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Question

Let f(x)=x^3-3 x^2-1, x2x \geq 2 Find the value of df^{-1} / dx at the point x=-1=f(3)

Solution

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Since f(3)=1f(3) = -1, a=3a = 3 and b=1b = -1. Since f(x)=3x26xf'(x) = 3x^2 - 6x, f(3)=3(3)26(3)=9f'(3) = 3(3)^2 - 6(3) = 9 and hence

df1dxx=1=1f(f1(1))=1f(3)=19\left.\dfrac{df^{-1}}{dx}\right|_{x=-1} = \dfrac{1}{f'(f^{-1}(-1))} = \dfrac{1}{f'(3)} = \dfrac{1}{9}

If f1(x)f^{-1}(x) is the inverse function of f(x)f(x) and f(a)=bf(a) = b, then df1/dxdf^{-1}/dx at x=bx = b is

df1dxx=b=1f(f1(b))=1f(a)\left.\dfrac{df^{-1}}{dx}\right|_{x=b} = \dfrac{1}{f'(f^{-1}(b))} = \dfrac{1}{f'(a)}

In other words, (f1)(b)=1f(a)(f^{-1})'(b) = \dfrac{1}{f'(a)} since a=f1(b)a = f^{-1}(b).

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