f(x)=x³+3sinx+2cosx, a=2 ,find (f－¹）′（a) ?∫(cosx/2+sinx) dx=? evaluate the integral.
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(a) Let f^-1(x) = y then f(y) = x
y^3 + 3 sin y + 2 cos y = x
y^3 + 3 sin y + 2 cos y = 2
Upon solving, we get y = 0
Therefore, f^-1(2) = 0
(b) Let 2 + sin x = u then upon differentiation, cos x dx = du
The integral becomes ∫ du/u
= ln u + C
= ln (2 + sin x) + C
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Hey, Did you mean you wanted the DERIVATIVE of the inverse function at x = 2? In that case, the answer will be different.guru200941080.0773047801
Answer for Q1 if the question meant to ask derivative of [f^-1(2)] is here ...
Let f^-1 (x) = g
From the Inverse Function Theorem, we have
g'(a) = 1/[f'[g(a)]
So g'(2) = 1/[f'[g(2)]
Let g(2) = b
Since g is the inverse of f, g(2) = b means f(b) = 2
b^3 + 3 sin b + 2 cos b = 2
b = 0 satisfies the above equation, therefore b = 0
So g(2) = 0
g'(2) = 1/[f'(0)]
= 1/[3 * 0^2 + 3 cos 0 - 2 sin 0]
= 1/3