# How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for y=x^4-2x^3?

Jul 2, 2016

intercept / stationary point at x= 0 is an inflexion point

stationary point at $x = \frac{3}{2}$ is a relative minimum

#### Explanation:

$y = {x}^{4} - 2 {x}^{3} = {x}^{3} \left(x - 2\right)$ so $y = 0$ at $x = 0 , 2$

$y ' = 4 {x}^{3} - 6 {x}^{2} = {x}^{2} \left(4 x - 6\right)$ so $y ' = 0$ at $x = 0 , \frac{3}{2}$

$y ' ' = 12 {x}^{2} - 12 x = 12 x \left(x - 1\right)$

$y ' ' \left(0\right) = 0$ and $y ' ' \left(\frac{3}{2}\right) = 9 \left[> 0\right]$

so the intercept and stationary point at x= 0 is an inflexion point

and the stationary point at $x = \frac{3}{2}$ is a relative minimum

globally, the dominant term in the expression is ${x}^{4}$ so ${\lim}_{x \to \pm \infty} y = + \infty$

you can see all of this in the plot
graph{x^4 - 2x^3 [-10, 10, -5, 5]}