Question

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

(b)$\left(f·g\right)\left(-1\right)$

Short answer

The solution is (a) $3x^3+4x^2+8$

(b)The value at x=-1 is$9$

Step-by-step solution

Step 1. Given information

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

Part (a) Step 1. Finding (f.g)(x)

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

Part (b) Step 1. Substitute x = -1 

$$\begin{gathered} \left(f·g\right)\left(x\right)=3x^3+4x^2+8 \\ \left(f·g\right)\left(-1\right)=3{\left(-1\right)}^3+4{\left(-1\right)}^2+8 \\ =9 \end{gathered}$$