Question

How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for `3x^5 - 5x^3`?

Answer 1

Take the first derivative...

Explanation:

`f'(x) = 15x^4 - 15x^2`

Minima and maxima occur at places where the above equation evaluates to zero. Right away, you should be able to see that `x = 0` is one of these places.

But where else?

`f'(x) = 15x^2(x^2 - 1)`

`=15x^2(x+ 1)(x - 1)`

...which gives you the other points: `+1 and -1`

To determine whether these might be maxima or minima, you must take the second derivative:

`f''(x) = 60x^3 - 30x`

and evaluate at x = -1, 0, and +1.

`f''(-1) = -60 + 30 = -30`, which is negative, so `x = -1` is a relative maxima.

`f''(1) = 30`, which is positive, so `x = 1` is a relative minima.

`f''(0) = 0`, so this point is neither minima nor maxima.

Finding the regions where the original function is increasing or decreasing requires a little more analysis:

Examine the first derivative equation:

( Eq. 1) `f'(x) = 15x^2(x+ 1)(x - 1)`

Note that if x < -1, then the terms x+1 and x-1 are both negative, and term `15x^2` is positive, since any number squared is positive.

Therefore, in the region `x < -1`, the first derivative evaluates to a positive * negative * negative, and is therefore positive. So the original function is increasing where `x < -1`.

In the region x > 1, the second derivative is obviously positive, so the function is increasing where `x > 1`.

In the region `-1 < x < 0`, the `x + 1` term is positive, and the `x-1` term is negative. The first derivative is therefore a positive * negative * positive number, which is negative. The original function is therefore DECREASING in the region `-1 < x < 0`.

(a similar line of reasoning applies to the region `0 < x < 1`).

Always helps to have a graph of the function to serve as a "sanity check".
graph{3x^5 - 5x^3 [-10, 10, -5, 5]}

GOOD LUCK!