Question

find the maximum and minimum value of Y = x³-3x

Answer 1

Answer:

Maximum = 2

Minimum = -2

Explanation:

Given: $y=x^3-3x$

We need to find the maximum or minimum of y

First we find the derivative and critical point.

$y'=3x^2-3$

For critical point: y'=0

$3x^2-3=0$

$x=-1,1$

Second derivative of y

$y''=6x$

At x=-1

y''=-6

If y''<0 then y is maximum at x=-1

Thus, The maximum value of y is 2

At x=1

y''=6

If y''>0 then y is minimum at x=1

Thus, The minimum value of y is -2

Hence, The maximum is 2 and minimum is -2