Question

find f(x) and find the value(s) of x where f(x)=0.
f(x)=\frac{x}{x^{2}+16}
f(x)=□

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}$. Here, $u(x) = x$, so $u^{\prime}(x)=1$, and $v(x)=x^{2}+16$, so $v^{\prime}(x) = 2x$. Then $f^{\prime}(x)=\frac{1\cdot(x^{2}+16)-x\cdot(2x)}{(x^{2}+16)^{2}}$.

Step2: Simplify the derivative

Expand the numerator: $f^{\prime}(x)=\frac{x^{2}+16 - 2x^{2}}{(x^{2}+16)^{2}}=\frac{16 - x^{2}}{(x^{2}+16)^{2}}$.

Step3: Set $f^{\prime}(x)=0$

Set $\frac{16 - x^{2}}{(x^{2}+16)^{2}} = 0$. Since the denominator $(x^{2}+16)^{2}>0$ for all real $x$, we only need to solve $16 - x^{2}=0$.

Step4: Solve for $x$

Rewrite $16 - x^{2}=0$ as $x^{2}=16$. Then $x=\pm4$.

Answer:

$f^{\prime}(x)=\frac{16 - x^{2}}{(x^{2}+16)^{2}}$; $x = 4$ or $x=-4$