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Precalculus : Relations and Functions

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Example Questions

Example Question #1 : Evaluate Functions

Evaluate f(x) for $x=\frac{3\pi}{4}$ .

f(x)=sin^2(x)

Possible Answers:

$\frac{3}{4}$

$\frac{1}{2}$

$-\frac{1}{2}$

$\frac{\sqrt2}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

We evaluate the function when $x=\frac{3\pi}{4}$ .

$f(\frac{3\pi}{4})=sin^2(\frac{3\pi}{4})=(\frac{\sqrt{2}}{2})^2=\frac{2}{4}=\frac{1}{2}$

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Example Question #41 : Functions

Find the domain of f(x) below

$\frac{\sqrt{x-7}}{x^3+1}$

Possible Answers:

$(-\infty,-1) \cup (7,\infty)$

$(-\infty,7)$

$[7,\infty)$

[-1,7]

Correct answer:

$[7,\infty)$

Explanation:

We have x^3+1=(x+1)(x^2-x+1). We have $x^2-x+1\ >0$ for all real numbers. x+1=0 when x=-1 . The denominator is undefined when x=-1 .

The nominator is defined if $x\ge 7$ .

The domain is: $\{x\in \mathbb{R}: x\ne -1 \}\cap \{x\in \mathbb{R}: x\ge 7\}=\{x\in \mathbb{R}: x\ge 7\}$

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Example Question #31 : Relations And Functions

The domain of the following function

$f(x)=\frac{1}{\sqrt{x-1}-1}$

is:

Possible Answers:

(1,2)

$(1,2)\cup (2,\infty)$

$(-\infty,1)$

$(1,2]\cup (2,\infty)$

$(2,\infty)$

Correct answer:

$(1,2)\cup (2,\infty)$

Explanation:

$\sqrt{x-1}$ is defined when $x\ge 1$ . Since we do not want $\sqrt{x-1}$ to be 0 in the denominator we must have $x\ >1$ . $\sqrt{x-1}-1=0$ when x=2.

Thus we need to exclude 2 also. Therefore the domain is:

$$\{x\in \mathbb{R} : x>1\cap x\neq 2\}=(1,2)\cup (2,\infty)$

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Example Question #32 : Relations And Functions

Find the range of the following function:

f(x)=7

Possible Answers:

$\text{The set of real numbers}$

$\text{Range of } f= {7}$

$\text{The set of negative integers}$

$\text{The set of positive integers}$

{1}

Correct answer:

$\text{Range of } f= {7}$

Explanation:

Every element of the domain has as image 7.This means that the function is constant . Therefore,

the range of f is :{7}.

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Example Question #44 : Functions

What is the domain of the following function:

$f(x)=\frac{1}{1+\sqrt x}$

Possible Answers:

$\text{The domain is :}$

$\{ x\in \mathbb{R}}| x \ge 0\}$

$\text{The set of real numbers}$

$\text{The set of real numbers}$ $\{-1,1\}$

$\text{The set of real numbers}$ $\mathbb{R}-\{ 0,1\}$

$\text{The set of real numbers except 0}$

Correct answer:

$\text{The domain is :}$

$\{ x\in \mathbb{R}}| x \ge 0\}$

Explanation:

Note that in the denominator, we need to have $x\ge 0$ to make the square root of x defined. In this case $1+\sqrt x$ is never zero. Hence we have no issue when dividing by this number. Therefore the domain is the set of real numbers that are $\ge 0$

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Example Question #45 : Functions

Find the domain of the following function f(x) given below:

$f(x)=\sqrt{ x^3-1}$

Possible Answers:

$(-\infty,-1)\cup (1,\infty)$

$\{x \in \mathbb{R}: x\ge 1\}$

$\mathbb{R}$

$(-\infty, -1)$

$(-\infty, 1)$

Correct answer:

$\{x \in \mathbb{R}: x\ge 1\}$

Explanation:

x^3-1=(x-1)(x^2-x+1) . Since $(x^2-x+1)\ >0$ for all real numbers. To make the square root positive we need to have $x-1\ge 0$ .

Therefore the domain is :

$\{x\in\mathbb{R}: x\ge 1\}=[1,\infty)$

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Example Question #46 : Functions

What is the range of :

f(x)=1+2cos(x)

Possible Answers:

[-1,1]

[0,3]

$\text{The set of all the reals}$

[-1,3]

[-3,3]

Correct answer:

[-1,3]

Explanation:

We know that $-1\le cos(x)\le 1$ . So $-2\le 2cos(x)\le 2$ .

Therefore:

$1-2\le 1+2cos(x)\le 1+2$ .

This gives:

$-1\le 1+2cos(x)\le 3$ .

Therefore the range is:

[-1,3]

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Example Question #47 : Functions

Find the range of f(x) given below:

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

Possible Answers:

$(0,\infty)$

${}(-\infty,\infty)$

$[2,\infty)$

$[0,\infty)$

$(-\infty, 2)$

Correct answer:

$[2,\infty)$

Explanation:

Note that: we can write f(x) as :

x^2+2x+3=(x+1)^2+2 .

Since,

$(x+1)^2\ge 0 $ for all reals.$

Therefore,

$(x+1)^2+2\ge 2.$

So the range is $[2,\infty)$

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Example Question #51 : Functions

What is the range of :

f(x)=7+cos(3x)

Possible Answers:

[6,8]

[0,1]

[-7,7]

[0,7]

[-1,1]

Correct answer:

[6,8]

Explanation:

We have $-1\le cos(3x)\le 1$ .

Adding 7 to both sides we have:

$6\le cos(3x)\le 8$ .

Therefore $6\le f(x)\le 8$ .

This means that the range of f is [6,8]

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Example Question #52 : Functions

Find the domain of the following function:

$f(x)=\sqrt{121-x}$

Possible Answers:

[-121,121]

[0,121]

$[121,\infty)$

$(-\infty,121]$

${}(-\infty,\infty)$

Correct answer:

$(-\infty,121]$

Explanation:

The part inside the square root must be positive. This means that we must be $\ge 0$ . Thus

$121-x\ge0$ . Adding -121 to both sides gives $-x\ge -121$ . Finally multiplying both sides by (-1) give:

$x\le 121$ with x reals. This gives the answer.

Note: When we divide by a negative we need to flip our sign.

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Precalculus : Relations and Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Evaluate Functions

Example Question #41 : Functions

Example Question #31 : Relations And Functions

Example Question #32 : Relations And Functions

Example Question #44 : Functions

Example Question #45 : Functions

Example Question #46 : Functions

Example Question #47 : Functions

Example Question #51 : Functions

Example Question #52 : Functions

All Precalculus Resources

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