Question

Find h′(1) if h=f∘g, f(x)=−(3x^2)−7 and the equation of the tangent line of g at 1 is y=9x−8.

Answer 1

@e.mccormick

Answer 2

Do you know the chain rule for composition of functions?

Answer 3

mostly

Answer 4

The \(h=f\circ g\) means you have a composition. So I am thinking you need to chain rule that, then do the evaluation at 1.

Answer 5

So I did the product rule and got 6x(9x-8)+((3x^2)-7)(9)
What should I do from there - because what I did after got me the wrong answer =/

Answer 6

chain rule is this \(f(g(x))=f'(g(x))g'(x)\)

Answer 7

hmm.... but they never actually give you g. Just the equation of the tangent line there.

Answer 8

Hmm... have you done integrals or antiderivatives at this point?

Answer 9

Nope that's later apparently for my class

Answer 10

It is latter most of the time. I think all the g stuff will go to 1, based on what I see there.

Answer 11

So what would you suggest the answer be?

Answer 12

I think if you just work on the derivative of −(3x^2) it should get you the right answer. The -7 goes away at that point. So derive that, then evaluate it at 1.

Answer 13

Since they are asking for h'(1) and you are given g(1)
I think f(1) = -(3(9x-8)^2) - 7
then f'(1) = -3(2)(9x-8)(9)
f'(1) = -6(1)(9)
f'(1) = -54

Answer 14

@jkristia You are not given g(1). You are given g'(1)'s equation of the tangent line.

Answer 15

oh - right, my mistake

Answer 16

Yaah, well... I was making that mistake at the start too. LOL. And since this is pre-integral work, I think they went for something simple.

Answer 17

So is there any other way how to get -54 then? because that is apparently the right answer

Answer 18

I don't see how to get the composite function without taking the anti-derivative of g'(1) then

No actually it still works, you take derivate of f(x) - then insert g'(1), argh, no, you are missing the '9' then ... hmmm

f(1) = -(3(x)^2) - 7
f'(1) = -3(2)(x)
now insert g' instead of x
f'(1) = -3(2)(9x-8)(9) <-- missing '9'
f'(1) = -6(1)(9)
f'(1) = -54

Answer 19

Yah... with it pre-integration I was a tad stumped on that. LOL.

Answer 20

Wait... what if the fact that g is a tangent line is incidental!

Answer 21

.... no.... "tangent line OF g" so not incidental...

Answer 22

Hmm, I thought I had a solution, but didn't work..
I would be interested in knowing the correct solution (not just result) for this

Answer 23

Yah... same boat as me. It seems... odd... I am sure it is either something simple we missed, or a poorly written question, with no middle ground!

Answer 24

most likely something we (I) missed - it happens a lot to me haha

Answer 25

http://openstudy.com/study#/updates/4ff259c3e4b03c0c488af448

Answer 26

The g'(1)=9 doh! And I even thought about that before I got lost trying to find g()

Answer 27

are u done this question or u want solution?

Answer 28

We found the solution and our mistake in finding it. LOL @Invizen Did you get the point from that link to a related problem, or still need more?

Answer 29

ok :)

Answer 30

I'm all good now thanks so much!

Answer 31

>> round about way of telling you that g ′ (3)=−3
ahhh, that Is what I was missing, so in this case
So
g(1) = 9x-8 = 1
g'(1) = 9
f'(x) = -6x
h'(1) = f'(g(1))g'(1)
h'(1) = -6(1)(9)
h'(1) = -54

Answer 32

Yah, "equation of"... slope... LOL. y=mx+b came and bit us on the anatomy!