Question

attempt 1: 2 attempts remaining. calculate ( g(b) ) and ( g(b) ), where ( g ) is the inverse of ( f = x+sin(x) ) where ( b = 0 ). ( g(b)= ) ( g(b)= ) submit answer next item

Explanation:

Step1: Recall inverse - function derivative formula

If \(y = g(x)\) is the inverse of \(y = f(x)\), then \(g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))}\). First, find the derivative of \(f(x)=x+\sin(x)\). Using the sum - rule and the derivative of \(\sin(x)\), we have \(f^{\prime}(x)=1 + \cos(x)\).

Step2: Use the fact that if \(b = f(a)\) (since \(g\) is the inverse of \(f\), \(g(b)=a\)), then \(g^{\prime}(b)=\frac{1}{f^{\prime}(a)}\)

We know that \(b = 0\). We need to find \(a\) such that \(f(a)=0\), i.e., \(a+\sin(a)=0\). The solution of the equation \(a+\sin(a)=0\) is \(a = 0\) (because when \(a = 0\), \(0+\sin(0)=0\)).

Step3: Calculate \(f^{\prime}(a)\)

Since \(a = 0\) and \(f^{\prime}(x)=1+\cos(x)\), then \(f^{\prime}(0)=1+\cos(0)=1 + 1=2\). So \(g^{\prime}(0)=\frac{1}{f^{\prime}(0)}=\frac{1}{2}\).

Step4: Calculate \(g(0)\)

Since \(g\) is the inverse of \(f\) and \(f(0)=0\), then \(g(0)=0\).

Answer:

\(g(0)=0\), \(g^{\prime}(0)=\frac{1}{2}\)