Q. 279 - Intermediate Algebra

Question

For function $f\left(x\right)=x-1$ and $g\left(x\right)=4x^2+3x-5$ find (a) $\left(f·g\right)\left(x\right)$    (b ) $\left(f·g\right)\left(-2\right)$

Step-by-Step Solution

Answer

The solution is (a) $4x^3-x^2-8x+5$ (b) $-15$

Step 1. Given information

$f\left(x\right)=x-1$ and $g\left(x\right)=4x^2+3x-5$

Part (a) Step 1. Find f · g x

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

$$\begin{gathered} \left(f·g\right)\left(x\right)=f\left(x\right)·g\left(x\right) \\ \left(f·g\right)\left(x\right)=\left(x-1\right)\left(4x^2+3x-5\right) \\ \left(f·g\right)\left(x\right)=x\left(4x^2+3x-5\right)-1\left(4x^2+3x-5\right) \end{gathered}$$

distribute the parenthesis 

$$\begin{gathered} \left(f·g\right)\left(x\right)=x\left(4x^2+3x-5\right)-1\left(4x^2+3x-5\right) \\ \left(f·g\right)\left(x\right)=x·4x^2+x·3x+x\left(-5\right)-1·3x-1·\left(-5\right) \end{gathered}$$

Simplify

(a) $\left(f·g\right)\left(x\right)=4x^3-x^2-8x+5$

Part (b) Step 1. Substitute x = - 2

Substitute $x=-2$ and simplify

In part (a) we found $\left(f·g\right)\left(x\right)$ and now are asked to find $\left(f·g\right)\left(-2\right)$

To find $\left(f·g\right)\left(-2\right)$ substitute $x=-2$

$$\begin{gathered} \left(f·g\right)\left(x\right)=4x^3-x^2-8x+5 \\ \left(f·g\right)\left(-2\right)=4{\left(-2\right)}^3-{\left(-2\right)}^2-8\left(-2\right)+5 \end{gathered}$$

(b) $\left(f·g\right)\left(-2\right)=-15$