Question

Really having some trouble with this

Answer 1

Hint: Through trial and error, you should find that f(pi/2) = 5pi/2

Answer 2

I used input values of angles found on the unit circle (0, pi/6, pi/4, etc)

Answer 3

Another hint:
\[\Large f\ ' (x) = 5-\sin(x)\]

Answer 4

Sorry jim, just saw this

Answer 5

that's fine

Answer 6

Do you mean its co-terminal?

Answer 7

no I mean if you plug in x = pi/2, you get the output 5pi/2

Answer 8

Ah ha, gotcha

Answer 9

you'll have to think in reverse

if the input into the inverse is 5pi/2, then the output must be pi/2

Answer 10

right.

So I need to solve \[5-\sin(\frac{ \pi }{ 2 })\]

Answer 11

which is 4

Answer 12

But it will be 1/4 because of that formula thing

Answer 13

correct
\[\Large (f^{-1})'(x)=\frac{1}{f \ '(f^{-1}(x))}\]
\[\Large (f^{-1})'\left(\frac{5\pi}{2}\right)=\frac{1}{f \ ' \left(f^{-1}\left(\frac{5\pi}{2}\right)\right)}\]
\[\Large (f^{-1})'\left(\frac{5\pi}{2}\right)=\frac{1}{f \ '\left(\frac{\pi}{2}\right)}\]
\[\Large (f^{-1})'\left(\frac{5\pi}{2}\right)=\frac{1}{4}\]

Answer 14

Beautiful, thank you

Answer 15

no problem