Continuity and DifferentiabilityHard
Question
If g is inverse of f and f(x) = x2 + 3x - 3 (x > 0) then g′(1) equals -
Options
A.
B.- 1
C.1/5
D.
Solution
f-1(x) = g(x) ⇒ x = f(g(x))
Differentiating both sides,
1 = f′(g(x)) g′(x) ⇒ g′(x) =
Now f′(x) = 2x + 3
So g′(x) =
⇒ g′(1) = 
gof(x) = x g′(f(x)) f′(x) = 1
f(x) = 1 at x = 1 & f′(1) = 5
g′(1)f′(1) = 1 ⇒ g′(1) = 1/5
Differentiating both sides,
1 = f′(g(x)) g′(x) ⇒ g′(x) =
Now f′(x) = 2x + 3
So g′(x) =
⇒ g′(1) = gof(x) = x g′(f(x)) f′(x) = 1
f(x) = 1 at x = 1 & f′(1) = 5
g′(1)f′(1) = 1 ⇒ g′(1) = 1/5
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