Question

Need calc help (updated)

Answer 1

What's the question?

Answer 2

Any ideas @zepdrix ?

Answer 3

or this question

Answer 4

Well recall that when you take a composition of a function and it's inverse,
\(\large\rm f(f^{-1}(x))=x\)
you simply get the argument of the function back as a result.
Let's take the derivative of each side of this equation, with respect to x.
We have some chain rule business going on,
\(\large\rm f'(f^{-1}(x))\cdot \left(f^{-1}\right)'(x)=1\)

If we divide through by f'(f^(-1)(x)),
we see that it establishes an identity for finding derivative of inverse function,\[\large\rm (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}\]

Answer 5

So there's our formula we can use, ummm thinking :d

Answer 6

Yea, its the formula where I get mixed up I think.

Answer 7

Still here, just working through the question again

Answer 8

Hmmm :\

Answer 9

@jim_thompson5910

Answer 10

This is one of those problems where you'll need to use a graphing calculator.

I used geogebra to find that the intersection of
y = 4x+cos(x)
y = 2pi
occurs at (pi/2, 2pi)

So this means that \(\Large f^{-1}(2\pi) = \frac{\pi}{2}\)


Using the formula @zepdrix wrote out, we would get

\[\Large (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}\]
\[\Large (f^{-1})'(2\pi)=\frac{1}{f'(f^{-1}(2\pi))}\]
\[\Large (f^{-1})'(2\pi)=\frac{1}{f'\left(\frac{\pi}{2}\right)}\]
\[\Large (f^{-1})'(2\pi)=\frac{1}{3}\]
So you have the correct answer

Answer 11

hm, ok. there should be a way to do it without though, this is a pretest for calc II and we are not allowed to use calculators for a while at my school. hm

Answer 12

Thank you for your help though, wish i could medal you both

Answer 13

If you can't use a calc, then I guess the next best way is to just guess and check various values on the unit circle. There are only so many to check so that's the good news. Hopefully your teacher wouldn't throw you values that aren't found on the unit circle.

Answer 14

You'll use the chain rule and then substitution for the next problem
\[\Large g(x) = 3*\left(f(x) - 2\right)^2 - \frac{2}{f(x)}\]
\[\Large g'(x) = 6*f'(x)*\left(f(x) - 2\right) + \frac{2}{\left[f(x)\right]^2}*f'(x)\]
\[\Large g'(2) = 6*f'(2)*\left(f(2) - 2\right) + \frac{2}{\left[f(2)\right]^2}*f'(2)\]
\[\Large g'(2) = 6*4*\left(3 - 2\right) + \frac{2}{\left[3\right]^2}*4\]
\[\Large g'(2) = \frac{224}{9}\]
I skipped a bunch of steps, but it's really arithmetic and not calculus that I skipped over