Question

math 104 3.6 absolute extrema question 8. consider the function ( g(x)=4x+\frac{1}{x} ). (a) find the derivative of ( g(x) ). (b) find the critical values of ( g(x) ). (c) find the absolute extrema of ( g(x) ) on the interval ( (0,infty) ). (d) find the absolute extrema of ( g(x) ) on the interval ( (-infty,0) ).

Explanation:

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

Using the power rule \((x^n)^\prime = nx^{n - 1}\), for \(g(x)=4x+\frac{1}{x}=4x + x^{-1}\), then \(g^\prime(x)=(4x)^\prime+(x^{-1})^\prime\).
\(g^\prime(x)=4- \frac{1}{x^{2}}\)

Step2: Find the critical values

Set \(g^\prime(x) = 0\), so \(4-\frac{1}{x^{2}}=0\).

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Step3: Analyze the sign of \(g^\prime(x)\) for \(x\in(0,\infty)\)

Let's choose a test - point in \((0,\frac{1}{2})\), say \(x = \frac{1}{4}\). Then \(g^\prime(\frac{1}{4})=4-\frac{1}{(\frac{1}{4})^{2}}=4 - 16=- 12<0\).
Choose a test - point in \((\frac{1}{2},\infty)\), say \(x = 1\). Then \(g^\prime(1)=4-\frac{1}{1^{2}}=3>0\).
So \(g(x)\) has a local minimum at \(x=\frac{1}{2}\). \(g(\frac{1}{2})=4\times\frac{1}{2}+\frac{1}{\frac{1}{2}}=2 + 2=4\).
As \(x
ightarrow0^{+}\), \(g(x)
ightarrow\infty\) and as \(x
ightarrow\infty\), \(g(x)
ightarrow\infty\). So the absolute minimum of \(g(x)\) on \((0,\infty)\) is \(4\) at \(x = \frac{1}{2}\) and there is no absolute maximum.

Step4: Analyze the sign of \(g^\prime(x)\) for \(x\in(-\infty,0)\)

Let's choose a test - point in \((-\infty,-\frac{1}{2})\), say \(x=-1\). Then \(g^\prime(-1)=4-\frac{1}{(-1)^{2}}=3>0\).
Choose a test - point in \((-\frac{1}{2},0)\), say \(x =-\frac{1}{4}\). Then \(g^\prime(-\frac{1}{4})=4-\frac{1}{(-\frac{1}{4})^{2}}=4 - 16=-12<0\).
So \(g(x)\) has a local maximum at \(x =-\frac{1}{2}\). \(g(-\frac{1}{2})=4\times(-\frac{1}{2})+\frac{1}{-\frac{1}{2}}=-2-2=-4\).
As \(x
ightarrow0^{-}\), \(g(x)
ightarrow-\infty\) and as \(x
ightarrow-\infty\), \(g(x)
ightarrow-\infty\). So the absolute maximum of \(g(x)\) on \((-\infty,0)\) is \(-4\) at \(x=-\frac{1}{2}\) and there is no absolute minimum.

Answer:

(a) \(g^\prime(x)=4-\frac{1}{x^{2}}\)
(b) Critical values \(x = \pm\frac{1}{2}\)
(c) Absolute minimum is \(4\) at \(x=\frac{1}{2}\), no absolute maximum
(d) Absolute maximum is \(-4\) at \(x =-\frac{1}{2}\), no absolute minimum