Q 280 - Intermediate Algebra

Question

For function $f\left(x\right)=2x-7$ and $g\left(x\right)=2x+7$

find (a) $\left(f·g\right)\left(x\right)$  (b) $\left(f·g\right)\left(-3\right)$

Step-by-Step Solution

Answer

The solution is (a) $4x^2-49$  (b)$-13$

Step 1. Given information

$f\left(x\right)=2x-7$ and $g\left(x\right)=2x+7$

Part (a) Step 1. Find f · g x

Substitute for,  $f\left(x\right)$ and $g\left(x\right)$ and multiply the polynomials.

$$\begin{gathered} \left(f·g\right)\left(x\right)=f\left(x\right)·g\left(x\right) \\ \left(f·g\right)\left(x\right)=\left(2x-7\right)\left(2x+7\right) \\ \left(f·g\right)\left(x\right)=2x\left(2x+7\right)-7\left(2x+7\right) \end{gathered}$$

apply the difference of two squares formula

$\left(f·g\right)\left(x\right)={\left(2x\right)}^2-7^2$

Simplify

(a) $\left(f·g\right)\left(x\right)=4x^2-49$

Part (b) Step 1. Substitute x = - 3

Substitute $x=-3$ and simplify.

In part (a) we found $\left(f·g\right)\left(x\right)$ and now are asked to find $\left(f·g\right)\left(-3\right)$

To find $\left(f·g\right)\left(-3\right)$ substitute  $x=-3$

$$\begin{gathered} \left(f·g\right)\left(x\right)=4x^2-49 \\ \left(f·g\right)\left(-3\right)=4{\left(-3\right)}^2-49 \end{gathered}$$

(b) $\left(f·g\right)\left(-3\right)=-13$