Question

let ( h(x)=5 - x - x^{3} ) and let ( f ) be the inverse function of ( h ).
notice that ( h(-1)=7 ).

( f^{prime}(7)=square )

Explanation:

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

If \(y = h(x)\) and \(f\) is the inverse of \(h\), then \(f^{\prime}(y)=\frac{1}{h^{\prime}(x)}\) where \(y = h(x)\).
Given \(y = 7\) and \(h(-1)=7\), so \(x=-1\).

Step2: Find the derivative of \(h(x)\)

Differentiate \(h(x)=5 - x - x^{3}\) using the power rule.
The power rule states that if \(y = ax^{n}\), then \(y^{\prime}=nax^{n - 1}\).
For \(h(x)\), \(h^{\prime}(x)=\frac{d}{dx}(5)-\frac{d}{dx}(x)-\frac{d}{dx}(x^{3})\).
Since \(\frac{d}{dx}(c)=0\) (where \(c\) is a constant), \(\frac{d}{dx}(x)=1\), and \(\frac{d}{dx}(x^{3}) = 3x^{2}\), we have \(h^{\prime}(x)=-1-3x^{2}\).

Step3: Evaluate \(h^{\prime}(x)\) at \(x = - 1\)

Substitute \(x=-1\) into \(h^{\prime}(x)\):
\(h^{\prime}(-1)=-1-3(-1)^{2}\).
First, calculate \((-1)^{2}=1\), then \(h^{\prime}(-1)=-1 - 3\times1=-4\).

Step4: Use the formula for \(f^{\prime}(y)\)

Since \(f^{\prime}(7)=\frac{1}{h^{\prime}(-1)}\) (because \(h(-1) = 7\)), and \(h^{\prime}(-1)=-4\).

Answer:

\(-\frac{1}{4}\)