# How do you find the maximum, minimum and inflection points and concavity for the function y=1/5(x^4-4x^3)?

Nov 5, 2016

Inflection point at x=0, Minima at x=3
concave up in $\left(- \infty , 0\right)$ and $\left(3 , \infty\right)$
concave down in (0,3)

#### Explanation:

First get critical points by making $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{5} \left(4 {x}^{3} - 12 {x}^{2}\right)$

=$\frac{4}{5} {x}^{2} \left(x - 3\right)$=0 gives x=0, 3

Now using second derivative test, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{1}{5} \left(12 {x}^{2} - 24 x\right)$

At x=0, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$=0 hence it is a inflection point.

At x=3, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{1}{5} \left(108 - 72\right) = \frac{36}{5}$ which is Positive. Hence there is a minima at x=3

For concavity second derivative test can be used as shown below