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f(x) = {x}^(2) + 4x - 5 ; >-2

Find
\frac{d {f}^( - 1) }{dx} at x=16​

Please show solving

User Chollier
by
9.2k points

1 Answer

2 votes

The inverse function theorem says


(\mathrm df^(-1))/(\mathrm dx)(16)=\frac1{(\mathrm df)/(\mathrm dx)(f^(-1)(16))}

We have


f(x)=x^2+4x-5

defined on
x>-2, for which we get


f^(-1)(x)=-2+√(x+9)

and


f^(-1)(16)=-2+√(16+9)=3

The derivative of
f(x) is


f'(x)=2x+4

So we end up with


(\mathrm df^(-1))/(\mathrm dx)(16)=\frac1{(\mathrm df)/(\mathrm dx)(3)}=\frac1{10}

User Bharanitharan
by
7.8k points

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