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

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Answer 1 · 2015-08-30T15:07:26

Refer Explanation section

Given -

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

$\frac{d y}{d x} = - 2 x$ $dy/dx = - 2x$

$\frac{d y}{d x} = 0 \Rightarrow - 2 x = 0$ $dy/dx = 0 => - 2x = 0$

$x = 0$ $x = 0$

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

$\frac{d^{2} y}{{d x}^{2}} = - 2 < 0$ $(d^2y)/dx^2 = - 2 < 0$

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

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

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