question 8 of 13 > consider the series. $sum_{n = 1}^{infty}\frac{7n}{(n^{2}+1)^{\frac{3}{5}}}$ let…

question 8 of 13 > consider the series. $sum_{n = 1}^{infty}\frac{7n}{(n^{2}+1)^{\frac{3}{5}}}$ let $f(x)=\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}$. find $f(x)$. (express numbers in exact form. use symbolic notation and fractions where needed.) $f(x)=$ evaluate $int_{3}^{infty}\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}dx$. (express numbers in exact form. use symbolic notation and fractions where needed.) $int_{3}^{infty}\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}dx=$ identify the true statement(s) about the series and the integral. the integral test does not apply because $f$ is not continuous on $(1,infty)$. the integral test does not apply because $f$ is only increasing when $xgeq1$. $f(x)>0$ and is continuous for all $xgeq3$. $f(x)<0$ for all $xgeq3$, so $f$ is eventually decreasing. the integral $int_{3}^{infty}f(x)dx$ is finite, so the series converges by the integral test. the integral $int_{3}^{infty}f(x)dx$ is infinite, so the series diverges by the integral test.

question 8 of 13 > consider the series. $sum_{n = 1}^{infty}\frac{7n}{(n^{2}+1)^{\frac{3}{5}}}$ let $f(x)=\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}$. find $f(x)$. (express numbers in exact form. use symbolic notation and fractions where needed.) $f(x)=$ evaluate $int_{3}^{infty}\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}dx$. (express numbers in exact form. use symbolic notation and fractions where needed.) $int_{3}^{infty}\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}dx=$ identify the true statement(s) about the series and the integral. the integral test does not apply because $f$ is not continuous on $(1,infty)$. the integral test does not apply because $f$ is only increasing when $xgeq1$. $f(x)>0$ and is continuous for all $xgeq3$. $f(x)<0$ for all $xgeq3$, so $f$ is eventually decreasing. the integral $int_{3}^{infty}f(x)dx$ is finite, so the series converges by the integral test. the integral $int_{3}^{infty}f(x)dx$ is infinite, so the series diverges by the integral test.

Answer

Explanation:

Step1: Use quotient - rule to find $f'(x)$

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 7x$, $u'=7$, $v=(x^{2}+1)^{\frac{3}{5}}$, and using the chain - rule $v'=\frac{3}{5}(x^{2}+1)^{-\frac{2}{5}}\cdot2x=\frac{6x}{5(x^{2}+1)^{\frac{2}{5}}}$. [ \begin{align*} f'(x)&=\frac{7(x^{2}+1)^{\frac{3}{5}}-7x\cdot\frac{6x}{5(x^{2}+1)^{\frac{2}{5}}}}{(x^{2}+1)^{\frac{6}{5}}}\ &=\frac{7(x^{2}+1)^{\frac{3}{5}}\cdot5(x^{2}+1)^{\frac{2}{5}}-42x^{2}}{5(x^{2}+1)^{\frac{6}{5}}}\ &=\frac{35(x^{2}+1)-42x^{2}}{5(x^{2}+1)^{\frac{8}{5}}}\ &=\frac{35x^{2}+35 - 42x^{2}}{5(x^{2}+1)^{\frac{8}{5}}}\ &=\frac{35 - 7x^{2}}{5(x^{2}+1)^{\frac{8}{5}}} \end{align*} ]

Step2: Evaluate the improper integral $\int_{3}^{\infty}\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}dx$

Let $u = x^{2}+1$, then $du = 2xdx$. When $x = 3$, $u=10$, and as $x\rightarrow\infty$, $u\rightarrow\infty$. [ \begin{align*} \int_{3}^{\infty}\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}dx&=\frac{7}{2}\int_{10}^{\infty}u^{-\frac{3}{5}}du\ &=\frac{7}{2}\lim_{b\rightarrow\infty}\int_{10}^{b}u^{-\frac{3}{5}}du\ &=\frac{7}{2}\lim_{b\rightarrow\infty}\left[\frac{u^{-\frac{3}{5}+1}}{-\frac{3}{5}+1}\right]{10}^{b}\ &=\frac{7}{2}\lim{b\rightarrow\infty}\left[\frac{u^{\frac{2}{5}}}{\frac{2}{5}}\right]{10}^{b}\ &=\frac{35}{4}\lim{b\rightarrow\infty}(b^{\frac{2}{5}}-10^{\frac{2}{5}})=\infty \end{align*} ]

Step3: Analyze the statements

  • The function $f(x)=\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}$ is continuous for all $x\geq3$ since the denominator is non - zero for $x\geq3$. So the statement "The Integral Test does not apply because $f$ is not continuous on $(1,\infty)$" is false.
  • $f'(x)=\frac{35 - 7x^{2}}{5(x^{2}+1)^{\frac{8}{5}}}$. When $x\geq3$, $35 - 7x^{2}=7(5 - x^{2})\lt0$, so $f'(x)\lt0$ for all $x\geq3$, and $f$ is eventually decreasing. The statement "The Integral Test does not apply because $f$ is only increasing when $x\geq1$" is false.
  • Since $f(x)=\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}\gt0$ for $x\geq3$ and is continuous for $x\geq3$, and $f'(x)\lt0$ for $x\geq3$.
  • Since $\int_{3}^{\infty}f(x)dx=\infty$, by the Integral Test, the series $\sum_{n = 1}^{\infty}\frac{7n}{(n^{2}+1)^{\frac{3}{5}}}$ diverges.

Answer:

$f'(x)=\frac{35 - 7x^{2}}{5(x^{2}+1)^{\frac{8}{5}}}$ $\int_{3}^{\infty}\frac{7x}{(x^{2}+1)^{\frac{3}{5}}}dx=\infty$ The true statements are:

  • $f(x)>0$ and is continuous for all $x\geq3$.
  • $f'(x)<0$ for all $x\geq3$, so $f$ is eventually decreasing.
  • The integral $\int_{3}^{\infty}f(x)dx$ is infinite, so the series diverges by the Integral Test.