- /1 points details my notes tanapcalc10 4.2.068.mi. find the relative extrema, if any, of the function. use…

Explanation:

Step1: Find the first - derivative

Rewrite $f(t)=5t + 3t^{-1}$. Then $f^\prime(t)=5-3t^{-2}=5-\frac{3}{t^{2}}$. Set $f^\prime(t) = 0$, so $5-\frac{3}{t^{2}}=0$. Then $\frac{3}{t^{2}}=5$, and $t^{2}=\frac{3}{5}$, which gives $t=\pm\sqrt{\frac{3}{5}}$.

Step2: Find the second - derivative

Differentiate $f^\prime(t)=5 - 3t^{-2}$ with respect to $t$. $f^{\prime\prime}(t)=6t^{-3}=\frac{6}{t^{3}}$.

Step3: Apply the second - derivative test

When $t = \sqrt{\frac{3}{5}}$, $f^{\prime\prime}(\sqrt{\frac{3}{5}})=\frac{6}{(\sqrt{\frac{3}{5}})^{3}}>0$. So $f(t)$ has a relative minimum at $t=\sqrt{\frac{3}{5}}$. $f(\sqrt{\frac{3}{5}})=5\sqrt{\frac{3}{5}}+\frac{3}{\sqrt{\frac{3}{5}}}=\sqrt{15}+\sqrt{15}=2\sqrt{15}$. When $t=-\sqrt{\frac{3}{5}}$, $f^{\prime\prime}(-\sqrt{\frac{3}{5}})=\frac{6}{(-\sqrt{\frac{3}{5}})^{3}}<0$. So $f(t)$ has a relative maximum at $t = -\sqrt{\frac{3}{5}}$. $f(-\sqrt{\frac{3}{5}})=-5\sqrt{\frac{3}{5}}-\frac{3}{\sqrt{\frac{3}{5}}}=-\sqrt{15}-\sqrt{15}=-2\sqrt{15}$.

Answer:

relative maximum $(t,f)=\left(-\sqrt{\frac{3}{5}},-2\sqrt{15}\right)$ relative minimum $(t,f)=\left(\sqrt{\frac{3}{5}},2\sqrt{15}\right)$