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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The domain of \(g(x)\) is \(x
eq - 1\). 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 \(g^{\prime\prime}(x)\)

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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.