Calculus 1 Quiz: Increasing Decreasing Intervals

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Increasing Decreasing IntervalsQuestion 1 of 10

Let $f(x)$ be a piecewise function defined as $f(x) = \begin{cases} x^2 + 2x + 2 & \text{if } x \le 0 \\ x^3 - 4x + 2 & \text{if } x > 0 \end{cases}$ . On which of the following disjoint intervals is the function $f(x)$ decreasing?

(-1, 0)

and

(\frac{2}{\sqrt{3}}, \infty)

(-\infty, -1)

and

(0, \frac{2}{\sqrt{3}})

(-1, \frac{2}{\sqrt{3}})

(-\infty, -1)

only

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Calculus 1 Quiz

Calculus 1 Quiz: Increasing Decreasing Intervals

Practice Increasing Decreasing Intervals 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 Increasing Decreasing Intervals, 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

$(-1, 0)$ and $(\frac{2}{\sqrt{3}}, \infty)$
$(-\infty, -1)$ and $(0, \frac{2}{\sqrt{3}})$ (correct answer)
$(-1, \frac{2}{\sqrt{3}})$
$(-\infty, -1)$ only

Explanation: We analyze the derivative

f'(x)

on each piece. For

x < 0

f'(x) = 2x + 2

. This is negative when

2x+2 < 0

, i.e.,

x < -1

. So

f

is decreasing on

(-\infty, -1)

. For

x > 0

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

. This is negative when

3x^2 - 4 < 0

, i.e.,

x^2 < 4/3

, which for

x>0

means

0 < x < 2/\sqrt{3}

. So

f

is decreasing on

(0, 2/\sqrt{3})

. The function is continuous at

x=0

. Combining the results,

f(x)

is decreasing on

(-\infty, -1)

and

(0, 2/\sqrt{3})

Question 2

Let $f(x)$ be a twice-differentiable function such that $f'(x) > 0$ for all real numbers $x$ . Let $g(x) = f(1-x^2)$ . On which interval is $g(x)$ increasing?

$(0, \infty)$
$(-1, 1)$
$(-\infty, \infty)$
$(-\infty, 0)$ (correct answer)

Explanation: To determine where

g(x)

is increasing, we must find where its derivative

g'(x)

is positive. Using the chain rule,

g'(x) = f'(1-x^2) \cdot \frac{d}{dx}(1-x^2) = f'(1-x^2) \cdot (-2x)

. We are given that

f'(y) > 0

for any input

y

. Therefore, the term

f'(1-x^2)

is always positive. The sign of

g'(x)

is thus determined by the sign of the other factor,

-2x

. The function

g(x)

is increasing when

g'(x) > 0

, which means

-2x > 0

. Dividing by -2 and reversing the inequality sign gives

x < 0

. So,

g(x)

is increasing on the interval

(-\infty, 0)

Question 3

Let $f(x) = \frac{\ln(x)}{x^2}$ . On which of the following intervals is $f(x)$ increasing?

$(\sqrt{e}, \infty)$
$(0, 1)$
$(0, \sqrt{e})$ (correct answer)
$(e, \infty)$

Explanation: First, note the domain of

f(x)

x > 0

. We use the quotient rule to find the derivative:

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

. The function is increasing where

f'(x) > 0

. Since

x > 0

x^3

is positive, so the sign of

f'(x)

is determined by the sign of the numerator,

1 - 2\ln(x)

. We need

1 - 2\ln(x) > 0

, which simplifies to

1 > 2\ln(x)

, or

\frac{1}{2} > \ln(x)

. Exponentiating both sides gives

e^{1/2} > x

, so

x < \sqrt{e}

. Combining with the domain

x>0

, the interval of increase is

(0, \sqrt{e})

Question 4

Find the interval on which the function $f(x) = (x^2 - 4x)^{1/3}$ is increasing.

$(2, \infty)$ (correct answer)
$(0, 4)$
$(4, \infty)$
$(2, 4)$ and $(4, \infty)$

Explanation: First, we find the derivative using the chain rule:

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

. Critical points occur when the derivative is zero or undefined.

f'(x)=0

when

x=2

f'(x)

is undefined when

x=0

and

x=4

. To find where

f(x)

is increasing, we find where

f'(x) > 0

. The denominator involves a square,

(...)^{2/3}

, so it is always non-negative. Thus, the sign of

f'(x)

is determined by the sign of the numerator,

2(x-2)

. The expression

2(x-2)

is positive when

x > 2

. Therefore, the function is increasing on the interval

(2, \infty)

. Although the derivative is undefined at

x=4

, the function is continuous there, and since it is increasing on

(2, 4)

and

(4, \infty)

, we can conclude it is increasing on

(2, \infty)

Question 5

On which of the following disjoint intervals is the function $f(x) = \frac{x^2+3}{x-1}$ increasing?

$(-1, 3)$
$(-\infty, 1)$ and $(1, \infty)$
$(-1, 1)$ and $(3, \infty)$
$(-\infty, -1)$ and $(3, \infty)$ (correct answer)

Explanation: To find where

f(x)

is increasing, we must find where

f'(x) > 0

. Using the quotient rule,

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

. The denominator

(x-1)^2

is always positive for

x \neq 1

. Therefore, the sign of

f'(x)

is determined by the numerator,

(x-3)(x+1)

. This is a parabola opening upwards with roots at

x=-1

and

x=3

. The numerator is positive when

x < -1

x > 3

. Thus,

f(x)

is increasing on the intervals

(-\infty, -1)

and

(3, \infty)

Question 6

Determine the interval on which the function $f(x) = \arctan(x) - \frac{1}{2}\ln(1+x^2)$ is increasing.

