4. a. find the critical number(s) of ( g(x)=\frac{e^{x}}{x + 1} ).\nb. use the first derivative test to find…

Explanation:

Step1: Find the derivative of (g(x))

Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = e^{x}), (u^\prime=e^{x}), (v=x + 1), (v^\prime=1). [ \begin{align*} g^\prime(x)&=\frac{e^{x}(x + 1)-e^{x}\times1}{(x + 1)^{2}}\ &=\frac{e^{x}(x+1 - 1)}{(x + 1)^{2}}\ &=\frac{xe^{x}}{(x + 1)^{2}} \end{align*} ] The domain of (g(x)) is (x\neq - 1). Set (g^\prime(x)=0), since (e^{x}>0) for all (x), then (x = 0) is the critical number.

Step2: First Derivative Test

• Choose test intervals: ((-\infty,-1)), ((-1,0)), ((0,\infty))
• For (x\in(-\infty,-1)), let (x=-2), (g^\prime(-2)=\frac{-2e^{-2}}{(-2 + 1)^{2}}=-2e^{-2}<0)
• For (x\in(-1,0)), let (x =-\frac{1}{2}), (g^\prime(-\frac{1}{2})=\frac{-\frac{1}{2}e^{-\frac{1}{2}}}{(-\frac{1}{2}+1)^{2}}=-2e^{-\frac{1}{2}}<0)
• For (x\in(0,\infty)), let (x = 1), (g^\prime(1)=\frac{1\times e^{1}}{(1 + 1)^{2}}=\frac{e}{4}>0)

Since (g^\prime(x)) changes sign from negative to positive at (x = 0), (g(x)) has a local minimum at (x = 0). (g(0)=\frac{e^{0}}{0 + 1}=1)

Step3: Second Derivative Test

First, find (g^{\prime\prime}(x)) [ \begin{align*} g^{\prime\prime}(x)&=\frac{(e^{x}+xe^{x})(x + 1)^{2}-xe^{x}\times2(x + 1)}{(x + 1)^{4}}\ &=\frac{e^{x}(x + 1)[(x + 1)+x]-2xe^{x}(x + 1)}{(x + 1)^{4}}\ &=\frac{e^{x}(x^{2}+x+1)}{(x + 1)^{3}} \end{align*} ] Evaluate (g^{\prime\prime}(0)=\frac{e^{0}(0^{2}+0 + 1)}{(0 + 1)^{3}}=1>0)

Since (g^{\prime\prime}(0)>0), (g(x)) has a local minimum at (x = 0) and (g(0)=1)

Answer:

A. The critical number is (x = 0) B. Local minimum at ((0,1)) C. Local minimum at ((0,1)) D. The second - derivative test is more straightforward when the second - derivative is not too complex to calculate.