Functions

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Pre-Calculus › Functions

Questions 1 - 10

1

Determine the value of of the function

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

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

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

Explanation

In order to determine the value of of the function we set The value becomes

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

As such

2

Find the inverse of .

$y = \frac{x+1}{10}$

$y = \frac{x-1}{10}$

$x = \frac{y - 1}{10}$

Explanation

To find the inverse of the function, we switch the switch the and variables in the function.

Switching and gives

Then, solving for gives our answer:

$\frac{x-1}{10} = y$

3

Determine the value of of the function

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

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

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

Explanation

In order to determine the value of of the function we set The value becomes

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

As such

4

Determine the value of of the function

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

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

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

Explanation

In order to determine the value of of the function we set The value becomes

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

As such

5

Find the inverse of .

$y = \frac{x+1}{10}$

$y = \frac{x-1}{10}$

$x = \frac{y - 1}{10}$

Explanation

To find the inverse of the function, we switch the switch the and variables in the function.

Switching and gives

Then, solving for gives our answer:

$\frac{x-1}{10} = y$

6

Find the inverse of .

$y = \frac{x+1}{10}$

$y = \frac{x-1}{10}$

$x = \frac{y - 1}{10}$

Explanation

To find the inverse of the function, we switch the switch the and variables in the function.

Switching and gives

Then, solving for gives our answer:

$\frac{x-1}{10} = y$

7

Find the inverse of the follow function:

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

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

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

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

Explanation

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

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

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

8

Find the inverse function of .

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

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

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

None of the other answers.

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

Explanation

To find the inverse you must reverse the variables and solve for y.

Reverse the variables:

Solve for y:

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

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

9

and . Find $(f\circ g)(x)$ .

$(f\circ g)(x)=4x^2+12x+6$

$(f\circ g)(x)=2x^2+12x+12$

$(f\circ g)(x)=2x^2-6$

$(f\circ g)(x)=4x^2-12x+9$

$(f\circ g)(x)=2x^2-3$

Explanation

${\color{Blue} f(x)=x^2-3}$ and ${\color{Red} g(x)=2x+3}$ .

To find $(f\circ g)(x)$ we plug in the function everywhere there is a variable in the function .

This is the composition of "f of g of x".

$(f\circ g)(x)=f(g(x))={\color{Blue} (}{\color{Red} 2x+3}{\color{Blue} )}{\color{Blue} ^2-3}$

Foil the square and simplify:

$(f\circ g)(x)=(4x^2+12x+9)=\mathbf{4x^2+12x+6}$

10

Find if and .

Explanation

Replace and substitute the value of into so that we are finding .

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