for f(x)=2√(x - 1)(x - 4)+1, use the second - derivative test to determine the local minimum of f. as your…

Explanation:

Step1: Expand the function

First, expand $f(x)=2\sqrt{x - 1}(x - 4)+1=2(x - 1)^{\frac{1}{2}}(x - 4)+1=2(x^{\frac{3}{2}}-4x^{\frac{1}{2}}-x^{\frac{1}{2}} + 4)+1=2x^{\frac{3}{2}}-10x^{\frac{1}{2}}+9$.

Step2: Find the first - derivative

Using the power rule $(x^n)^\prime=nx^{n - 1}$, we have $f^\prime(x)=2\times\frac{3}{2}x^{\frac{1}{2}}-10\times\frac{1}{2}x^{-\frac{1}{2}}=3\sqrt{x}-\frac{5}{\sqrt{x}}=\frac{3x - 5}{\sqrt{x}}$, for $x>1$.

Step3: Find the critical points

Set $f^\prime(x)=0$, then $\frac{3x - 5}{\sqrt{x}}=0$. Since $\sqrt{x}>0$ for $x > 1$, we solve $3x-5 = 0$, getting $x=\frac{5}{3}$.

Step4: Find the second - derivative

$f^{\prime\prime}(x)=\frac{3\sqrt{x}-\frac{3x - 5}{2\sqrt{x}}}{x}=\frac{6x-(3x - 5)}{2x\sqrt{x}}=\frac{3x + 5}{2x\sqrt{x}}$.

Step5: Evaluate the second - derivative at the critical point

Substitute $x = \frac{5}{3}$ into $f^{\prime\prime}(x)$: $f^{\prime\prime}(\frac{5}{3})=\frac{3\times\frac{5}{3}+5}{2\times\frac{5}{3}\sqrt{\frac{5}{3}}}=\frac{10}{\frac{10}{3}\sqrt{\frac{5}{3}}}>0$. So $x=\frac{5}{3}$ is a local minimum.

Step6: Find the value of the function at the local minimum

Substitute $x=\frac{5}{3}$ into $f(x)$: $f(\frac{5}{3})=2\sqrt{\frac{5}{3}-1}(\frac{5}{3}-4)+1=2\sqrt{\frac{2}{3}}(-\frac{7}{3})+1=1-\frac{14\sqrt{6}}{9}$.

Answer:

$1-\frac{14\sqrt{6}}{9}$