Question

$f(x) = {x}^{2} + 4x - 5$
when $x > - 2$
find$\frac{d {f}^{ - 1} }{dx} at \: x = 16$
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Answer 1

Answer:

$\dfrac{df^{1}(16)}{dx} = \pm \dfrac{1}{10}$

Step-by-step explanation:

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

First we find the inverse function.

$y = x^2 + 4x - 5$

$x = y^2 + 4y - 5$

$y^2 + 4y - 5 = x$

$y^2 + 4y = x + 5$

$y^2 + 4y + 4 = x + 5 + 4$

$(y + 2)^2 = x + 9$

$y + 2 = \pm\sqrt{x + 9}$

$y = -2 \pm\sqrt{x + 9}$

$f^{-1}(x) = -2 \pm\sqrt{x + 9}$

$f^{-1}(x) = -2 \pm (x + 9)^{\frac{1}{2}}$

Now we find the derivative of the inverse function.

$\dfrac{df^{-1}(x)}{dx} = \pm \dfrac{1}{2}(x + 9)^{-\frac{1}{2}}$

$\dfrac{df^{-1}(x)}{dx} = \pm \dfrac{1}{2\sqrt{x + 9}}$

Now we evaluate the derivative of the inverse function at x = 16.

$\dfrac{df^{-1}(16)}{dx} = \pm \dfrac{1}{2\sqrt{16 + 9}}$

$\dfrac{df^{-1}(16)}{dx} = \pm \dfrac{1}{2\sqrt{25}}$

$\dfrac{df^{-1}(16)}{dx} = \pm \dfrac{1}{2 \times 5 }$

$\dfrac{df^{-1}(16)}{dx} = \pm \dfrac{1}{10}$