Unit 4 - Objective 1 - Increasing/Decreasing

 A function is increasing if the graph is rising as you move from left to right. For y = f(x), this means that y increases as x increases. A function is decreasing if the graph is falling as you move from left to right. For y = f(x), this means that y decreases as x increases.

The sign of the first derivative indicates whether a function is increasing or decreasing.

• If f '(a) > 0, then y = f(x) is increasing when x = a
• If f '(a) < 0, then y = f(x) is decreasing when x = a

Determine whether the curves below are increasing or decreasing at each point given.

y = f(x) = 6x² - 8x + 1 at x = -5 [or at the point (-5,191)].

 solution: f '(x) = 12x - 8 f '(-5) = -60 - 8 = -68 < 0 The derivative is negative whenx = -5 so the curve is decreasing there.

Instead of plotting a lot of points to determine the shape of a curve, use the derivative to find where the function is increasing and decreasing.

 The first derivative is the test function for increasing and decreasing.
If f '(x) is positive, f(x) is increasing.
If f '(x) is negative, f(x) is decreasing

Given y = f(x) = 2x³ - 3x² - 72x -4 , find the intervals where f(x) is increasing and where it is decreasing.

Steps
Example
1.Find f '(x)
1. f '(x) = 6x² - 6x - 72
2.Find the critical values, where f '(x) = 0 or f '(x) does not exist.
2. f '(x) = 6x² - 6x - 72 = 6(x² -x -12)
f '(x) = 6 (x-4) (x+3) = 0

c.v.: x = 4, x = -3
(no x values where f '(x) does not exist)

3.Put those critical on the x-axis and write the intervals.
3.
4.Pick one test value in each interval and put it in the derivative to determine if it is positive or negative. (I'm substituting into the factored form of the derivative.)

4.
test x = -4test x = 0test x = 5
f ' (-4) = 6 (-8)(-1)
f ' (-4) = 6 (-8) (-1)
f (x) positive

f (x) increasing
for x < -3

f ' (0) = -72

f ' (x) negative

f (x) decreasing
-3 < x < 4

f ' (5) = 6(1)(8)
f ' (5) = +48
f ' (x) positive

f (x) increasing
x > 4

Note: From the increasing and decreasing information above we can tell that this function has a relative maximum (where the curve changes from increasing to decreasing) when x = -3 and a relative minimum (where the curve changes from decreasing to increasing) when x = 4.

• The relative maximum is f(-3) = 2(-3)³ - 3(-3)² - 72(-3) - 4 = 131 and
• The relative minimum is f(4) = 2(4)³ - 3(4)² - 72(4) - 4 = -212

So the minimum point is (4,-212) and the maximum point is (-3,131). The high point is a relative maximum of y = 131 when x = -3, and the low point is a relative minimum of y = -212 when x = 4.