Algebra of Functions

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Pre-Calculus › Algebra of Functions

Questions 1 - 10

1

Find if and .

Explanation

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

2

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:

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

3

Find given

and

Explanation

To evaluate, first evaluate and then plug in that answer into . Thus,

Then, is

4

Find if and .

Explanation

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

5

Find given

and

Explanation

To evaluate, first evaluate and then plug in that answer into . Thus,

Then, is

6

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:

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

7

Given $f(x)=\sqrt{x-2}$ and $g(x)=x^2-2\sqrt{x}$ , find $(g\circ f)(x)$ .

$(g\circ f)(x)=x-2-2\sqrt[4]{x-2}$

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

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

$(g\circ f)(x)=x-2-2\sqrt{x-2}$

None of the other answers.

Explanation

$(g\circ f)(x)= g(f(x))$ and is read as "g of f of x" and is equivalent to plugging the function f(x) into the variables in the function g(x).

${\color{Red} f(x)=\sqrt{x-2}}$

${\color{Blue} g(x)=x^2-2\sqrt{x}}$

$(g\circ f)(x)= g(f(x))={\color{Blue}{\color{Red} \sqrt{x-2}}^2-2\sqrt{{\color{Red} \sqrt{x-2}}}}=\mathbf{x-2-2\sqrt[4]{x-2}}$

8

Simplify the following expression:

$\frac{9x^2+18x+9}{3x+3}+(x-1)$ .

Explanation

First, we can start off by factoring out constants from the numerator and denominator.

$\frac{9(x^2+2x+1)}{3(x+1)}$

The 9/3 simplifies to just a 3 in the numerator. Next, we factor the top numerator into , and simplify with the denominator.

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

We now have

9

If and , what does $\frac{f(x)}{g(x)}$ equal?

Explanation

We begin by factoring and we get .

Now, When we look at $\frac{f(x)}{g(x)}$ it will be $\frac{(3x-4)(x+7)}{(x+7)}$ .

We can take out from the numerator and cancel out the denominator, leaving us with .

10

Simplify the following expression:

$\frac{9x^2+18x+9}{3x+3}+(x-1)$ .

Explanation

First, we can start off by factoring out constants from the numerator and denominator.

$\frac{9(x^2+2x+1)}{3(x+1)}$

The 9/3 simplifies to just a 3 in the numerator. Next, we factor the top numerator into , and simplify with the denominator.

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

We now have

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