Q. 282 - Intermediate Algebra

Question

For function $f\left(x\right)=x^2-5x+2$ and $g\left(x\right)=x^2-3x-1$ 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)$\left(f·g\right)\left(x\right)=x^4-8x^3+16x^2-x-2$

(b) $24$

Step 1. Given Information

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

Part (a) Step 1. Find f · g x

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

(a) $\left(f·g\right)\left(x\right)=x^4-8x^3+16x^2-x-2$

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

$$\begin{gathered} \left(f·g\right)\left(x\right)=x^4-8x^3+16x^2-x-2 \\ \left(f·g\right)\left(-1\right)={\left(-1\right)}^4-8{\left(-1\right)}^3+16{\left(-1\right)}^2-\left(-1\right)-2 \end{gathered}$$

(b) $\left(f·g\right)\left(-1\right)=24$