# How do you find the end behavior of  f(x) = 4x- 5x^3?

Apr 3, 2016

Take the derivative, look at the signs.

#### Explanation:

$f ' \left(x\right) = 4 - 15 {x}^{2}$.
This equation shows the rate of change of $f \left(x\right)$ at certain x value.
From the equation you can see that $f ' \left(x\right) \ge 0$ when $- \frac{2}{\sqrt{15}} \le x \le \frac{2}{\sqrt{15}}$. For all other values, $f ' \left(x\right) < 0$.

The end behavior of $f \left(x\right) = 4 x - 5 {x}^{3}$ is that $f \left(x\right)$ approaches $- \infty$ as $x \to \infty$ and $\infty$ as $x \to \infty$.

Note: $f \left(x\right)$ approaches $\infty$ as $x$ decreases because negative $f ' \left(x\right)$ means $f \left(x\right)$ decreases as $x$ increases.