Tridip C. answered 08/09/20
2+ years of Experience in Tutoring Multivariable Calculus
f(x) = x3 + 7x + 2
∴ f(-47/330) = 1
∴ f-1(1) = -47/330 ——————— (1)
f(x) = x3 + 7x + 2
∴ f'(x) = 3x2 + 7
∴ f'(-47/330) = 256309/36300 ——————— (2)
Let, f-1(x) = g(x)
We know, f(f-1(x)) = x
∴ f(g(x)) = x
d[f(g(x))]
⇒ ----------- = 1
dx
⇒ f'(g(x)) · g'(x) = 1
Putting x = 1, we get
f'(g(1)) · g'(1) = 1
f'(f-1(1)) · (f-1)'(1) = 1
f'(-47/330) · (f-1)'(1) = 1 [from equation (1)]
256309/36300 · (f-1)'(1) = 1 [from equation (2)]
(f-1)'(1) = 36300/256309