Question

Find (f^-1)'(a):
f(x) = 3+x^2+tan(pi(x)/2),
-1 13 years ago

Answer 1

\[(f^{-1})'(a)=\frac{ 1 }{f'(f ^{-1}(a)) }\]

Answer 2

f(a);
\[3=3+x ^{2}+\tan(\frac{ pix }{2 })\]

Answer 3

\[0=x ^{2}+\tan(\frac{ pix }{ 2 })\]

Answer 4

stuck

Answer 5

If I am even on the right path.

Answer 6

I'm a little confused how you took the derivative.

Answer 7

Actually one sec.

Answer 8

Did you first take the inverse and then the derivative?

Answer 9

Have not done the derivative yet:
doing:
\[(f ^{-1})(x) = y <-> f(y) = x\]

Answer 10

so if I place 3 in for my f(x), I should find out what my \[(f ^{-1})\] is correct?

Answer 11

After taking the derivative.

Answer 12

so
f'(x) = 2x+sec^2(pix/2)?

Answer 13

poop did I do an integral there with the sec^2?

Answer 14

\[f'(x) = 2x+\frac{ 1 }{ 2 }\pi[\sec ^{2}(\frac{ x \pi }{ 2 })]\]

Answer 15

then .....
3 = the lovely right side above?

Answer 16

@jim_thompson5910
If you have time.

Answer 17

Hello

Answer 18

these inverses are going to be the death of me.

Answer 19

Was either path an ok way to go? Me initially and then what z suggested.

Answer 20

hint: what is f(0)

Answer 21

f(0) = 3

Answer 22

take the inverse of both sides

f(0) = 3

f^-1[f(0)] = f^-1(3)

0 = f^-1(3)

f^-1(3) = 0

Answer 23

do you see what to do now?

Answer 24

man my head is spinning.

Answer 25

not sure.
put 0 into the f'(x) and solve?

Answer 26

If I did the derivative correct.

Answer 27

You are correct

(f^-1)'(a) = 1/[f ' ( f^-1(a) )]

(f^-1)'(3) = 1/[f ' ( f^-1(3) )]

(f^-1)'(3) = 1/[f ' ( 0 )]

(f^-1)'(3) = ???

Answer 28

\[f'(3) = 2(3)+\frac{ 1 }{ 2 }\pi[\sec ^{2}(\frac{ 3 \pi }{ 2 })]\]

Answer 29

or am i missing yet another simple step and am killing myself with the algebra?

Answer 30

you want f'(0), not f'(3)

Answer 31

\[0+\frac{ 1 }{ 2 }\pi(1)\]
\[=\frac{ \pi }{ 2 }\]

Answer 32

suppose i should do a couple hundred and then it may start to get easier.

Answer 33

I miss college algebra.

Answer 34

Thank you both for your help!

Answer 35

np