Given that f(x)=2x+cos(x) is one-to-one, use the formula

Question

(f^−1)′(x)=1 / f′(f^−1(x))

to find (f^−1)′(1).

(f^−1)′(1)=

Answer 1

Let g(x) = f^-1(x)
f(0) = 1
So, g'(1) = 1/f'(0) = 1/(-sin(1))

Answer 2

Good

Answer 3

f(x) = 2x + cos x
Hence, f ' ( x ) = 2 - sin x ----- (2)

suppose f ' (x) = g (x) ----- (1)
Then,
Let g(1) = w
==> 1 = f (w) (Cause (1) )
==> 1 = 2w + cos w
==> Solving this equation we get w = 0 and so w = g(1) = f ' (1) = 0

Using the formula,
(f^−1)′(1)= 1 / f′(f^−1(x))
= 1/ f' (g(1))
= 1/f'(0)
= 1/ 2 (Using equation 2)