Question

1. a. find the critical number(s) of ( g(x)=\frac{e^{x}}{x + 1} ).

b. use the first derivative test to find any local min/max of ( g(x) ).
c. use the second derivative test to find any local min/max of ( g(x) ).
d. which test did you prefer to use?

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 \(u = e^{x}\), \(u^\prime=e^{x}\), \(v=x + 1\), \(v^\prime = 1\).

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Set \(g^\prime(x)=0\), since \(e^{x}>0\) for all \(x\in R\) and \((x + 1)^{2}>0\) for \(x
eq - 1\), then \(x = 0\) is a critical number. Also, \(x=-1\) makes \(g(x)\) undefined, so it is not in the domain of \(g(x)\) and not a critical number.

Step2: First - Derivative Test

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

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

Step3: Second - Derivative Test

First, find the second derivative. Using the quotient rule again on \(g^\prime(x)=\frac{xe^{x}}{(x + 1)^{2}}\)

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Evaluate \(g^{\prime\prime}(0)=\frac{e^{0}(1-0)}{(0 + 1)^{3}}=1>0\)

By the Second - Derivative Test, since \(g^\prime(0) = 0\) and \(g^{\prime\prime}(0)>0\), \(g(x)\) has a local minimum at \(x = 0\) with \(g(0)=1\)

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 in this case as the second - derivative calculation is manageable and gives a quick result (comparing to analyzing multiple intervals for the First - Derivative Test). But it depends on personal preference and the complexity of the function.