Question

an invertible function $f(x)$ is given along with a point that lies on its graph. using theorem 2.7.7, evaluate $(f^{-1})(x)$ at the indicated value.
the point $(3,46)$ is on the graph of $f(x)=x^{2}+10x + 7,xgeq - 5$. find $(f^{-1})(46)$.

Explanation:

Step1: Use the formula for the derivative of an inverse function

If \(y = f(x)\) is invertible, then \((f^{-1})'(y)=\frac{1}{f'(x)}\), where \(y = f(x)\).
Given \(f(x)=x^{2}+10x + 7,x\geq - 5\). First, find the derivative of \(f(x)\) using the power rule. The power rule states that if \(y=x^{n}\), then \(y'=nx^{n - 1}\).
For \(f(x)=x^{2}+10x + 7\), \(f'(x)=\frac{d}{dx}(x^{2})+\frac{d}{dx}(10x)+\frac{d}{dx}(7)\).
Since \(\frac{d}{dx}(x^{2}) = 2x\), \(\frac{d}{dx}(10x)=10\) and \(\frac{d}{dx}(7) = 0\), we have \(f'(x)=2x + 10\).

Step2: Find the \(x\) - value corresponding to \(y = 46\)

We know that if \(y = f(x)\) and \(y = 46\), then \(x^{2}+10x+7 = 46\).
Rearrange the equation to \(x^{2}+10x-39 = 0\).
Factor the quadratic equation: \(x^{2}+10x - 39=(x + 13)(x - 3)=0\).
Since \(x\geq - 5\), we reject \(x=-13\) and take \(x = 3\).

Step3: Evaluate \(f'(x)\) at \(x = 3\)

Substitute \(x = 3\) into \(f'(x)=2x + 10\).
\(f'(3)=2\times3+10=6 + 10=16\).

Step4: Use the formula for \((f^{-1})'(y)\)

Since \((f^{-1})'(y)=\frac{1}{f'(x)}\) and \(y = 46,x = 3,f'(3)=16\)
\((f^{-1})'(46)=\frac{1}{f'(3)}\)

Answer:

\(\frac{1}{16}\)