# How do use the first derivative test to determine the local extrema 3-x^2?

Aug 30, 2015

Refer Explanation section

#### Explanation:

Given -

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

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

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

$x = 0$

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

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 2 < 0$

As the second derivative is less than zero , the function has a maximum at (0, 3)

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