Q 283 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The solution is (a) $x^4+6x^3+9x^2+10x-12$ 

(b) $14$ 

Step 1. Given information

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

Part (a) Step 1. Find f · g x

$$\begin{gathered} \left(f·g\right)\left(x\right)=f\left(x\right)·g\left(x\right) \\ \left(f·g\right)\left(x\right)=\left(x^2+4x-3\right)\left(x^2+2x+4\right) \\ \left(f·g\right)\left(x\right)=x^2\left(x^2+2x+4\right)+4x\left(x^2+2x+4\right)-3\left(x^2+2x+4\right) \\ \left(a\right) \left(f·g\right)\left(x\right)=x^4+6x^3+9x^2+10x-12 \end{gathered}$$

Part (b) Step 1. Substitute x = 1

$$\begin{gathered} \left(f·g\right)\left(x\right)=x^4+6x^3+9x^2+10x-12 \\ \left(f·g\right)\left(1\right)={\left(1\right)}^4+6{\left(1\right)}^3+9{\left(1\right)}^2+10\left(1\right)-12 \\ \left(b\right) \left(f·g\right)\left(1\right)=14 \end{gathered}$$