Question

given the function f, find the slope of the line tangent to the graph of ( f^{-1} ) at the specified point on the graph of ( f^{-1} ).
( f(x)=sqrt{5 x},left(4, \frac{16}{5}
ight) )
the slope of the tangent line to the graph ( f^{-1} ) is

Explanation:

Step1: Find the derivative of \(f(x)\)

Given \(f(x)=\sqrt{5x}=(5x)^{\frac{1}{2}}\).
Using the power rule \((u^n)^\prime = nu^{n - 1}u^\prime\), where \(u = 5x\), \(n=\frac{1}{2}\), and \(u^\prime=5\).
\(f^\prime(x)=\frac{1}{2}(5x)^{-\frac{1}{2}}\times5=\frac{5}{2\sqrt{5x}}\).

Step2: Use the formula for the derivative of the inverse function

The formula is \((f^{-1})^\prime(a)=\frac{1}{f^\prime(f^{-1}(a))}\).
We are given the point \((a,b)=(4,\frac{16}{5})\) on the graph of \(f^{-1}\), so \(f^{-1}(4)=\frac{16}{5}\).
First, find \(f^\prime(f^{-1}(4))\): substitute \(x = \frac{16}{5}\) into \(f^\prime(x)\)
\(f^\prime(\frac{16}{5})=\frac{5}{2\sqrt{5\times\frac{16}{5}}}=\frac{5}{2\sqrt{16}}=\frac{5}{2\times4}=\frac{5}{8}\).
Then, \((f^{-1})^\prime(4)=\frac{1}{f^\prime(f^{-1}(4))}\).

Answer:

\(\frac{8}{5}\)