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Precalculus : Functions

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Example Questions

Example Question #1 : Absolute Value Functions

Which of the following is a point on the following function?

$y=\left | x^2-5x-56\right |+27$

Possible Answers:

(14,36)

(-27,-27)

(0,0)

(5,83)

(5,-43)

Correct answer:

(5,83)

Explanation:

One way to approach this problem would be to plug in each answer and see what works. However, I would be a little more strategic and eliminate any options that don't make sense.

Our y value will never be negative, so eliminate any options with a negative y-value.

Try (0,0) really quick, since it's really easy

$y=\left | x^2-5x-56\right |+27 \rightarrow 0=\left|0^2-5(0)-56\right|+27=83 \rightarrow 0\neq 83$

The only point that makes sense is (5,83), therefore it is the correct answer

$y=\left | x^2-5x-56\right |+27 \rightarrow 83=\left|5^2-5(5)-56\right|+27=83 \rightarrow 83=83$

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Example Question #3 : Special Functions

Evaluate: $\left | x-3\right |<15$

Possible Answers:

$x\leq18$

-12<x<18

-12,18

$-12\leq x\leq18$

x<18

Correct answer:

-12<x<18

Explanation:

Cancel the absolute value sign by separating the function $\left | x-3\right |<15$ into its positive and negative counterparts.

x-3<15

-(x-3)<15

Evaluate the first scenario.

x-3<15

x<18

Evaluate the second scenario.

-(x-3)<15

x-3>-15

x>-12

The correct answer is:

-12<x<18

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Example Question #1 : Absolute Value Functions

If $y=\left |x^3-5x^2+5 \right |$ , then what is the value of when x=3 ?

Possible Answers:

-13

-7

Correct answer:

Explanation:

We evaluate for x=3

y=|(3)^3-5(3)^2+5|

y=|27-45+5|

y=|-13|

Since the absolute value of any number represents its magnitude from and is therefore always positive, the final answer would be y=13

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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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Precalculus : Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Absolute Value Functions

Example Question #3 : Special Functions

Example Question #1 : Absolute Value 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

All Precalculus Resources

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