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Precalculus : Relations and Functions

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Example Questions

Example Question #21 : Functions

What is the domain of the following function?

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

Possible Answers:

$\mathbb{R}-\{\pi\}$

$\text{The set of all reals}$

- $\mathbb{R}-\{\pi ,3\pi\}$

$\mathbb{R}-\{-\pi\}$

$\mathbb{R}-\{{(2k+1)\pi\, ,$ k\in 0,1,2,3,4,.......\}$

Correct answer:

$\mathbb{R}-\{{(2k+1)\pi\, ,$ k\in 0,1,2,3,4,.......\}$

Explanation:

The denominator becomes zero whenever cos(x) becomes -1 and this happens when x is an odd multiple of $\pi$ . To avoid division by zero, we exclude all these numbers. Therefore the domain is:

$\mathbb{R}-\{{(2k+1)\pi\, ,$ k\in 0,1,2,3,4,.......\}$

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Example Question #31 : Functions

Given f(x) and h(x) , evaluate f(h(7)) .

f(x)=5x^2-13x+8

h(x)=x^3-343

Possible Answers:

343

Correct answer:

Explanation:

We are given two functions and asked to find f(h(7)).

I would begin by finding h(7)

h(x)=x^3-343

h(7)=7^3-343=343-343=0

Now that we know h(7), we go ahead and plug it into f(x).

f(x)=5x^2-13x+8

$f(0)=5\cdot 0^2-13\cdot 0+8=8$

So our final answer is simply 8.

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Example Question #32 : Functions

Solve the function. $\frac{10x^2-10}{x-1}=0$

Possible Answers:

$\pm \frac{1}{10}$

$\pm1$

Correct answer:

Explanation:

Determine the domain of $\frac{10x^2-10}{x-1}=0$ .

$x\neq1$

Multiply x-1 on both sides of the equation to cancel the denominator and divide the equation 10x^2-10=0 by ten.

x^2-1=0

x^2=1

$x= \pm1$

Since the domain states that $x\neq1$ , the only possible answer is x= -1 .

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Example Question #3 : Evaluate Functions

Solve for : 9^x+1=4

Possible Answers:

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{4}{9}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

Rewrite 9^x+1=4 so that the exponential variable is isolated.

9^x=3

Reconvert 9^x=3 to a similar base. Use exponents to redefine the terms.

3^2^x=3^1

Since all terms are of the same base, use the property of log to eliminate the base on both sides of the equation.

2x=1

Solve for .

$x=\frac{1}{2}$

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Example Question #1 : Evaluate Functions

Evaluate the following function when t=6

$f(t)=5^{3t-18}+5^t-t^2$

Possible Answers:

15590

15662

15589

Correct answer:

15590

Explanation:

Evaluate the following function when t=6

$f(t)=5^{3t-18}+5^t-t^2$

To evaluate this function, simply plug-in 6 for t and simplify:

$f(6)=5^{3(6)-18}+5^6-(6)^2=1+15625-36=15590$

So our answer is:

15590

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Example Question #1 : Evaluate Functions

Determine the value of f(7) of the function

$f(x)=\sqrt{71-x}$

Possible Answers:

$f(7)=\sqrt{78}$

$f(7)\,\, \textsc{dne}$

f(7)=8

f(7)=10

Correct answer:

f(7)=8

Explanation:

In order to determine the value of f(7) of the function we set x=7. The value becomes

$f(7)=\sqrt{71-7}=\sqrt{64}=8$

As such

f(7)=8

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Example Question #1 : Evaluate Functions

Find the difference quotient of the function f(x)=3x^2-2x-5 .

Possible Answers:

3h^2-2h-5

3(x+h)^2-2(x+h)-5

3h+6x-2

3h^2+3x^2-5

3h-3x-5

Correct answer:

3h+6x-2

Explanation:

Difference quotient equation is

$\frac{f(x+h)-f(x)}{h}$ and

f(x+h)=3(x+h)^2-2(x+h)-5 .

Now plug in the appropriate terms into the equation and simplify:

$\\ \frac{f(x+h)-f(x)}{h}\\ \\=\frac{3(x+h)^2-2(x+h)-5-(3x^2-2x-5)}{h}\\ \\=\frac{3(x^2+2xh+h^2)-2x-2h-5-3x^2+2x+5}{h}\\ \\=\frac{3x^2+6xh+3h^2-2x-2h-5-3x^2+2x+5}{h}\\ \\=\frac{6xh+3h^2-2h}{h}\\ \\=\frac{h(3h+6x-2)}{h}\\ \\=3h+6x-2$

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Example Question #1 : Evaluate Functions

Find the value of the following function when x=2

$g(x)=\frac{2x^2-4x+5}{x^4-16}$

Possible Answers:

The function is undefined at x=2

Correct answer:

The function is undefined at x=2

Explanation:

Find the value of the following function when x=2

$g(x)=\frac{2x^2-4x+5}{x^4-16}$

To evaluate this function, plug in 2 for x everywhere it arises and simplify:

$g(x)=\frac{2(2)^2-4(2)+5}{(2)^4-16}=\frac{5}{0}=\mbox{undefined}$

So our answer must be undefined, because we cannot divide by

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Example Question #8 : Evaluate Functions

Find the value of g(t) when $t=\frac{3\pi}{4}$

$g(t)=5\cos(t)-5\sin(t)$

Possible Answers:

$-5\sqrt{2}$

Correct answer:

$-5\sqrt{2}$

Explanation:

Find the value of g(t) when $t=\frac{3\pi}{4}$

$g(t)=5\cos(t)-5\sin(t)$

To evaluate this expression, plug in the given value of everywhere you see a and simplify:

$g(\frac{3\pi}{4})=5\cos(\frac{3\pi}{4})-5\sin(\frac{3\pi}{4})=5(\frac{-\sqrt{2}}{2}-\frac{\sqrt{2}}{2})=5(\frac{-2\sqrt{2}}{2})=-5\sqrt{2}$

$5(\frac{-2\sqrt{2}}{2})=-5\sqrt{2}$

So our answer is:

$-5\sqrt{2}$

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Example Question #9 : Evaluate Functions

If z(x) =-( -2x-3) , what is z(4) ?

Possible Answers:

Correct answer:

Explanation:

In order to evaluate this function, simply substitute the value of as a replacement of .

z(4) =-( -2(4)-3)

Simplify using the order of operations.

z(4) =-( -2(4)-3) = -(-8-3) = -(-11) = 11

The answer is:

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Precalculus : Relations and Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #21 : Functions

Example Question #31 : Functions

Example Question #32 : Functions

Example Question #3 : Evaluate Functions

Example Question #1 : Evaluate Functions

Example Question #1 : Evaluate Functions

Example Question #1 : Evaluate Functions

Example Question #1 : Evaluate Functions

Example Question #8 : Evaluate Functions

Example Question #9 : Evaluate Functions

All Precalculus Resources

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