MATH SOLVE

5 months ago

Q:
# Let f(x)=5−8x+15x^2. Calculate the following values:f(a)= _____f(a+h)= ____f(a+h)−f(a)/h= ____ for h≠0

Accepted Solution

A:

Answer:[tex]f(a)=5-8a+15a^2[/tex][tex]f(a+h)=5-8a-8h+15a^2+30ah+15h^2[/tex][tex]\frac{f(a+h)-f(a)}{h}=-8+30a+15h[/tex]Step-by-step explanation:We are given [tex]f(x)=5-8x+15x^2[/tex].We want to find [tex]f(a)[/tex] so we just replace the x there with a giving us:[tex]f(a)=5-8a+15a^2[/tex].We want to find [tex]f(a+h)[/tex] so we just replace x with (a+h) now giving us:[tex]f(a+h)=5-8(a+h)+15(a+h)^2[/tex].We will need to distribute and multiply things out here for later use so let's go ahead and do that:[tex]f(a+h)=5-8(a+h)+15(a+h)^2[/tex][tex]f(a+h)=5-8a-8h+15(a+h)(a+h)[/tex][tex]f(a+h)=5-8a-8h+15(a^2+2ah+h^2)[/tex][tex]f(a+h)=5-8a-8h+15a^2+30ah+15h^2[/tex]We want to find [tex]\frac{f(a+h)-f(a)}{h}[/tex] where h is not 0.This requires the parts we found above:[tex]\frac{f(a+h)-f(a)}{h}[/tex][tex]\frac{(5-8a-8h+15a^2+30ah+15h^2)-(5-8a+15a^2)}{h}[/tex]There are some thing that will zero out (cancel out) in the numerator.You have 5-8a+15a^2 in both parenthesis and you are subtracting so that part zero's out so you have this now:[tex]\frac{-8h+30ah+15h^2}{h}[/tex]Now you can divide h from top and bottom giving you:[tex]\frac{-8+30a+15h}{1}[/tex][tex]-8+30a+15h[/tex]