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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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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Example Question #991 : Pre Calculus

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

f(x)=sin^2(x)

Possible Answers:

$\frac{\sqrt2}{2}$

$-\frac{1}{2}$

$\frac{1}{2}$

$\frac{3}{4}$

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 #991 : Pre Calculus

Find the domain of f(x) below

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

Possible Answers:

$(-\infty,7)$

$[7,\infty)$

[-1,7]

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

$(-\infty,7)$

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 #992 : Pre Calculus

The domain of the following function

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

is:

Possible Answers:

$(2,\infty)$

$(-\infty,1)$

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

(1,2)

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

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 #993 : Pre Calculus

Find the range of the following function:

f(x)=7

Possible Answers:

{1}

$\text{The set of negative integers}$

$\text{The set of positive integers}$

$\text{The set of real numbers}$

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

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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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

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

Example Question #991 : Pre Calculus

Example Question #991 : Pre Calculus

Example Question #992 : Pre Calculus

Example Question #993 : Pre Calculus

All Precalculus Resources

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