locate the critical points of the following function. then use the second derivative test to determine whet…

Explanation:

Step1: Find the first - derivative

Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 2x^{2}$ and $v=\ln x$. The derivative of $y = 2x^{2}\ln x-27x^{2}$ is $y^\prime=(2x^{2})^\prime\ln x + 2x^{2}(\ln x)^\prime-(27x^{2})^\prime$. The derivative of $2x^{2}$ is $4x$, the derivative of $\ln x$ is $\frac{1}{x}$, and the derivative of $27x^{2}$ is $54x$. So $y^\prime = 4x\ln x+2x - 54x=4x\ln x - 52x=4x(\ln x - 13)$.

Step2: Find the critical points

Set $y^\prime = 0$. Since $4x(\ln x - 13)=0$, we have two cases: $4x = 0$ or $\ln x-13 = 0$. The domain of $y = 2x^{2}\ln x-27x^{2}$ is $x>0$, so we ignore $x = 0$. Solving $\ln x-13 = 0$, we get $\ln x=13$, then $x = e^{13}$.

Step3: Find the second - derivative

Differentiate $y^\prime = 4x\ln x - 52x$ using the product rule on $4x\ln x$. The derivative of $4x\ln x$ is $4\ln x+4$ and the derivative of $-52x$ is $-52$. So $y^{\prime\prime}=4\ln x + 4-52=4\ln x - 48$.

Step4: Apply the second - derivative test

Evaluate $y^{\prime\prime}$ at $x = e^{13}$. Substitute $x = e^{13}$ into $y^{\prime\prime}$: $y^{\prime\prime}(e^{13})=4\ln(e^{13})-48=4\times13 - 48=52 - 48 = 4>0$. Since $y^{\prime\prime}(e^{13})>0$, the function has a local minimum at $x = e^{13}$ and no local maximum.

Answer:

A. The local minimum/minima of f is/are at $x = e^{13}$. B. There is no local maximum of f.