$(1, \infty)$
$(-\infty, 1)$ (correct answer)
$(0, \infty)$
$(-\infty, 0)$

Explanation: To find the interval of increase, we compute the derivative

f'(x)

and find where it is positive.

f'(x) = \frac{d}{dx}(\arctan(x)) - \frac{1}{2}\frac{d}{dx}(\ln(1+x^2)) = \frac{1}{1+x^2} - \frac{1}{2} \cdot \frac{2x}{1+x^2} = \frac{1-x}{1+x^2}

. The function is increasing when

f'(x) > 0

. Since the denominator

1+x^2

is always positive, the sign of

f'(x)

is determined by the numerator

1-x

. The inequality

1-x > 0

implies

1 > x

, or

x < 1

. Therefore, the function is increasing on the interval

(-\infty, 1)

Question 7

For which range of values for the constant $k$ is the function $f(x) = x^3 - kx^2 + 3x + 1$ always increasing for all real numbers $x$ ?

$k \le -3$ or $k \ge 3$
$k = 3$ or $k = -3$
$-9 \le k \le 9$
$-3 \le k \le 3$ (correct answer)

Explanation: For

f(x)

to be always increasing, its derivative

f'(x)

must be greater than or equal to zero for all

x

. The derivative is

f'(x) = 3x^2 - 2kx + 3

. This is a quadratic function that opens upwards. For it to be non-negative for all

x

, it can have at most one real root. This condition is met when the discriminant is less than or equal to zero. The discriminant is

\Delta = b^2 - 4ac = (-2k)^2 - 4(3)(3) = 4k^2 - 36

. Setting

\Delta \le 0

, we get

4k^2 - 36 \le 0

, which simplifies to

k^2 \le 9

. The solution to this inequality is

-3 \le k \le 3

Question 8

Let $f(x) = e^x \cos(x)$ . On which of the following intervals is the function $f$ decreasing?

$(0, \frac{\pi}{4})$
$(\frac{\pi}{4}, \frac{5\pi}{4})$ (correct answer)
$(\frac{5\pi}{4}, 2\pi)$
$(\frac{\pi}{2}, \frac{3\pi}{2})$

Explanation: To find where the function is decreasing, we need to find where its derivative is negative. Using the product rule,

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

. Since

e^x

is always positive, the sign of

f'(x)

is determined by the sign of

\cos(x) - \sin(x)

. The function

f

is decreasing when

f'(x) < 0

, which means

\cos(x) - \sin(x) < 0

, or

\cos(x) < \sin(x)

. This inequality holds for

x

in the interval

(\frac{\pi}{4}, \frac{5\pi}{4})

Question 9

The function $f(x) = (x^2-1)|x-2|$ is decreasing on which of the following intervals?

$(\frac{2 - \sqrt{7}}{3}, \frac{2 + \sqrt{7}}{3})$
$(\frac{2 + \sqrt{7}}{3}, \infty)$
$(\frac{2 + \sqrt{7}}{3}, 2)$ (correct answer)
$(-1, 1)$

Explanation: We split the function into two cases. For

x < 2

f(x) = (x^2-1)(-(x-2)) = -x^3 + 2x^2 + x - 2

. Then

f'(x) = -3x^2 + 4x + 1

. The roots of

f'(x)=0

are

x = \frac{-4 \pm \sqrt{16-4(-3)(1)}}{-6} = \frac{2 \pm \sqrt{7}}{3}

. Since

f'(x)

is a downward-opening parabola, it is negative (and

f

is decreasing) on

(-\infty, \frac{2 - \sqrt{7}}{3})

and

(\frac{2 + \sqrt{7}}{3}, 2)

. For

x > 2

f(x) = (x^2-1)(x-2) = x^3 - 2x^2 - x + 2

. Then

f'(x) = 3x^2 - 4x - 1

. For

x>2

f'(x)

is always positive, so

f

is increasing on

(2, \infty)

. Thus, one of the intervals where

f

is decreasing is

(\frac{2 + \sqrt{7}}{3}, 2)

Question 10

On which interval is the function $h(x) = x^2 e^{-x}$ increasing?

$(0, 2)$ (correct answer)
$(-\infty, 0)$ and $(2, \infty)$
$(-\infty, 1)$
$(2, \infty)$

Explanation: When you're asked where a function is increasing, you need to find where its derivative is positive. A function increases when its slope is positive and decreases when its slope is negative. For

h(x) = x^2 e^{-x}

, you'll need the product rule since this is the product of

x^2

and

e^{-x}

. Taking the derivative:

h'(x) = 2x \cdot e^{-x} + x^2 \cdot (-e^{-x}) = e^{-x}(2x - x^2) = xe^{-x}(2 - x)

Since

e^{-x}

is always positive, the sign of

h'(x)

depends only on

x(2-x)

. This expression equals zero when

x = 0

x = 2

, giving you critical points that divide the number line into three intervals. Testing the sign on each interval: when

x < 0

, you have (negative)(positive) = negative; when

0 < x < 2

, you have (positive)(positive) = positive; when

x > 2

, you have (positive)(negative) = negative. Therefore,

h'(x) > 0

only on

(0, 2)

, making this where the function increases. This confirms answer A is correct. Answer B suggests the function increases on

(-\infty, 0)

and

(2, \infty)

, but these are exactly where

h'(x) < 0

. Answer C claims increasing on

(-\infty, 1)

, but the function decreases for

x < 0

. Answer D suggests increasing on

(2, \infty)

, but the derivative is negative there. Remember: always find the derivative first, then determine where it's positive. The critical points help you identify the intervals to test.