Calculus 1 Problem

Question

Find (f^-1)(a)

F(x)= x - 6/x , x<0 , a=1

Find (f^-1)(a)

F(x)=tan^-1(x) + 8x^2, a=0

Answer 1

here is how I proceed,

x=y^-1-6/y^-1

sets up quadradic in y^-1

and since y^-1<0 requirement (same as x<0 in original function)

y^-1=f^-1(x)=(x/2)-(√(x²+24))/2

at x=1,,,,,,y^-1= -2

using same procedure,

x=tan^-1(y^-1)+8*(y^-1)²

at x=0 must find when tan^-1(y^-1)+8*(y^-1)²=0

graphing and by observation y^-1=0

it is kind of neat though

I can compute d(f^-1(x))dx for any given value of f^-1(x)

and at f^-1(x)=0 [also x=0] d(f^-1(x))/dx=1

which by looking at graph would assume it is zero.