Question

$\text{Forfunctions}f\left(x\right)=5x^2-4x-1\text{and}g\left(x\right)=x^2+3x+8,\text{find}\left(a\right)(f+g)\left(x\right)\left(b\right)(f+g)\left(3\right)\left(c\right)(f-g)\left(x\right)\left(d\right)(f-g)(-2).$

Short answer

The value is

$\begin{array}{l}\left(a\right)(f+g)\left(x\right)=6x^2-x+7\\ \left(b\right)(f+g)\left(3\right)=58\\ \left(c\right)(f-g)\left(x\right)=4x^2-7x-9\\ \left(d\right)(f-g)(-2)=21\end{array}$

Step-by-step solution

Step 1. Given information 

$\begin{array}{l}\text{Thegivenfunctionsare}\\ f\left(x\right)=5x^2-4x-1\\ g\left(x\right)=x^2+3x+8\\ \left(a\right)(f+g)\left(x\right)\left(b\right)(f+g)\left(3\right)\left(c\right)(f-g)\left(x\right)\left(d\right)(f-g)(-2)\end{array}$

Step 2. Solve for part (a) 

$\begin{array}{l}\text{Weknow}(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)\\ \text{Substitutethefunctions}f\left(x\right)\text{and}g\left(x\right)\\ \Rightarrow (f+g)\left(x\right)=\left(5x^2-4x-1\right)+\left(x^2+3x+8\right)\\ =5x^2-4x-1+x^2+3x+8\left[\text{Rewritewithoutparentheses}\right]\\ =5x^2+x^2-4x+3x-1+8\left[\text{Putliketermstogether}\right]\\ =6x^2-x+7\end{array}$

Step 3. Solve for part (b) 

$\begin{array}{l}\text{Wehave}(f+g)\left(x\right)=6x^2-x+7\\ \Rightarrow (f+g)\left(3\right)=6{\left(3\right)}^2-\left(3\right)+7\left[\text{Substitute}x=3\right]\\ =6\left(9\right)-3+7\\ =54-3+7\\ =58\end{array}$

Step 4. Solve for part (c) 

$\begin{array}{l}\text{Weknow}(f-g)\left(x\right)=f\left(x\right)-g\left(x\right)\\ \text{Substitutethefunctions}f\left(x\right)\text{and}g\left(x\right)\\ \Rightarrow (f-g)\left(x\right)=\left(5x^2-4x-1\right)-\left(x^2+3x+8\right)\\ =5x^2-4x-1-x^2-3x-8\left[\text{Rewritewithoutparentheses}\right]\\ =5x^2-x^2-4x-3x-1-8\left[\text{Putliketermstogether}\right]\\ =4x^2-7x-9\end{array}$

Step 4. Solve for part (d) 

$\begin{array}{l}\text{Wehave}(f-g)\left(x\right)=4x^2-7x-9\\ \Rightarrow (f-g)(-2)=4x^2-7x-9\left[\text{Substitute}x=2\right]\\ =4{\left(2\right)}^2-7\left(-2\right)-9\\ =4\left(4\right)+14-9\\ =16+14-9\\ =21\end{array}$