Question

let f be the function defined by f(x)=2x + e^x. if g(x)=f^(-1)(x) for all x and the point (0,1) is on the graph of f, what is the value of g(1)? a) 1/(2 + e) b) 1/3 c) 1/2 d) 3 e) 2 + e

Explanation:

Step1: Find the derivative of \(f(x)\)

Given \(f(x)=2x + e^{x}\), by the sum - rule and basic derivative rules, \(f^\prime(x)=\frac{d}{dx}(2x)+\frac{d}{dx}(e^{x})=2 + e^{x}\).

Step2: Use the formula for the derivative of an inverse function

The formula for the derivative of the inverse function \(g = f^{-1}(x)\) is \(g^\prime(x)=\frac{1}{f^\prime(g(x))}\). We want to find \(g^\prime(1)\). Since the point \((0,1)\) is on the graph of \(f\) (i.e., \(f(0)=2\times0 + e^{0}=1\)), then \(g(1) = 0\).

Step3: Evaluate \(f^\prime\) at \(g(1)\)

Substitute \(x = g(1)=0\) into \(f^\prime(x)\). We get \(f^\prime(0)=2+e^{0}=2 + 1=3\).

Step4: Calculate \(g^\prime(1)\)

Since \(g^\prime(1)=\frac{1}{f^\prime(g(1))}\) and \(f^\prime(g(1))=f^\prime(0) = 3\), then \(g^\prime(1)=\frac{1}{3}\).

Answer:

B. \(\frac{1}{3}\)