Question

f(x)=7x^3-3x^2+3 using f(a+h)-f(a)/h

Answer 1

I started with 7(a+h)^3-3(a+h)^2+3/h

But from there I do not know how to FOIL out (a+h)^3 properly. I would normally foil it out as if it were squared and then multiply the last (a+h) to what I got when it was expanded from squared form

Answer 2

@jim_thompson5910

Answer 3

f(x)=7x^3-3x^2+3

f(a+h)=7(a+h)^3-3(a+h)^2+3

f(a+h)=7(a+h)(a+h)^2-3(a+h)^2+3

f(a+h)=7(a+h)(a^2+2ah+h^2)-3(a^2+2ah+h^2)+3

f(a+h)=7[a(a^2+2ah+h^2)+h(a^2+2ah+h^2)]-3(a^2+2ah+h^2)+3

f(a+h)=7[a^3+2a^2h+ah^2+a^2h+2ah^2+h^3]-3(a^2+2ah+h^2)+3

f(a+h)=7a^3+14a^2h+7ah^2+7a^2h+14ah^2+7h^3-3a^2-6ah-3h^2+3

f(a+h)=7a^3+14a^2h+7ah^2+7a^2h+14ah^2+7h^3-3a^2-6ah-3h^2+3

f(a+h) = 7a^3+21a^2h+21ah^2+7h^3-3a^2-6ah-3h^2+3

Answer 4

from there I factor out an ah from the numerator which then the h i pulled out cancels with the h from the denominator and I am left with a(7a^2+21a-21h+7h^2-3a-3h-3

Answer 5

*+21a+21h*

Answer 6

well,

f(a+h) - f(a)

turns into

(7a^3+21a^2h+21ah^2+7h^3-3a^2-6ah-3h^2+3) - (7a^3-3a^2+3)

7a^3+21a^2h+21ah^2+7h^3-3a^2-6ah-3h^2+3 - 7a^3+3a^2-3

21a^2h+21ah^2+7h^3-6ah-3h^2

then you factor out h to get

h(21a^2+21ah+7h^2-6a-3h)

Answer 7

this is all in the numerator of course

Answer 8

- (7a^3-3a^2+3) where does this part come from?

Answer 9

that's -f(a)

well after you replace f(a) with 7a^3-3a^2+3

Answer 10

ohhhhh I was leaving that off...hang on a moment let me back up a little bit

Answer 11

ok

Answer 12

@jimthompson_5910 > so the answer over all is: 21a^2+21ah+7h^2-6a-3h???

Answer 13

that is correct

Answer 14

wonderful! Thank you

Answer 15

yw