# How do you find the intervals of increasing and decreasing using the first derivative given y=(x^5-5x)/5?

Jan 16, 2017

If $y = f \left(x\right)$, then:

$f$ is increasing, when $x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$

$f$ is decreasing, when $x \in \left(- 1 , 1\right)$

#### Explanation:

Let $y = f \left(x\right) = \frac{{x}^{2} - 5 x}{5}$. Then $f$ is differentiable, since we can break it up like $\frac{1}{5} {x}^{5} - x$, which is a polynomial, and infinitely differentiable. Its first derivative is:

$\frac{d}{\mathrm{dx}} \left(\frac{1}{5} {x}^{5} - x\right) = {x}^{4} - 1$. We want to find the roots of this expression, or where the derivative is zero:

${x}^{4} - 1 = 0 \implies \left({x}^{2} - 1\right) \left({x}^{2} + 1\right) = 0 \implies \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 1\right) = 0$

We can see that for $x = - 1$ or $x = 1$, $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$.

${x}^{4} - 1$ is a polynomial with two roots. Since the coefficient of ${x}^{4}$ is $1 > 0$, it holds that:

${x}^{4} - 1$ is positive, when $x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$

${x}^{4} - 1$ is negative, when $x \in \left(- 1 , 1\right)$

When $\frac{\mathrm{dy}}{\mathrm{dx}} > 0$, $f$ is increasing.

When $\frac{\mathrm{dy}}{\mathrm{dx}} < 0$, $f$ is decreasing.

Therefore,

$f$ is increasing, when $x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$

$f$ is decreasing, when $x \in \left(- 1 , 1\right)$

(Since $f$'s domain is $\mathbb{R}$, we can safely extend our analysis to the entirety of $\left(- \infty , + \infty\right)$).