Question

f(-1)=5 and f'(-1)=8
Let h(x)=x^11(f(x))
Evaluate h'(x) at x=-1
h'(-1)=?

Answer 1

You can consider using the product rule \[h(x)=g(x)*f(x)\]\[h'(x) = g'(x)f(x)+g(x)f'(x)\]
you don't actually need to know f(x) or f'(x) since you're only trying to find h'(-1) and f(-1) and f'(-1) is given.

Answer 2

ahh i'm really not getting it. are you saying the answer is 80 because 40+40?
please help and thanks!

Answer 3

thanks @2le

Answer 4

In your situation, your g(x) (from my example) is \[x^{11}\]
so \[h'(x)=11x^{10}*f(x) + x^{11}*f'(x)\]
now the answer is asking for h'(-1) so you just have to plug the (-1) for every x value.
\[h'(-1)=11(-1)^{10}*f(-1)+(-1)^{-1}*f'(-1)\]
The exponent is a simple even power is positive, odd power is negative. The f(-1) and f'(-1) is already given in the original problem.

Answer 5

ahh! im getting 352 or-352. not sure because of the exponents and it's still not saying it's right!

Answer 6

can you possibly explain any further @2le ?! pleaseeee! i'm not getting the final answer!

Answer 7

Are you understanding the steps up to that point? f(-1) = 5 and f'(-1)=8 is given.
h'(-1) = 11*1*5+(-1)*8 = 55-8=47. I don't know how you got in the hundreds.

Answer 8

\[(-1)^{10}=1 \]
\[(-1)^{11}=-1 \]

Answer 9

OHHHHH! HAHAHA omg thank you! THANK YOU!