use the first derivative test to find the location of all local extrema for the function given below. enter…

Explanation:

Step1: Find the derivative of $f(x)$

Use the quotient - rule. If $f(x)=-\frac{10x}{2x^{2}+4}$, and the quotient - rule states that if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u=- 10x$, $u^\prime=-10$, $v = 2x^{2}+4$, $v^\prime = 4x$. Then $f^\prime(x)=-\frac{-10(2x^{2}+4)-(-10x)(4x)}{(2x^{2}+4)^{2}}=\frac{- 20x^{2}-40 + 40x^{2}}{(2x^{2}+4)^{2}}=\frac{20x^{2}-40}{(2x^{2}+4)^{2}}$.

Step2: Find the critical points

Set $f^\prime(x) = 0$. So $\frac{20x^{2}-40}{(2x^{2}+4)^{2}}=0$. Since the denominator $(2x^{2}+4)^{2}>0$ for all real $x$, we set the numerator equal to zero: $20x^{2}-40 = 0$. Then $x^{2}=2$, which gives $x=\pm\sqrt{2}$.

Step3: Use the first - derivative test

Choose test points in the intervals $(-\infty,-\sqrt{2})$, $(-\sqrt{2},\sqrt{2})$, and $(\sqrt{2},\infty)$. For the interval $(-\infty,-\sqrt{2})$, let $x=-2$. Then $f^\prime(-2)=\frac{20\times(-2)^{2}-40}{(2\times(-2)^{2}+4)^{2}}=\frac{80 - 40}{(8 + 4)^{2}}=\frac{40}{144}>0$. For the interval $(-\sqrt{2},\sqrt{2})$, let $x = 0$. Then $f^\prime(0)=\frac{20\times0^{2}-40}{(2\times0^{2}+4)^{2}}=\frac{-40}{16}<0$. For the interval $(\sqrt{2},\infty)$, let $x = 2$. Then $f^\prime(2)=\frac{20\times2^{2}-40}{(2\times2^{2}+4)^{2}}=\frac{80 - 40}{(8 + 4)^{2}}=\frac{40}{144}>0$.

Since $f(x)$ changes from increasing ($f^\prime(x)>0$) to decreasing ($f^\prime(x)<0$) at $x =-\sqrt{2}$, $f(x)$ has a local maximum at $x =-\sqrt{2}$. Since $f(x)$ changes from decreasing ($f^\prime(x)<0$) to increasing ($f^\prime(x)>0$) at $x=\sqrt{2}$, $f(x)$ has a local minimum at $x=\sqrt{2}$.

Answer:

Local maxima occur at $x =-\sqrt{2}$. Local minima occur at $x=\sqrt{2}$.