How do you find the maximum or minimum of f(x)=3x^2?

Nov 14, 2016

The function has a minimum at $x = 0$

Explanation:

Given -

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 x$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 6 x = 0$

$x = \frac{0}{6} = 0$

At x=0 ; (d^2y)/(dx^2)=6>0

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

Hence the function has a minimum at $x = 0$