Question

For the functions $f\left(x\right)=2x^2+1$and $g\left(x\right)=3x-2$, find

the following and simplify:

(a) $f-g$

(b) $f·g$

(c) $f(x+h)-f\left(x\right)$

Short answer

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

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

(b) $(f·g)\left(x\right)=6x^3-4x^2+3x-2$

(c)$f(x+h)-f\left(x\right)=4xh+2h^2$

Step-by-step solution

Step 1. given data

The given functions are

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

Step 2. Part (a)

The difference of function$f\left(x\right)\&g\left(x\right)$is

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

Step 3. Part (b)

The product of functions$f\left(x\right)\&g\left(x\right)$is

$$\begin{gathered} (f·g)\left(x\right)=f\left(x\right)·g\left(x\right) \\ (f·g)\left(x\right)=(2x^2+1)(3x-2) \\ (f·g)\left(x\right)=\left(2x^2\right)\left(3x\right)-\left(2x^2\right)\left(2\right)+\left(1\right)\left(3x\right)-\left(1\right)\left(2\right) \\ (f·g)\left(x\right)=6x^3-4x^2+3x-2 \end{gathered}$$

Step 4. Part (c)

Substitute $x+h$ for x in the function $f\left(x\right)$

$$\begin{gathered} f\left(x\right)=2x^2+1 \\ f(x+h)=2{(x+h)}^2+1f(x+h)=2{(x}^2+2xh+h^2)+1 \\ f(x+h)=2x^2+4xh+2h^2+1 \end{gathered}$$

sorole="math" localid="1645903721224" $$\begin{gathered} f(x+h)-f\left(x\right)=(2x^2+4xh+2h^2+1)-(2x^2+1) \\ f(x+h)-f\left(x\right)=2x^2+4xh+2h^2+1-2x^2-1 \\ f(x+h)-f\left(x\right)=4xh+2h^2 \end{gathered}$$