Question

How do you find the intervals of increasing and decreasing using the first derivative given `y=x^3-4x`?

Answer 1

Increasing: `(-oo, -2/sqrt(3))` and `(2/sqrt(3), oo)`
Decreasing: `(-2/sqrt(3), 2/sqrt(3))`

Explanation:

We start by finding the derivative (use the power rule).

`y' = 3x^2 - 4`

Critical numbers will occur when the derivative is `0`, that's to say the tangent line is horizontal. Critical numbers occur when the function goes from increasing to decreasing or vice versa.

`0 = 3x^2 - 4 -> x^2 = 4/3 -> x = +-2/sqrt(3)`

We now select a test point in one of the intervals, let it be `x = 0` for `(-2/sqrt(3), 2/sqrt(3))`.

`y'(0) = 3(0)^2 - 4 = -4`

Since this value is negative, the function is decreasing here. This means it must be increasing on `(-oo, -2/sqrt(3))` and `(2/sqrt(3), oo)`.

Finally, notice how I write the intervals as open intervals (aka the round parentheses instead of the closed parentheses). This is because at the critical values the function is neither increasing nor decreasing.

Hopefully this helps!

Answer 2

The intervals of increasing are `x in(-oo, -2/sqrt3) uu (2/sqrt3,+oo)` and the interval of decreasing is `(-2/sqrt3, 2/sqrt3)`

Explanation:

We calculate the first derivative

`y=x^3-4x`

`y'=3x^2-4`

Critical points are when `y'=0`

`3x^2-4=0`, `=>`, `x=+-2/sqrt3`

We build a variation chart

`Interval` `(-oo, -2/sqrt3)` `color(white)(aa)` `(-2/sqrt3, 2/sqrt3)` `color(white)(aa)` `(2/sqrt3,+oo)`

`sign y'` `color(white)(aaaaaaaa)` `+` `color(white)(aaaaaaaaaaaaaa)` `-` `color(white)(aaaaaaaaaaaa)` `+`

`y` `color(white)(aaaaaaaaaaaaa)` `↗` `color(white)(aaaaaaaaaaaaaa)` `↘` `color(white)(aaaaaaaaaaaa)` `↗`

graph{x^3-4x [-10, 10, -5, 5]}