What is the value of (f^-1)'(x) when x=3?

Question

F(x)=(1/4)x^3 + x -1

What is the value of (f^-1)'(x) when x=3?

Answer 1

f(x)= x^3/4 +x-1=y

switch x and y

x=y^3/4+y-1

y=f^-1(x)

y(3)= f^-1(3)=y when y^3/4 +y-1=3

.25y^3 +y-4=0

y^3+4y-16=0

y=2 =f^-1(3)

2^3+4(2)-16=0

8+8-16=0

16-16=0

but you want f'^-1(3), not f^-1(3)

hard to make out some small differences on some computer screens

possible small clue anyway

here's another attempt

y^3/4+y-1=x

take the derivative with respect to x, y'=dy/dx

and maybe y=2 when x=3, from above, so

(3/4)y^2(y')+y'= 1

y'(x)=1/(3y^2/4 +1)

y'(3)=1/(3(2^2)/4 +1)=1/(3+1)=1/4

and, for whatever it's worth, you can factor y^3+4y-16

= (y-2)(y^2 +2y+8)

2 is the only real root,

the other 2 roots are imaginary: -1+/-isqr7