Question

consider the function (f(x)=\frac{x^{2}}{5x^{2}+3}). list the (x) values of the inflection points of (f). if there are no inflection points, enter none.

Explanation:

Step1: Rewrite the function

First, rewrite $f(x)=\frac{5x^{2}}{x^{2}+3}=5-\frac{15}{x^{2}+3}$.

Step2: Find the first - derivative

Using the quotient rule, if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. For $y=\frac{15}{x^{2}+3}$, $u = 15$, $u^\prime=0$, $v=x^{2}+3$, $v^\prime = 2x$. So $y^\prime=\frac{0\times(x^{2}+3)-15\times2x}{(x^{2}+3)^{2}}=-\frac{30x}{(x^{2}+3)^{2}}$. Then $f^\prime(x)=\frac{30x}{(x^{2}+3)^{2}}$.

Step3: Find the second - derivative

Using the quotient rule again, for $y = f^\prime(x)=\frac{30x}{(x^{2}+3)^{2}}$, $u = 30x$, $u^\prime=30$, $v=(x^{2}+3)^{2}$, $v^\prime = 2(x^{2}+3)\times2x=4x(x^{2}+3)$. Then $f^{\prime\prime}(x)=\frac{30(x^{2}+3)^{2}-30x\times4x(x^{2}+3)}{(x^{2}+3)^{4}}=\frac{30(x^{2}+3)-120x^{2}}{(x^{2}+3)^{3}}=\frac{30x^{2}+90 - 120x^{2}}{(x^{2}+3)^{3}}=\frac{90 - 90x^{2}}{(x^{2}+3)^{3}}$.

Step4: Set the second - derivative equal to zero

Set $f^{\prime\prime}(x)=0$, so $\frac{90 - 90x^{2}}{(x^{2}+3)^{3}}=0$. Since $(x^{2}+3)^{3}\gt0$ for all real $x$, we solve $90 - 90x^{2}=0$.
$90 - 90x^{2}=0$ implies $1 - x^{2}=0$, which gives $x=\pm1$.

Answer:

$x = 1,x=-1$