# How do you describe the concavity of the graph and find the points of inflection (if any) for f(x) = x^3 - 3x + 2?

Oct 18, 2015

The function has a minimum at >$x = 1$ and the curve is concave upwards.

The function has a maximum at >$x = - 1$ and the curve is concave downwards

#### Explanation:

Given -

$y = {x}^{3} - 3 x + 2$

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

$\frac{{d}^{2} x}{{\mathrm{dx}}^{2}} = 6 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 3 {x}^{2} - 3 = 0$

$3 {x}^{2} = 3$

${x}^{2} = \frac{3}{3} = 1$

$\sqrt{{x}^{2}} = \pm \sqrt{1}$

$x = 1$
$x = - 1$

At >$x = 1$ ;

$\frac{{d}^{2} x}{{\mathrm{dx}}^{2}} = 6 \left(1\right) = 6 > 0$

At >x=1 ; dy/dx=0;(d^2x)/(dx^2)>0

Hence the function has a minimum at >$x = 1$ and the curve is concave upwards.

At >$x = - 1$ ;

$\frac{{d}^{2} x}{{\mathrm{dx}}^{2}} = 6 \left(- 1\right) = - 6 < 0$

At >x=-1 ; dy/dx=0;(d^2x)/(dx^2)<0

Hence the function has a maximum at >$x = - 1$ and the curve is concave downwards .

graph{3x^3-3x+2 [-10, 10, -5, 5]}

Watch this lesson also'on Maxima / Minima'