# How do you find the intervals of increasing and decreasing given y=-x^3+2x^2+2?

Aug 18, 2017

Well, the first derivative represents the slope of the graph of that equation at every point.

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2} + 4 x$

And the function is increasing wherever this is > 0.

We can factor out x:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \left(- 3 x + 4\right)$

...and this tells us that the slope of the graph of this eq. is zero when x = 0 and $- 3 x + 4 = 0$, or $x = \frac{4}{3}$

These turn out to be the endpoints of the interval where the function is increasing, but you can do a little more analysis:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \left(- 3 x + 4\right) > 0$

Note that if x is < 0, then the term (-3x + 4) has to be > 0, since -3x is > 0 (since a negative number times a negative number is positive), and any positive number + 4 is also positive.

So then x (-3x+ 4) must be NEGATIVE, since a negative number (remember that x < 0) times a positive number is always negative.

So x is DECREASING in the interval where x < 0.

Now, look again at the term (-3x + 4).

if x is > 0, but less than 4/3, then this term is positive. If x is > 4/3, then $- 3 x + 4$ is negative, so therefore the slope ($x \left(- 3 x + 4\right)$)
will also be negative.

So, therefore 0 < x < 4/3 is the only interval where the original function $- {x}^{3} + 2 {x}^{2} + 2$ is increasing.

Or, you can cheat, by graphing the function, and picking out the increasing interval by eye.