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

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}})