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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 #21 : Inverse Functions

Find the inverse of this function: $h(x)=\sqrt{\frac{1}{x}}+1$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to have the inverse of a function, the new function must perform the inverse opperations in the opposite order. One way to ensure that is true is to consider the case of y=h(x) , switch x and y, then solve for y.

In this case, $y=\sqrt{\frac{1}{x}}+1$ becomes $x=\sqrt{\frac{1}{y}}+1$ and we solve for y.

$x=\sqrt{\frac{1}{y}}+1$ subtract 1 from both sides

$x-1 = \sqrt{\frac{1}{y}}$ square both sides

$(x-1)^2 = \frac{1}{y}$ now we will take both sides to the power of -1, in other words flip each side to the reciprocal.

We can consider (x-1)^2 to be $\frac{(x-1)^2}{1}$

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

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Example Question #102 : Functions

Find the inverse of the follow function:

y=3x^2-9

Possible Answers:

$y=\pm\sqrt{\frac{x+9}{3}}$

$y=\frac{1}{3x^2+9}$

$y=\frac{1}{3x^2-9}$

$y=\sqrt{\frac{x-9}{3}}$

Correct answer:

$y=\pm\sqrt{\frac{x+9}{3}}$

Explanation:

To find the inverse, substitute all x's for y's and all y's for x's and then solve for y.

x=3y^2-9

x+9=3y^2

$\frac{x+9}{3}=y^2$

$y=\pm\sqrt{\frac{x+9}{3}}$

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Example Question #103 : Functions

Find the inverse function.

Find the inverse function $f^{^{-1}}(x)$ of the function $f(x)=\frac{3x-4}{2}$

Possible Answers:

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

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

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

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

None of these answers are correct.

Correct answer:

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

Explanation:

To find an inverse function, switch x and y variables and solve again for y. The new function is the inverse. f(x) can be called y. To check your answer, you can insert either function into the x variable of the other, and the equations should both solve to equal x.

$f(x)= \frac{3x-4}{2}$

$y= \frac{3x-4}{2}$

$x= \frac{3y-4}{2}$

2x= 3y-4

2x+4=3y

$\frac{2x+4}{3}=y$

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

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Example Question #104 : Functions

Find the inverse of this function:

y=x^2 +6x + 9

Possible Answers:

y= (x+3)^2

y=(x-3)^2

$y=-3\pm \sqrt{x}$

$y=3+\sqrt{x}$

y=9+x

Correct answer:

$y=-3\pm \sqrt{x}$

Explanation:

To find the inverse of a function like this, switch the x and y variables (thereby "inverting" the function) and then solve for x.

First we would set up:

x=y^2 +6y + 9

And then let's factor:

x=(y+3)^2

And take the square root of both sides:

$\pm \sqrt{x} = y+3$

Leaving us with the final answer:

$y=-3\pm \sqrt{x}$

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Example Question #111 : Functions

If $f(x)=\frac{1+3x}{2-x}$ , find $f^{-1}(1)$ .

Possible Answers:

$-\frac{3}{4}$

$\frac{1}{3}$

$\frac{4}{3}$

$\frac{2}{3}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{4}$

Explanation:

Set f(x)=y , thus $y=\frac{1+3x}{2-x}$ .

Now switch with .

So now,

$x=\frac{1+3y}{2-y}$ .

Simplify to isolate by itself.

$\\ x=\frac{1+3y}{2-y}\rightarrow 2x-xy\\ =1+3y\rightarrow 3y+xy\\=2x-1$

Therefore,

$\\ y(3+x)=2x-1\\ \\ y=\frac{2x-1}{3+x}$ .

Now substitute with $f^{-1}(x)$ ,

$f^{-1}(x)=\frac{2x-1}{3+x}$ , and

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

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Example Question #112 : Functions

Find the inverse of this function: $f(x)=\frac{8}{3}x-8$

Possible Answers:

$f^{-1}(x)=\frac{8}{3}x-3$

$f^{-1}(x)=\frac{3}{8}x-3$

$f^{-1}(x)=\frac{3}{8}x+3$

$f^{-1}(x)=\frac{8}{3}x+8$

$f^{-1}(x)=\frac{8}{3}x+3$

Correct answer:

$f^{-1}(x)=\frac{3}{8}x+3$

Explanation:

$f(x)=\frac{8}{3}x-8$

Write the equation in terms of x and y:

