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Precalculus : Pre-Calculus

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Example Questions

Example Question #41 : Pre Calculus

Given the equation $\small xy+3x^{2}-y-7=0$ , determine the horizontal asymptote.

Possible Answers:

$\small \small y=0$

$\small \small x=1$

$\small y=1$

$\text{No horizontal asymptote exists}$

$\small \small y=3$

Correct answer:

$\text{No horizontal asymptote exists}$

Explanation:

Change the equation into the y=x form.

$\small xy+3x^{2}-y-7=0$

$\small xy-y=7-3x^{2}$

$\small \small y(x-1)=7-3x^{2}$

$\small \small \small\frac{y(x-1)}{(x-1)}=\frac{7-3x^{2}}{(x-1)}$

$\small \small \small \small \small\ y=\frac{7-3x^{2}}{x-1}$

When the degree of the exponent in the numerator is higher than the degree of the exponent in the denominator, there is no horizontal asymptote. Thus the answer is no horizontal asymptote.

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Example Question #8 : Rational Functions

At what value(s) of are the vertical asymptote(s) of the function $f(x)=\frac{x^2+6x+8}{x^2+x-12}$ located?

Possible Answers:

x=3

x=-4;x=3

x=-4

x=-4;x=2

Correct answer:

x=3

Explanation:

Asymptotes can occur where the function is undefined, or when the denominator is equal to . Simplfying our function gives us,

$f(x)=\frac{x^2+6x+8}{x^2+x-12}=\frac{(x+2)(x+4)}{(x-3)(x+4)}=\frac{(x+2)}{(x-3)}$ .

Now solving for when our denominator is zero gives us,

x-3=0

x=3

Therefore, there is a vertical asymptote at x=3

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Example Question #1 : Triangles

Let ABC be a right triangle with sides = 3 inches, = 4 inches, and = 5 inches. In degrees, what is the $\sin \Theta$ where $\Theta$ is the angle opposite of side ?

Possible Answers:

Correct answer:

Explanation:

3-4-5_triangle

We are looking for $\sin(\theta )$ . Remember the definition of sin in a right triangle is the length of the opposite side divided by the length of the hypotenuse.

So therefore, without figuring out $\Theta$ we can find

$\sin \Theta =\frac{3}{5}=.6$

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Example Question #2 : Trigonometric Functions And Graphs

Simplify the function below:

$f(x)=cos^2(\frac{1}{2}cos^{-1}x)$

Possible Answers:

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

$\frac{x}{2}$

cosx

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

Correct answer:

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

Explanation:

We need to use the following formulas:

a) $cos^2x=\frac{1+cos2x}{2}$

and

b) $cos(cos^{-1}x)=x$

We can simplify f(x) as follows:

$f(x)=cos^2(\frac{1}{2}cos^{-1}x)=\frac{1+cos(cos^{-1}x)}{2}=\frac{1+x}{2}$

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Example Question #1 : Using Pythagorean Identities

Simplify

$\cot(x)\times{\sin(x) }$

Possible Answers:

$\arcsin(\pi )$

$\cos(x)$

$\csc(x)$

$\tan(x)$

$\arcsin(x)$

Correct answer:

$\cos(x)$

Explanation:

$\cot(x) = \cos(x)/\sin(x)$ . Thus: $[\cos(x)/\sin(x)]\times\sin(x)=\cos(x)$

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Example Question #2 : Using Pythagorean Identities

Simplify

$\frac{cot (x)\cdot tan(x)}{cos(x)}$

Possible Answers:

$\cos(x)$

$\sec(x)$

$\cot(x)/sec(x)$

$\csc(x)$

$\cos(x)\times\tan(x)$

Correct answer:

$\sec(x)$

Explanation:

$\cot(x)\times\tan(x) = 1$

and

$1/(\cos(x)) = \sec(x)$ .

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Example Question #2 : Solving Trigonometric Equations

If $sinx-cosx=\frac{1}{3}$ , find the value of the function $f(x)=sin^{3}x+cos^{3}x$ .

