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

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Answer 1 · 2017-01-04T09:57:09

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

$y=2+3x-x^3$
$dy/dx = 3-3x^2$

the local may be maximum or minimum when
$dy/dx=0$

$0=3-3x^2$
$3x^2 = 3$
$x^2=1$
$x=1 and -1$ .

when x = 1,
$y=2+3(1)-(1)^3 = 2+3-1 = 4$

$(d^2x) /dy^2=-6x$
$when x = 1, (dx^2)/dy^2 = -6(1) = -6$
since $(d^2x)/dy^2 < 0$ , it is maximum point at (1,4)

when x = -1,
$y=2+3(-1)-(-1)^3 = 2-3+1 = 0$

$(d^2x) /dy^2=-6x$
$when x = -1, (dx^2)/dy^2 = -6(-1) = 6$
since $(d^2x)/dy^2 > 0$ , it is minimum point at (-1,0)

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