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

Explanation:

Step1: Find the domain

For $y = \sqrt{6x}\ln(4x)$, we need $6x\geq0$ and $4x>0$. So the domain is $x > 0$.

Step2: Use product - rule to find the first - derivative

If $y = u\cdot v$ where $u=\sqrt{6x}=(6x)^{\frac{1}{2}}$ and $v = \ln(4x)$. $u^\prime=\frac{1}{2}(6x)^{-\frac{1}{2}}\cdot6=\frac{3}{\sqrt{6x}}$ and $v^\prime=\frac{1}{x}$. By the product - rule $y^\prime=u^\prime v+uv^\prime=\frac{3\ln(4x)}{\sqrt{6x}}+\frac{\sqrt{6x}}{x}=\frac{3\ln(4x)}{\sqrt{6x}}+\frac{\sqrt{6}}{\sqrt{x}}$.

Step3: Use quotient - rule and product - rule to find the second - derivative

Let $y^\prime=\frac{3\ln(4x)}{\sqrt{6x}}+\frac{\sqrt{6}}{\sqrt{x}}$. For $y_1=\frac{3\ln(4x)}{\sqrt{6x}}$, by the quotient - rule $\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = 3\ln(4x)$, $u^\prime=\frac{3}{x}$, $v=\sqrt{6x}$, $v^\prime=\frac{3}{\sqrt{6x}}$. $y_1^\prime=\frac{\frac{3}{x}\cdot\sqrt{6x}-3\ln(4x)\cdot\frac{3}{\sqrt{6x}}}{6x}=\frac{3\sqrt{6x}\cdot\sqrt{6x}-9x\ln(4x)}{6x\sqrt{6x}}=\frac{18x - 9x\ln(4x)}{6x\sqrt{6x}}=\frac{3-\frac{3}{2}\ln(4x)}{\sqrt{6x}}$. For $y_2=\frac{\sqrt{6}}{\sqrt{x}}=\sqrt{6}x^{-\frac{1}{2}}$, $y_2^\prime=-\frac{\sqrt{6}}{2}x^{-\frac{3}{2}}$. $y^{\prime\prime}=y_1^\prime + y_2^\prime=\frac{3-\frac{3}{2}\ln(4x)}{\sqrt{6x}}-\frac{\sqrt{6}}{2x\sqrt{x}}$. Set $y^{\prime\prime}=0$ to find inflection points. $\frac{3-\frac{3}{2}\ln(4x)}{\sqrt{6x}}-\frac{\sqrt{6}}{2x\sqrt{x}} = 0$. First, get a common denominator $2x\sqrt{6x}$. $\frac{2x(3-\frac{3}{2}\ln(4x))- 6}{2x\sqrt{6x}}=0$. $6x - 3x\ln(4x)-6 = 0$. $2x - x\ln(4x)-2 = 0$. Let $t = 4x$, then $x=\frac{t}{4}$, and the equation becomes $\frac{t}{2}-\frac{t}{4}\ln(t)-2 = 0$. $2t - t\ln(t)-8 = 0$. We can also rewrite $y^{\prime\prime}=\frac{3-\frac{3}{2}\ln(4x)}{\sqrt{6x}}-\frac{\sqrt{6}}{2x\sqrt{x}}=\frac{6x - 3x\ln(4x)-6}{2x\sqrt{6x}}$. Set $y^{\prime\prime}=0$, so $6x - 3x\ln(4x)-6 = 0$ or $2x - x\ln(4x)-2 = 0$. Using a graphing utility or numerical methods (e.g., Newton - Raphson method), we find that the root of $2x - x\ln(4x)-2 = 0$ is approximately $x\approx1.35$.

Step4: Test intervals

Choose a test point in the interval $(0,1.35)$, say $x = 1$. $y^{\prime\prime}(1)=\frac{3-\frac{3}{2}\ln(4)}{\sqrt{6}}-\frac{\sqrt{6}}{2}>0$. Choose a test point in the interval $(1.35,\infty)$, say $x = 2$. $y^{\prime\prime}(2)=\frac{3-\frac{3}{2}\ln(8)}{\sqrt{12}}-\frac{\sqrt{6}}{4\sqrt{2}}<0$.

Answer:

A. The function is concave up on $(0, \frac{e^{2}}{4})$ and concave down on $(\frac{e^{2}}{4},\infty)$.