Question

suppose f(x) = x^3+2x-3, use (f^-1)'(a) = 1/f'(f^-1(a)) to find (f^-1)'(0).... i need the work please

Answer 1

you aren't given the value of f^-1(0) ?

Answer 2

that is the value to find

Answer 3

confusing

Answer 4

so
f^-1'(0)=1/f'(f^-1(0))
first find f'(x)
f'(x)=3x^2+2
f'(f^-1(0))=3(f^-1(0))^2+2
now
f^-1'(0)=1/3(f^-1(0))^2+2
now we only need to find f^-1(0)
to do that, we need to remember what inverse functions really are
suppose if I have f(x) and f^-1(x). The values of x's of f(x) will be values of y's of f^-1(x) and the values of y's of f(x) will be values of x's of f^-1(x). Or in other words, if (a,b) is in f(x), then (b,a) is in f^-1(x). So the value of f^-1(x) is nothing but the value of x in f(x) if y=0. Now we find the value of x in f(x) if y=0 and that will also be the value of f^-1(0).
f(x)=x^3+2x-3
f(x)=0
x^3+2x-3=0
notice that if I plug one, f(x)=0. So x in f(x) is equal to 1 if f(x)=0 so that will be the value of f^-1(0)=1 as we establish before. plugging this in to f^-1'(0):
f^-1'(0)=1/(3(1)^2+2)=1/(3+2)=1/5 <answer

Answer 5

phew