# How do you sketch the graph by determining all relative max and min, inflection points, finding intervals of increasing, decreasing and any asymptotes given f(x)=(4x)/(x^2+1)?

Dec 28, 2016

$f \left(x\right)$ starts from $y = 0$ and keeps decreasing until $x = - 1$ where it has a minimum for $y = - 2$
$f \left(x\right)$ increases for $x \in \left(- 1 , 1\right)$ reaching a maximum for $x = 1$ at $y = 2$
Finally, for $x > 1$ $f \left(x\right)$ is strictly decreasing and approaches $y = 0$ from positive values.

#### Explanation:

We start observing that $f \left(x\right)$ is defined and continuous for all $x \in \mathbb{R}$ and calculate the first and second derivative:

$f \left(x\right) = \frac{4 x}{{x}^{2} + 1}$

$f ' \left(x\right) = \frac{4 \left({x}^{2} + 1\right) - 8 {x}^{2}}{{\left({x}^{2} + 1\right)}^{2}} = - 4 \frac{{x}^{2} - 1}{{\left({x}^{2} + 1\right)}^{2}}$

$f ' ' \left(x\right) = - 4 \frac{2 x {\left({x}^{2} + 1\right)}^{2} - 4 x \left({x}^{2} - 1\right) \left({x}^{2} + 1\right)}{{\left({x}^{2} + 1\right)}^{4}} = - 4 \frac{2 x \left({x}^{2} + 1\right) - 4 x \left({x}^{2} - 1\right)}{{\left({x}^{2} + 1\right)}^{3}} = - 8 \frac{{x}^{3} + x - 2 {x}^{3} + 2 x}{{\left({x}^{2} + 1\right)}^{3}} = \frac{8 x \left({x}^{2} - 3\right)}{{\left({x}^{2} + 1\right)}^{3}}$

Now we can determine:

1) ${\lim}_{x \to \pm \infty} \frac{4 x}{{x}^{2} + 1} = 0$

So $f \left(x\right)$ has $y = 0$ as horizontal asymptote on both sides.

2) $f ' \left(x\right) = 0$ for $x = \pm 1$ so there are two critical points.
We can also see that the denominator of $f ' \left(x\right)$ is always positive, so the sign of the derivative os the same of the numerator $- 4 \left({x}^{2} - 1\right)$ that is:

$f ' \left(x\right) < 0$ for $x \in \left(- \infty , - 1\right)$
$f ' \left(x\right) > 0$ for $x \in \left(- 1 , 1\right)$
$f ' \left(x\right) < 0$ for $x \in \left(1 , + \infty\right)$

3) $f ' ' \left(x\right) = 0$ for $x = 0$ and $x = \pm \sqrt{3}$ so there are three inflection points, and:

$f ' ' \left(x\right) < 0$ for $x \in \left(- \infty , - \sqrt{3}\right)$
$f ' ' \left(x\right) > 0$ for $x \in \left(- \sqrt{3} , 0\right)$
$f ' ' \left(x\right) < 0$ for $x \in \left(0 , \sqrt{3}\right)$
$f ' ' \left(x\right) > 0$ for $x \in \left(\sqrt{3} , + \infty\right)$

We can conclude that:

$f \left(x\right)$ starts from $y = 0$ and keeps decreasing until $x = - 1$ where it has a minimum for $y = - 2$
$f \left(x\right)$ increases for $x \in \left(- 1 , 1\right)$ reaching a maximum for $x = 1$ at $y = 2$
Finally, for $x > 1$ $f \left(x\right)$ is strictly decreasing and approaches $y = 0$ from positive values.
$f \left(x\right)$ is concave down for $x \in \left(- \infty , - \sqrt{3}\right)$ and $x \in \left(0 , \sqrt{3}\right)$, concave up elsewhere and has three inflection points in
$x = 0$ and $x = \pm \sqrt{3}$

graph{(4x)/(x^2+1) [-10, 10, -5, 5]}