If f(x) = 1/(x-2), (f circle g)'(1) = 6 and g'(1) = -1, then g(1) =

Question

anonymous · Answer

Here I'll post what I've tried...

anonymous · Answer

I did $$f'(g(1)) * g'(1) = 6$$
$$-6 = f'(g(1))$$
$$f'(x) = -1/(x-2)^2$$
$$-6 = -1/(x-2)^2$$
$$(x-2)^2 = 1/6$$
$$x = 2 +-\sqrt{6}/6$$
and g(1) = x

ganeshie8 · Answer

Looks good!

anonymous · Answer

It does? But it doesn't match any of my answer choices

anonymous · Answer

They are: 


a. -7
b. -5
c. 5
d. 7
e. 8

jim_thompson5910 · Answer

I did it a slightly different way, but I'm getting the same result


f(x) = 1/(x-2)

f(g(x)) = 1/(g(x)-2)

(f o g)(x) = 1/(g(x)-2)

(f o g)'(x) = -(g'(x))/(g(x)-2)^2

(f o g)'(1) = -(g'(1))/(g(1)-2)^2

6 = -(-1)/(z-2)^2 ... let z = g(1)

6 = 1/(z-2)^2

6(z-2)^2 = 1

(z-2)^2 = 1/6

z-2 = +-sqrt(1/6)

z-2 = +-sqrt(6)/6

z = 2+-sqrt(6)/6

g(1) = 2+-sqrt(6)/6

anonymous · Answer

Then I think they put the wrong answer choices. It's from a practice AP exam so there could be a mistake in it

ganeshie8 · Answer

unique answer is not possible because of the quadratic expression in the denominator of f'(x) so most likely there is some mistake

anonymous · Answer

That's true. I'll just have to ask my teacher when I come back from break

anonymous · Answer

Thank you for helping guys!