How do you find all local maximum and minimum points using the second derivative test given y=2+3x-x^3?

Jan 4, 2017

maximum at (1,4) and minimum at (-1,0)

Explanation:

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

the local may be maximum or minimum when
$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$0 = 3 - 3 {x}^{2}$
$3 {x}^{2} = 3$
${x}^{2} = 1$
$x = 1 \mathmr{and} - 1$.

when x = 1,
$y = 2 + 3 \left(1\right) - {\left(1\right)}^{3} = 2 + 3 - 1 = 4$

(d^2x) /dy^2=-6x
$w h e n x = 1 , \frac{{\mathrm{dx}}^{2}}{\mathrm{dy}} ^ 2 = - 6 \left(1\right) = - 6$
since $\frac{{d}^{2} x}{\mathrm{dy}} ^ 2 < 0$, it is maximum point at (1,4)

when x = -1,
$y = 2 + 3 \left(- 1\right) - {\left(- 1\right)}^{3} = 2 - 3 + 1 = 0$

(d^2x) /dy^2=-6x
$w h e n x = - 1 , \frac{{\mathrm{dx}}^{2}}{\mathrm{dy}} ^ 2 = - 6 \left(- 1\right) = 6$
since $\frac{{d}^{2} x}{\mathrm{dy}} ^ 2 > 0$, it is minimum point at (-1,0)