What did I do wrong?

Question

f(x)= 5x^2 at x= -1

5(1-h)^2 = 5-2h+h^2
=f'(-1)=lim(h-0)= 5-2h+h^2-5/h
=f'(-1)=lim(h-0)= h(-2+h)/h
=-2

The answer is actually -10...what did I do wrong??
Thank you..

Answer 1

Well, your first error is that you should have computed f(x+h), not f(-x-h). Otherwise, you'll get the derivative of f(-x) at x = -(-1) = 1. Your second error is that you forgot to distribute the 5 over the entire quantity in the parentheses. 5(-1+h)² = 5(1 - 2h + h²) = 5 - 10h + 5h². Now, computing this correctly, we see that:

f'(-1) = [h→0]lim (5(-1+h)² - 5)/h
f'(-1) = [h→0]lim (5 - 10h + 5h² - 5)/h
f'(-1) = [h→0]lim (-10h + 5h²)/h
f'(-1) = [h→0]lim -10 + 5h
f'(-1) = -10

As your answer key states.

Answer 2

Evaluated the expression at x=1 and not x= -1?

Also, 5(-1-h)^2 = 5 + 10h + 5h^2

Answer 3

5(-1+ h)^2 = 5(1 -2h + h^2) = 5 -10h + 5h^2

Answer 4

What are you trying to find?
lim at x = -1, or f(-1)
in this case, both are 5
Where you getting -10 from?

Answer 5

f'(-1) = [h→0]lim (5(-1+h)² - 5)/h
f'(-1) = [h→0]lim (5 - 10h + 5h² - 5)/h
f'(-1) = [h→0]lim (-10h + 5h²)/h
f'(-1) = [h→0]lim -10 + 5h
f'(-1) = -10