# How do you find the critical points for y=x^4-3x^3+5?

Jan 11, 2017

$\left(0 , 5\right)$ and
$\left(2.25 , - 3.54\right)$ (2dp)

#### Explanation:

$y = {x}^{4} - 3 {x}^{3} + 5$

Differentiating wrt $x$ gives:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} - 9 {x}^{2}$

At a critical point, $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\implies {x}^{2} \left(4 x - 9\right) = 0$
$\therefore x = 0 , \frac{9}{4}$

When;

$x = 0 \implies y = 5$
$x = 2.25 \implies y \approx - 3.54$

So there are two critical points:

$\left(0 , 5\right)$ and $\left(2.25 , - 3.54\right)$ (2dp)

We we can see on the graph:
graph{y = x^4 - 3x^3 + 5 [-2, 5, -5.81, 9.99]}