math 104 3.6 absolute extrema question 8. consider the function ( g(x)=4x+\frac{1}{x} ). (a) find the…

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). [ \begin{align*} 4-\frac{1}{x^{2}}&=0\ \frac{1}{x^{2}}&=4\ x^{2}&=\frac{1}{4}\ x&=\pm\frac{1}{2} \end{align*} ]

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\rightarrow0^{+}), (g(x)\rightarrow\infty) and as (x\rightarrow\infty), (g(x)\rightarrow\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\rightarrow0^{-}), (g(x)\rightarrow-\infty) and as (x\rightarrow-\infty), (g(x)\rightarrow-\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