Question

determine the point at which the graph of the function has a horizontal tangent line.
f(x)=\frac{x^{2}}{x^{2}+6}
((x,y)=(quad))

Explanation:

Step1: Find the derivative using quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{2}$, $u^\prime=2x$, $v=x^{2}+6$, and $v^\prime = 2x$. So, $f^\prime(x)=\frac{2x(x^{2}+6)-x^{2}(2x)}{(x^{2}+6)^{2}}$.

Step2: Simplify the derivative

Expand the numerator: $2x(x^{2}+6)-x^{2}(2x)=2x^{3}+12x - 2x^{3}=12x$. So, $f^\prime(x)=\frac{12x}{(x^{2}+6)^{2}}$.

Step3: Set the derivative equal to zero

A horizontal tangent line occurs when $f^\prime(x) = 0$. Since $(x^{2}+6)^{2}>0$ for all real $x$, we set the numerator equal to zero. So, $12x = 0$, which gives $x = 0$.

Step4: Find the $y$ - value

Substitute $x = 0$ into the original function $f(x)$. $f(0)=\frac{0^{2}}{0^{2}+6}=0$.

Answer:

$(0,0)$