Possible Answers:

$\frac{11}{27}$

$-\frac{11}{27}$

$-\frac{13}{27}$

$\frac{13}{27}$

Correct answer:

$\frac{13}{27}$

Explanation:

We can write:

$sinx-cosx=\frac{1}{3}$

$\Rightarrow (sinx-cosx)^{2}=\frac{1}{9}$

$\Rightarrow sin^2x+cos^2x-2sinxcosx=\frac{1}{9}$

$\Rightarrow 2sinxcosx=1-\frac{1}{9}$

$\Rightarrow sinxcosx=\frac{4}{9}$

We can also write:

(sinx-cosx)^3=sin^3x+cos^3x+3sinxcos^2x-3cosxsin^2x

$\Rightarrow sin^3x+cos^3x=(sinx-cosx)^3+3sinxcosx(sinx-cosx)$

Now we can substitute the values of (sinx - cosx) and (sinx * cosx) in the obtained function:

$\Rightarrow sin^3x+cos^3x=(\frac{1}{3})^3+3*\frac{4}{9}*\frac{1}{3}=\frac{1}{27}+\frac{4}{9}=\frac{13}{27}$

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Example Question #11 : Functions

Identifying a function.

Which of the following is a function.

Possible Answers:

h(x)=-3x^4-2x+3sinx

$x=\left | y \right |$

(x-4)^2+y^2=25

$g(x)=\pm \sqrt{x^2-9}$

x=2y^2-14

Correct answer:

h(x)=-3x^4-2x+3sinx

Explanation:

The only relation listed that doesn't map more than one dependent variable value for some independent variable value is h(x). Another way of saying this is that each value in the domain of h corresponds to a distinct value in its range. h(x) is the only one-to-one relation, and so is the only option that is a function.

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Example Question #12 : Relations And Functions

Evaluating a function.

Evaluate f(-3) , when

$f(x)=\frac{3x^2-4x+1}{x+7}$ .

Possible Answers:

f(-3)=4

$f(-3)=-\frac{1}{2}$

f(-3)=1.6

f(-3)=10

$f(-3)=1\tfrac{3}{5}$

Correct answer:

f(-3)=10

Explanation:

Plug in -3 for x in the function rule:

$\frac{3(-3)^2-4(-3)+1}{-3+7}=\frac{3\cdot 9+12+1}{4}=\frac{27+12+1}{4}=\frac{40}{4}=10$ .

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Example Question #12 : Relations And Functions

Finding inverse functions.

Given,

$h(x)=\frac{5x+2}{x-3}$

find

$h^{-1}(x)$ .

Possible Answers:

$h^{-1}(x)=\frac{3x+2}{x-5}$

$h^{-1}(x)=\frac{-2}{3}$

$h^{-1}(x)=\frac{5x+2}{x-5}$

$h^{-1}(x)=\frac{3y+2}{y-5}$

$h^{-1}(x)=\frac{5x+2}{x-3}$

Correct answer:

$h^{-1}(x)=\frac{3x+2}{x-5}$

Explanation:

To find an inversefunction find the input in terms of the output. In other words, solve the function for the dependent variable.

First, set h(x) = y. Now solve for y.

$y=\frac{5x+2}{x+3}$

Multiply the denominator on both sides

y(x-3)=5x+2

Distribute y

xy-3y=5x+2

Rearrange to get only x terms on one side.

xy-5x=3y+2

Factor out x on the left side.

x(y-5)=3y+2

Divide by (y-5) on both sides to get x by itself.

$x=\frac{3y+2}{y-5}$

The final step is to recognize that the independent variable is now the dependent variable. To put this into proper notation switch x and y. The inverse function is

$h^{-1}(x)=\frac{3x+2}{x-5}$ .

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #41 : Pre Calculus

Example Question #8 : Rational Functions

Example Question #1 : Triangles

Example Question #2 : Trigonometric Functions And Graphs

Example Question #1 : Using Pythagorean Identities

Example Question #2 : Using Pythagorean Identities

Example Question #2 : Solving Trigonometric Equations

Example Question #11 : Functions

Example Question #12 : Relations And Functions

Example Question #12 : Relations And Functions

All Precalculus Resources

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