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

$ⓓ(f-g)(-3)$.

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

The given functions are$f\left(x\right)=2x^2-4x+1$and$g\left(x\right)=5x^2+8x+3$

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

$(f+g)\left(x\right)=(2x^2-4x+1)+(5x^2+8x+3)$

Rewrite without the parentheses

$(f+g)\left(x\right)=2x^2-4x+1+5x^2+8x+3$

Put like terms together

$(f+g)\left(x\right)=2x^2+5x^2+8x-4x+1+3$

Combine like terms

$(f+g)\left(x\right)=7x^2+4x+4$

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$(f+g)\left(x\right)$.

$(f+g)\left(x\right)=7x^2+4x+4$

put $x=2$

role="math" localid="1644579375257" $$\begin{gathered} (f+g)\left(2\right)=7{\left(2\right)}^2+4·2+4 \\ (f+g)\left(2\right)=7·4+4·2+4 \\ (f+g)\left(2\right)=28+8+4 \\ (f+g)\left(2\right)=40 \end{gathered}$$

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

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

$(f-g)\left(x\right)=(2x^2-4x+1)-(5x^2+8x+3)$

Rewrite without the parentheses

$(f-g)\left(x\right)=2x^2-4x+1-5x^2-8x-3$

Put like terms together

$(f-g)\left(x\right)=2x^2-5x^2-8x-4x+1-3$

Combine like terms

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$(f-g)\left(x\right)$.

$(f-g)\left(x\right)=-3x^2-12x-2$

put $x=-3$

$$\begin{gathered} (f-g)(-3)=-3{(-3)}^2-12·(-3)-2 \\ (f-g)(-3)=-3·9+12·3-2 \\ (f-g)(-3)=-27+36-2 \\ (f-g)(-3)=7 \end{gathered}$$