For each function, find$ⓐ(f+g)\left(x\right)ⓑ(f+g)\left(2\right)ⓒ(f-g)\left(x\right)$

$ⓓ(f-g)(-3)$

$f\left(x\right)=3x^3-x^2-2x+3$and$g\left(x\right)=3x^3-7x$

The given functions are$f\left(x\right)=3x^3-x^2-2x+3$and$g\left(x\right)=3x^3-7x$

We find $ⓐ(f+g)\left(x\right)ⓑ(f+g)\left(2\right)ⓒ(f-g)\left(x\right)ⓓ(f-g)(-3)$for the given function.

We find the value of $(f+g)\left(x\right)$for the given function.

We know$(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)$

$(f+g)\left(x\right)=(3x^3-x^2-2x+3)+(3x^3-7x)$

Rewrite without the parentheses

$(f+g)\left(x\right)=3x^3-x^2-2x+3+3x^3-7x$

Put like terms together

$(f+g)\left(x\right)=3x^3+3x^3-x^2-2x-7x+3$

Combine like terms

$(f+g)\left(x\right)=6x^3-x^2-9x+3$

In part (a) we found $(f+g)\left(x\right)$and now we have to find $(f+g)\left(2\right)$.

To find $(f+g)\left(2\right)$we substitute $x=2$in the value of $(f+g)\left(x\right)$.

$(f+g)\left(x\right)=6x^3-x^2-9x+3$

put $x=2$

$$\begin{gathered} (f+g)\left(2\right)=6{\left(2\right)}^3-{\left(2\right)}^2-9·2+3 \\ (f+g)\left(2\right)=6·8-4-9·2+3 \\ (f+g)\left(2\right)=48-4-18+3 \\ (f+g)\left(2\right)=29 \end{gathered}$$

We have to find the value of $(f-g)\left(x\right)$for the given function.

We know$(f-g)\left(x\right)=f\left(x\right)-g\left(x\right)$

$(f-g)\left(x\right)=(3x^3-x^2-2x+3)-(3x^3-7x)$

Rewrite without the parentheses

$(f-g)\left(x\right)=3x^3-x^2-2x+3-3x^3+7x$

Put like terms together

$(f-g)\left(x\right)=3x^3-3x^3-x^2-2x+7x+3$

Combine like terms

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

In part (c) we found $(f-g)\left(x\right)$now we have to find$(f-g)(-3)$

To find the value of $(f-g)(-3)$we put $x=-3$in the value of$(f-g)\left(x\right)$

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

put $x=-3$

$$\begin{gathered} (f-g)(-3)=-{(-3)}^2+5(-3)+3 \\ (f-g)(-3)=-9-15+3 \\ (f-g)(-3)=-21 \end{gathered}$$