Question

for the function $y = f(x) = x^{2}-x + 6,xgeq0.5$, find $\frac{df^{-1}}{dy}\big|_{y = 11}$
$(f^{-1})(11)=$

Explanation:

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

If \(y = f(x)\) is a one - to - one differentiable function with inverse function \(x = f^{-1}(y)\), then \((f^{-1})'(y)=\frac{1}{f'(x)}\), where \(y = f(x)\).

Step2: First, find \(f'(x)\)

Given \(f(x)=x^{2}-x + 6\), using the power rule \((x^n)'=nx^{n - 1}\), we have \(f'(x)=2x-1\).

Step3: Then, find \(x\) such that \(f(x)=11\)

Set \(x^{2}-x + 6=11\), which simplifies to \(x^{2}-x - 5=0\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=-1\), \(c=-5\)), we get \(x=\frac{1\pm\sqrt{1+20}}{2}=\frac{1\pm\sqrt{21}}{2}\). Since \(x\geq0.5\), we take \(x=\frac{1 + \sqrt{21}}{2}\).

Step4: Calculate \((f^{-1})'(11)\)

By the formula \((f^{-1})'(11)=\frac{1}{f'(x)}\) with \(f(x)=11\). Substitute \(x=\frac{1+\sqrt{21}}{2}\) into \(f'(x)\): \(f'(\frac{1+\sqrt{21}}{2})=2\times\frac{1+\sqrt{21}}{2}-1=1+\sqrt{21}-1=\sqrt{21}\). So \((f^{-1})'(11)=\frac{1}{\sqrt{21}}\).

Answer:

\(\frac{1}{\sqrt{21}}\)