# How do you find the range of f(x) = x^3 - 3x + 2?

Jul 7, 2015

Since the grade of the function is odd (3), there is no absolute maximum or minimum.

#### Explanation:

If you let $x$ grow, $f \left(x\right)$ will grow as well, without limit:
Or, in "the language":
${\lim}_{x \to \infty} f \left(x\right) = \infty$
The other way (down) works as well:
${\lim}_{x \to - \infty} f \left(x\right) = - \infty$

There are two local extremes though, which you can find by setting the derivative to zero:
$f ' \left(x\right) = 3 {x}^{2} - 3 = 0 \to 3 {x}^{2} = 3 \to {x}^{2} = 1 \to$
$x = 1 \mathmr{and} x = - 1$
Substituting in $f \left(x\right)$ you get $\left(- 1 , 4\right) \mathmr{and} \left(1 , 0\right)$
graph{x^3-3x+2 [-13.7, 14.78, -4.39, 9.85]}