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

locate the critical points of the following function. then use the second derivative test to determine whether they corre f(x)=2x² in x - 27x² a. the critical point(s) is(are) x = e¹³. (use a comma to separate answers as needed. type an exact answer in terms of e.) b. there are no critical points for f. what is/are the local minimum/minima of f? select the correct choice below and, if necessary, fill in the answer box to comp a. the local minimum/minima of f is/are at x = (use a comma to separate answers as needed. type an exact answer in terms of e.) b. there is no local minimum of f.

locate the critical points of the following function. then use the second derivative test to determine whether they corre f(x)=2x² in x - 27x² a. the critical point(s) is(are) x = e¹³. (use a comma to separate answers as needed. type an exact answer in terms of e.) b. there are no critical points for f. what is/are the local minimum/minima of f? select the correct choice below and, if necessary, fill in the answer box to comp a. the local minimum/minima of f is/are at x = (use a comma to separate answers as needed. type an exact answer in terms of e.) b. there is no local minimum of f.

Answer

Explanation:

Step1: Find the first - derivative

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 2x^{2}$ and $v=\ln x$ and the power - rule $(x^{n})^\prime=nx^{n - 1}$. The derivative of $y = 2x^{2}\ln x-27x^{2}$ is $y^\prime=(2x^{2}\cdot\frac{1}{x}+4x\ln x)-54x=2x + 4x\ln x-54x=4x\ln x-52x=4x(\ln x - 13)$. Set $y^\prime = 0$. Since $x>0$ (because $\ln x$ is defined for $x>0$), we solve $\ln x-13 = 0$. So $\ln x=13$ and $x = e^{13}$.

Step2: Find the second - derivative

Differentiate $y^\prime=4x\ln x-52x$ using the product rule on $4x\ln x$. $y^{\prime\prime}=4(\ln x + 1)-52=4\ln x-48$.

Step3: Apply the second - derivative test

Evaluate $y^{\prime\prime}$ at $x = e^{13}$. $y^{\prime\prime}(e^{13})=4\ln(e^{13})-48=4\times13 - 48=52 - 48 = 4>0$. Since $y^{\prime\prime}(e^{13})>0$, the function has a local minimum at $x = e^{13}$.

Answer:

The critical point(s) is(are) $x = e^{13}$. The local minimum/minima of $f$ is/are at $x = e^{13}$.