an invertible function $f(x)$ is given along with a point that lies on its graph. using theorem 2.7.7…

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.\nthe point $(3,46)$ is on the graph of $f(x)=x^{2}+10x + 7,xgeq - 5$. find $(f^{-1})(46)$.

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.\nthe point $(3,46)$ is on the graph of $f(x)=x^{2}+10x + 7,xgeq - 5$. find $(f^{-1})(46)$.

Answer

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})