 4.

Explanation: For a function to be differentiable at a point, it must first be continuous at that point. We check for continuity at

x=2

\lim_{x \to 2^-} f(x) = 2^2+1=5

. However,

\lim_{x \to 2^+} f(x) = 4(2)-5=3

. Since the left-hand and right-hand limits are not equal, the function is not continuous at

x=2

. Because the function is not continuous, it cannot be differentiable at

x=2

. Note that if one were to only check the derivatives of the pieces (

2x

and

4

), one would find that they both approach 4 as

x

approaches 2, which is a common trap.

Question 18

At which of the following values of $x$ is the function $f(x) = (x^2-9)|x-2|$ continuous but not differentiable?

$x=2$ only (correct answer)
$x=3$ and $x=-3$ only
$x=2, x=3$ , and $x=-3$
The function is differentiable for all real numbers.

Explanation: The function involves an absolute value,

|x-2|

, which can cause non-differentiability at

x=2

. We can write

f

as a piecewise function:

f(x) = \begin{cases} (x^2-9)(x-2) & x \ge 2 \\ -(x^2-9)(x-2) & x < 2 \end{cases}

. The function is a product of polynomials, so it is continuous everywhere. To check differentiability at

x=2

, we compare the left and right derivatives. For

x>2

f'(x) = 3x^2-4x-9

. For

x<2

f'(x) = -3x^2+4x+9

. The right-hand derivative at

x=2

\lim_{x \to 2^+} (3x^2-4x-9) = 3(4)-4(2)-9 = -5

. The left-hand derivative is

\lim_{x \to 2^-} (-3x^2+4x+9) = -3(4)+4(2)+9 = 5

. Since

-5 \ne 5

, the function is not differentiable at

x=2

. The points

x=3

and

x=-3

are roots of the polynomial but do not cause non-differentiability.

Question 19

Which statement best describes the function $f(x) = \sqrt[3]{(x-3)^2}$ at the point $x=3$ ?

It is continuous and differentiable at $x=3$ .
It is continuous but not differentiable at $x=3$ due to a corner.
It is continuous but not differentiable at $x=3$ due to a cusp. (correct answer)
It is not continuous at $x=3$ .

Explanation: The function can be written as

f(x) = (x-3)^{2/3}

. The function is continuous at

x=3

since

f(3) = (3-3)^{2/3} = 0

. To check for differentiability, we find the derivative:

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

. As

x

approaches 3, the denominator approaches 0, so the derivative approaches infinity. Specifically,

\lim_{x \to 3^+} f'(x) = +\infty

and

\lim_{x \to 3^-} f'(x) = -\infty

. Since the one-sided limits of the derivative are infinite and have opposite signs, the graph has a cusp at

x=3

. Therefore, the function is not differentiable at

x=3

Question 20

Which statement best describes the function $g(x) = (x+2)^{1/3} + 1$ at the point $x=-2$ ?

It is continuous but not differentiable at $x=-2$ due to a vertical tangent. (correct answer)
It is continuous but not differentiable at $x=-2$ due to a cusp.
It is continuous but not differentiable at $x=-2$ due to a corner.
It is neither continuous nor differentiable at $x=-2$ .

Explanation: The function

g(x) = \sqrt[3]{x+2} + 1

is continuous at

x=-2

because

g(-2) = \sqrt[3]{-2+2} + 1 = 1

. To check differentiability, we find the derivative:

g'(x) = \frac{1}{3}(x+2)^{-2/3} = \frac{1}{3(x+2)^{2/3}}

. As

x

approaches

-2

from either the left or the right, the term

(x+2)^{2/3}

in the denominator approaches 0 through positive values. Thus,

\lim_{x \to -2} g'(x) = +\infty

. When the limit of the derivative is infinite at a point, the function has a vertical tangent there and is not differentiable.