Posted on

Increasing/Decreasing & Relative Extrema

Q:  Find:

(a) the intervals where f(x) is increasing, the intervals where f(x) is decreasing, and

(b) the relative extrema of f(x) = -2x³ -9x² + 60x -8

A:

(a)  We know a function is increasing when it has a positive slope (when it is going up-hill).  The derivative of a function is the measure of the slope (at each point).  Therefore, to find where f(x) is increasing, we are looking for when the derivative is positive:  f'(x) > 0.  Similarly, to find when f(x) is decreasing, we look for where the derivative is negative: f'(x)<0.

First, let’s find the derivative of f(x):

f(x) = -2x³ -9x² + 60x -8

[using the power rule mainly, we get]:

f'(x) = -6x² – 18x + 60

Now, we need to find critical numbers: where f'(x) = 0:

-6x² – 18x + 60 = 0

Start by factoring:

-6(x² + 3x – 10) = 0

-6(x + 5)(x – 3) = 0

We know that -6(x + 5)(x – 3) = 0 when x = -5 and x = 3.

Therefore, we need to do some interval testing… We test a number on each of the intervals:

(-∞, -5), (-5, 3), and (3, ∞)

Interval 1: Pick any number from (-∞, -5) and plug it into: -6(x + 5)(x – 3)… I will pick -10:

-6(-10 + 5)(-10 – 3) = -6 * -5 * -13 = -390.  All I really care about is whether the answer is positive or negative.  I do not care what the actual number is.  Since the answer is negative, that means the derivative is negative on the interval (-∞, -5).  This means the function is decreasing from (-∞, -5).

Interval 2: Pick any number from (-5, 3) and plug it into: -6(x + 5)(x – 3)… I will pick 0:

Remember, I don’t care what the actual answer is… I just care if it is positive or negative… And, if you plug in 0, you get out a positive number.  This means the derivative is positive on the interval (-5, 3).  The function is increasing from (-5, 3).

Interval 3: Same process as above… When you plug in a number from the interval (3, ∞) into the derivative, you get a negative number.  This means the derivative is negative from (3, ∞).  The function is decreasing from (3, ∞).

(b)  Relative extrema (maximums or minimums) can occur when the derivative is 0.  We have already established that the derivative is zero when x = -5 and when x = 3.  This two numbers our the possible extrema candidates…

There are many ways to determine if -5 and 3 are maximums, minimums, or neither… Here is one way:

Consider the intervals we already discussed:

(-∞, -5)  (-5, 3)  (3, ∞)

Now, I am going to put arrows above the intervals to indicate whether the function is increasing or decreasing on that interval:

↓           ↑           ↓

(-∞, -5)  (-5, 3)  (3, ∞)

The function is essentially going down (to -5) then up (to 3) then down.  The turn-around points occur at -5 and 3.  Therefore, -5 will give a relative minimum since it lies at the bottom of the bowl and 3 will give a relative maximum since it lies at the top….

To find the actual values of the max and min, plug -5 into the original f(x) function:

f(x) = -2(-5)³ -9(-5)² + 60(-5) -8 = 250 – 225 – 300 – 8 = -283.

-283 is a relative minimum

Now, plug in 3 to get the relative maximum:

f(x) = -2(3)³ -9(3)² + 60(3) -8 = -54 – 81 + 180 – 8 = 37

37 is a relative maximum

One thought on “Increasing/Decreasing & Relative Extrema

Leave a Reply