## Related questions with answers

Question

Consider $f(x)=\dfrac{x^2}{x^2+3}$. Find the local maximum and minimum values of $f$.

Solution

VerifiedStep 1

1 of 2Using the solution for the exercise 10a, we have that $f'(x)=\frac{6x}{\left(x^2+3\right)^2}$ and the critical point is

$c=0,$

which can be local maximum or local minimum. We calculate values of $f'$ for some negative and some positive point:

$\begin{align*} f(-1)&=\frac{6 \cdot (-1)}{\left((-1)^2+3\right)^2}=\frac{-6}{16}<0,\\ f(1)&=\frac{6 \cdot 1}{\left(1^2+3\right)^2}=\frac{6}{16}>0. \end{align*}$

Since $f$ decreases at $(-\infty,0)$ and then increases at $(0, +\infty)$, we get that $\text{\textcolor{#4257b2}{ $f(0)=0$ is local minimum. Function $f$ has no local maximum}}$.

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