FunctionHard
Question
Let f be a real-valued function defined on the interval (- 1, 1) such that e-xf(x) = 2 + 
dt, for all x ∈ (-1, 1) and let f-1 be the inverse function of f. Then (f-1)′ (2) is equal to
dt, for all x ∈ (-1, 1) and let f-1 be the inverse function of f. Then (f-1)′ (2) is equal toOptions
A.1
B.1/3
C.1/2
D.1/e
Solution
e-x f(x) = 2 + dt .....(i)
f(f-1(x)) = x
⇒ f′(f-1(x)) (f-1(x))′ = 1 ⇒ (f-1(2))′ =
⇒ f(0) = 2 ⇒ f-1(2) = 0
(f-1(2))′
e-x(f′(x) - f(x)) =
Put x = 0 ⇒ f′(0) - 2 = 1 ⇒ f′(0) = 3
(f-1(2))′ = 1/3.
dt .....(i)f(f-1(x)) = x
⇒ f′(f-1(x)) (f-1(x))′ = 1 ⇒ (f-1(2))′ =
⇒ f(0) = 2 ⇒ f-1(2) = 0(f-1(2))′
e-x(f′(x) - f(x)) =
Put x = 0 ⇒ f′(0) - 2 = 1 ⇒ f′(0) = 3
(f-1(2))′ = 1/3.
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