How do you find increasing, decreasing intervals, local max mins, concave up and down for  f(x) = (x^2)/(x^2 +3) ?

Aug 3, 2015

$f \left(x\right)$ is concave up on the interval $\left(- 1 , 1\right)$ and concave down on $\left(- \infty , - 1\right) \cup \left(1 , \infty\right)$.

Explanation:

Start by calculating the first derivative of $f \left(x\right)$ - use the quotient rule

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left({x}^{2}\right)\right] \cdot \left({x}^{2} + 3\right) - {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 3\right)}{{x}^{2} + 3} ^ 2$

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

${f}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2 {x}^{3}}}} + 6 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{2 {x}^{3}}}}}{{x}^{2} + 3} ^ 2 = \frac{6 x}{{x}^{2} + 3} ^ 2$

Before calculating the second derivative, find the critical points of the function by having ${f}^{'} = 0$. These points will help you determine the local minimum and local maximum.

$\frac{6 x}{{x}^{2} + 3} ^ 2 = 0 \iff 6 x = 0 \implies x = \textcolor{g r e e n}{0} \to$ critical point

To determine where the function is increasing and where it's decreasing, examine the sign of the first derivative around the critical point $x = 0$.

Since the numerator of ${f}^{'}$ will always be positive, the sign of the first derivative will be determined by the numerator.

This means that you have $6 x < 0$ for $x < 0$, so the function is decreasing on $\left(- \infty , 0\right)$.

Likewise, because $6 x > 0$ for $x > 0$, the function is increasing on $\left(0 , \infty\right)$.

Now calculate the second derivative - use the quotient and the chain rules

$\frac{d}{\mathrm{dx}} \left({f}^{'}\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left(6 x\right)\right] \cdot {\left({x}^{2} + 3\right)}^{2} - 6 x \cdot \frac{d}{\mathrm{dx}} {\left({x}^{2} + 3\right)}^{2}}{{\left({x}^{2} + 3\right)}^{2}} ^ 2$

${f}^{' '} = \frac{6 \cdot {\left({x}^{2} + 3\right)}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} - 6 x \cdot 2 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\left({x}^{2} + 3\right)}}} \cdot 2 x}{{x}^{2} + 3} ^ \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}$

${f}^{' '} = \frac{6 \cdot \left({x}^{2} + 3\right) - 24 {x}^{2}}{{x}^{2} + 3} ^ 3$

${f}^{' '} = \frac{- 18 {x}^{2} + 18}{{x}^{2} + 3} ^ 3 = - \frac{18 \cdot \left({x}^{2} - 1\right)}{{x}^{2} + 3} ^ 3$

You can determine the local minimum and local maximum of the function by using the second derivative test, which tells you that if you have $c$ as a critical point of $f \left(x\right)$, then the sign of ${f}^{' '} \left(c\right)$ will tell you

• ${f}^{' '} \left(c\right) < 0 \to$ local maximum

• ${f}^{' '} \left(c\right) > 0 \to$ local minimum

So, your critical point is $c = 0$, which means that you have

${f}^{' '} \left(0\right) = - 18 \cdot \frac{{0}^{2} - 1}{{0}^{2} + 3} ^ 3$

${f}^{' '} \left(0\right) = - 18 \cdot \frac{\left(- 1\right)}{27} = \frac{18}{27} = \frac{2}{3}$

The point $\left(0 , f \left(0\right)\right)$ will be a local minimum. Moreover, the point $\left(0 , f \left(0\right)\right)$ will be an absolute minimum as well, since

$f \left(x\right) = {x}^{2} / \left({x}^{2} + 3\right) > 0 , \left(\forall\right) x \ne 0$ on $\left(- \infty , \infty\right)$

To determine where the function is concave up and where it's concave down, analyze the behavior of ${f}^{' '}$ around the Inflection points, where ${f}^{' '} = 0$.

${f}^{' '} = - \frac{18 \left({x}^{2} - 1\right)}{{x}^{2} + 3} ^ 2 = 0$

This implies that

$- 18 \left({x}^{2} - 1\right) = 0 \iff {x}^{2} = 1 \implies x = \pm 1$

Now you're going to see what the sign of the second derivative is for the intervals

• $\left(- \infty , - 1\right)$

It's clear that ${f}^{' '}$ will be negative on this interval, so $f$ will be concave down.

• $\left(- 1 , 1\right)$

On this interval, ${f}^{' '}$ will be positive, since $\left({x}^{2} - 1\right)$ will be negative. As a result, $f$ will be concave up.

• $\left(1 , \infty\right)$

Once again, ${f}^{' '}$ will be negative for this interval, so $f$ will be concave down.