Question

Find the maximum and minimum points of the curve y = `x^3-x^2-5x`?

Answer 1

relative maximum: `(-1, 3)`

relative minimum: `(5/3, -175/27)`

Explanation:

Given: `y = x^3 - x^2 -5x`

You can use the first derivative test or the second derivative test. The second derivative test is straight forward if the second derivative is easy to find. It does not require setting up intervals and testing a value in each interval.

Using the Second Derivative Test:

`f(x) = x^3 - x^2 -5x`

`f'(x) = 3x^2 -2x -5`

Find the critical values:

`f'(x) = 3x^2 -2x -5 = 0`

`f'(x) = (3x - 5)(x +1) = 0`

critical values: `x = 5/3, -1`

Find critical points:

`f(-1) = (-1)^3 -(-1)^2 -5(-1) = 3`; `" "(-1, 3)`

`f(5/3) = (5/3)^3 - (5/3)^2 -5(5/3)`

`f(5/3) = 125/27 - 25/9 - 25/3`

`f(5/3) = 125/27 - 75/27 - 225/27 = -175/27`; `" "(5/3, -175/27)`

#Find the second derivative and evaluate at the critical values:

`f''(x) = 6x - 2`

`f''(-1) < 0 => "relative max at "(-1, 3)`

`f''(5/3) < 0 => "relative min at "(5/3, -175/27)`

graph{x^3-x^2-5x [-20.28, 20.27, -10.14, 10.13]}