Question

answer the following true or false: if ( g(x) ) is the inverse of a differentiable function ( f(x) ) with derivative ( f^{prime}(x)=6+sin left(x^{2}
ight) ), then ( g^{prime}(0)=\frac{1}{6} ). true false

Explanation:

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

If \(g(x)\) is the inverse of \(f(x)\), then \(g^{\prime}(y)=\frac{1}{f^{\prime}(x)}\) where \(y = f(x)\).

Step2: Find \(x\) such that \(f(x)=0\)

We need to find \(x\) for which \(f(x) = 0\). Let's assume \(x = 0\). Then \(f(0)=\int_{0}^{0}(6+\sin(t^{2}))dt=0\) (by the fundamental theorem of calculus, \(\int_{a}^{a}h(t)dt = 0\) for any function \(h(t)\)).

Step3: Calculate \(f^{\prime}(x)\) at \(x = 0\)

Given \(f^{\prime}(x)=6+\sin(x^{2})\), when \(x = 0\), \(f^{\prime}(0)=6+\sin(0)=6\)

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

Since \(y = f(x)=0\) when \(x = 0\), and \(g^{\prime}(y)=\frac{1}{f^{\prime}(x)}\), substituting \(y = 0\) and \(x = 0\) gives \(g^{\prime}(0)=\frac{1}{f^{\prime}(0)}\)

Since \(f^{\prime}(0)=6\), then \(g^{\prime}(0)=\frac{1}{6}\)

Answer:

True