determine the intervals on which the following function is concave up or concave down. identify any…

Explanation:

Step1: Find the first - derivative

Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u =-\sqrt{2x}=-(2x)^{\frac{1}{2}}$ and $v=\ln(2x)$. $u^\prime=-\frac{1}{2}(2x)^{-\frac{1}{2}}\times2=-\frac{1}{\sqrt{2x}}$, $v^\prime=\frac{2}{2x}=\frac{1}{x}$. $f^\prime(x)=-\frac{\ln(2x)}{\sqrt{2x}}-\frac{1}{\sqrt{2x}}=-\frac{\ln(2x) + 1}{\sqrt{2x}}$.

Step2: Find the second - derivative

Use the quotient rule $(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u=-(\ln(2x)+1)$ and $v = \sqrt{2x}=(2x)^{\frac{1}{2}}$. $u^\prime=-\frac{2}{2x}=-\frac{1}{x}$, $v^\prime=\frac{1}{2}(2x)^{-\frac{1}{2}}\times2=\frac{1}{\sqrt{2x}}$. $f^{\prime\prime}(x)=\frac{\frac{1}{x}\sqrt{2x}+\frac{\ln(2x)+1}{\sqrt{2x}}}{2x}=\frac{\frac{2 + \ln(2x)+1}{ \sqrt{2x}}}{2x}=\frac{\ln(2x)+3}{2x\sqrt{2x}}$.

Step3: Find the inflection points

Set $f^{\prime\prime}(x) = 0$, then $\ln(2x)+3 = 0$. $\ln(2x)=-3$, so $2x=e^{-3}$, and $x=\frac{e^{-3}}{2}$. The domain of $f(x)$ is $x>0$.

Step4: Test the intervals

Choose a test - point in the interval $(0,\frac{e^{-3}}{2})$, say $x=\frac{1}{2e^{4}}$. $f^{\prime\prime}(\frac{1}{2e^{4}})=\frac{\ln(2\times\frac{1}{2e^{4}})+3}{2\times\frac{1}{2e^{4}}\sqrt{2\times\frac{1}{2e^{4}}}}=\frac{\ln(e^{-4}) + 3}{e^{-4}\times e^{-2}}=\frac{-4 + 3}{e^{-6}}=-e^{6}<0$. Choose a test - point in the interval $(\frac{e^{-3}}{2},\infty)$, say $x = 1$. $f^{\prime\prime}(1)=\frac{\ln(2)+3}{2\sqrt{2}}>0$.

Answer:

A. The function is concave down on $(0,\frac{e^{-3}}{2})$ and concave up on $(\frac{e^{-3}}{2},\infty)$