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 $u = \sqrt{6x}= \sqrt{6}x^{\frac{1}{2}}$ and $v=\ln(4x)$. $u^\prime=\frac{\sqrt{6}}{2}x^{-\frac{1}{2}}$ and $v^\prime=\frac{1}{x}$. $f^\prime(x)=\frac{\sqrt{6}}{2}x^{-\frac{1}{2}}\ln(4x)+\sqrt{6}x^{\frac{1}{2}}\cdot\frac{1}{x}=\frac{\sqrt{6}}{2}x^{-\frac{1}{2}}\ln(4x)+\sqrt{6}x^{-\frac{1}{2}}=\frac{\sqrt{6}}{2}x^{-\frac{1}{2}}(\ln(4x) + 2)$.

Step2: Find the second - derivative

Use the product rule again. Let $u=\frac{\sqrt{6}}{2}x^{-\frac{1}{2}}$ and $v=\ln(4x)+2$. $u^\prime=-\frac{\sqrt{6}}{4}x^{-\frac{3}{2}}$ and $v^\prime=\frac{1}{x}$. $f^{\prime\prime}(x)=-\frac{\sqrt{6}}{4}x^{-\frac{3}{2}}(\ln(4x)+2)+\frac{\sqrt{6}}{2}x^{-\frac{1}{2}}\cdot\frac{1}{x}=-\frac{\sqrt{6}}{4}x^{-\frac{3}{2}}(\ln(4x)+2)+\frac{\sqrt{6}}{2}x^{-\frac{3}{2}}=\frac{\sqrt{6}}{4}x^{-\frac{3}{2}}(2 - (\ln(4x)+2))=-\frac{\sqrt{6}}{4}x^{-\frac{3}{2}}\ln(4x)$.

Step3: Find the critical points of $f^{\prime\prime}(x)$

Set $f^{\prime\prime}(x) = 0$. $-\frac{\sqrt{6}}{4}x^{-\frac{3}{2}}\ln(4x)=0$. Since $x^{-\frac{3}{2}}\neq0$ for $x>0$ (the domain of $y = \sqrt{6x}\ln(4x)$ is $x>0$), then $\ln(4x)=0$. $\ln(4x)=0\Rightarrow4x = 1\Rightarrow x=\frac{1}{4}$.

Step4: Test the intervals

Choose a test - point in the interval $(0,\frac{1}{4})$, say $x=\frac{1}{8}$. $f^{\prime\prime}(\frac{1}{8})=-\frac{\sqrt{6}}{4}(\frac{1}{8})^{-\frac{3}{2}}\ln(\frac{4}{8})=-\frac{\sqrt{6}}{4}(8)^{\frac{3}{2}}\ln(\frac{1}{2})>0$. Choose a test - point in the interval $(\frac{1}{4},\infty)$, say $x = 1$. $f^{\prime\prime}(1)=-\frac{\sqrt{6}}{4}(1)^{-\frac{3}{2}}\ln(4)<0$.

Answer:

A. The function is concave up on $(0,\frac{1}{4})$ and concave down on $(\frac{1}{4},\infty)$.