$y=\frac{8}{3}x-8$

Switch the x and y (this inverts the relationship of the two variables):

$x=\frac{8}{3}y-8$

Solve for y:

$x+8=\frac{8}{3}y$

$\frac{3}{8}(x+8)=\frac{8}{3}y*\frac{3}{8}$

$\frac{3}{8}x+3=y$

Rewrite to indicate this is the inverse:

$f^{-1}(x)=\frac{3}{8}x+3$

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Example Question #6 : Inverses

Find $f^{-1}(x)$ for $f(x)=\frac{2x-7}{8x+3}$

Possible Answers:

$f^{-1}(x)=\frac{3x+7}{2-8x}$

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

$f^{-1}(x)=\frac{3x^{2}+4x}{8}$

$f^{-1}(x)=\sqrt{x^{2}+6}$

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

Correct answer:

$f^{-1}(x)=\frac{3x+7}{2-8x}$

Explanation:

$f(x)=\frac{2x-7}{8x+3}$

To find the inverse of a function, first swap the x and y in the given function.

$x=\frac{2y-7}{8y+3}$

Solve for y in this re-written form.

x(8y+3)=2y-7

8xy+3x=2y-7

8xy-2y=-7-3x

y(8x-2)=-7-3x

$y=\frac{-7-3x}{8x-2}$

$y=\frac{-(7+3x)}{-(-8x+2)}$

$y=\frac{3x+7}{2-8x}$

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Example Question #113 : Functions

Find the inverse of: 4x-y=5

Possible Answers:

y=4x-5

$y= \frac{1}{4}x + \frac{5}{4}$

y=4x+5

$y=\frac{4}{5}, \frac{1}{4}$

$y= \frac{1}{4}x + 5$

Correct answer:

$y= \frac{1}{4}x + \frac{5}{4}$

Explanation:

Interchange the variables and solve for .

4y-x=5

Add on both sides.

4y=x+5

Divide by four on both sides.

$\frac{4y}{4}=\frac{x+5}{4}$

The answer is: $y= \frac{1}{4}x + \frac{5}{4}$

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Example Question #114 : Functions

Find the inverse function ( $f^{-1}$ ) of the function $f(x)=\sqrt{6x-4}$

Possible Answers:

None of these answers are correct.

$f^{-1}=\frac{x^2+4}{6}$

$f^{-1}=\frac{x^2-4}{6}$

$f^{-1}=\frac{(x-2)^2}{4}$

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

Correct answer:

$f^{-1}=\frac{x^2+4}{6}$

Explanation:

$f(x)=\sqrt{6x-4}$

f(x) can be called y. Switch x and y, and solve for y. The resulting new equation is the inverse of f(x).

$y=\sqrt{6x-4}$

$x=\sqrt{6y-4}$

$x^2=(\sqrt{6y-4})^2$

x^2=6y-4

x^2+4=6y

$\frac{x^2+4}{6}=y$

$\frac{x^2+4}{6}=f^{-1}(x)$

To double check your work, substitute f(x) into its inverse or vice versa. Both substitutions should equal x.

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Example Question #11 : Matrices

Which of the following is the inverse of h(x) ?

h(x)=3lnx

Possible Answers:

$y=e^{\frac{x}{3}}$

$y=\frac{e^x}{3}$

$y=3e^{x}$

$y=9e^{x}$

Correct answer:

$y=e^{\frac{x}{3}}$

Explanation:

Which of the following is the inverse of h(x) ?

h(x)=3lnx

To find the inverse of a function, we need to swap x and y, and then rearrange to solve for y. The inverse of a function is basically the function we get if we swap the x and y coordinates for every point on the original function.

So, to begin, we can replace the h(x) with y.

y=3lnx

Next, swap x and y

x=3lny

Now, we need to get y all by itself; we can to begin by dividng the three over.

$\frac{1}{3}x=lny$

Now, recall that

And that we can rewrite any log as an exponent as follows:

log_b(a)=x

b^x=a

So with that in mind, we can rearrange our function to get y by itself:

$\frac{1}{3}x=lny$

Becomes our final answer:

$y=e^{\frac{x}{3}}$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #21 : Inverse Functions

Example Question #102 : Functions

Example Question #103 : Functions

Example Question #104 : Functions

Example Question #111 : Functions

Example Question #112 : Functions

Example Question #6 : Inverses

Example Question #113 : Functions

Example Question #114 : Functions

Example Question #11 : Matrices

All Precalculus Resources

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