## Increasing And Decreasing Intervals

In this page increasing and decreasing intervals we are going to discuss about how to find increasing and decreasing-interval for any function.

Procedure to find where the function is increasing or decreasing:

• First we need to find the first derivative
• Then set f ' (x ) = 0
• Put solutions on the number line
• Separate the intervals
• choose random value from the interval and check them in the first derivative
• If we get positive number for the chosen values,we can say that the function is increasing in that particular interval
• If we get negative number for the chosen values,we can say that the function is decreasing in that particular interval.

Example 1:

Find the intervals in which f (x) = 2x³ + x² - 20 x is increasing or decreasing

Solution:

f (x) = 2x³ + x² - 20 x

f '(x) = 2(3x²) + 2 x - 20

f '(x) = 6x² + 2 x - 20

÷ by 2 ⇒ 3x² +  x - 10

f '(x) = 0

3x² +  x - 10 = 0

(3x - 5) (x + 2) = 0

( 3x - 5) = 0            ( x + 2) = 0

3 x = 5                   x = - 2

x = 5/3

We can split this into three intervals (-∞,-2) (-2,5/3) (5/3,∞). Now let us see the given function is increasing or decreasing in which intervals.

Interval (3x-5) (x+2) f ' (x) Intervals of increasing/decreasing
-∞ < x < -2
-
-
+
increasing on (-∞,-2]
-2 < x < 5/3
+
-
-
decreasing on [-2,5/3]
5/3 < x < -∞
+
+
+
increasing on [5/3,-∞)

The given is increasing on (-∞,-2] ∪ [5/3,-∞) and decreasing on [-2,5/3]

Example 2:

Find the intervals in which f (x) = x³ - 3 x + 1 is increasing or decreasing

Solution:

f (x) = x³ - 3 x + 1

f '(x) = (3x²) - 3 (1)

f '(x) = 3x² - 3

÷ by 3 ⇒ x² - 1

f '(x) = 0

x² - 1 = 0

(x + 1) (x - 1) = 0

(x + 1) = 0            (x - 1) = 0

x = -1                  x = 1

We can split this as three intervals (-∞,-1) (-1,1) (1,∞). Now let us see the given function is increasing or decreasing in which intervals.

Interval (x+1) (x-1) f ' (x) Intervals of increasing/decreasing
-∞ < x < -1
-
-
+
increasing on (-∞,-1]
-1 < x < 1
+
-
-
decreasing on [-1,1]
1 < x < ∞
+
+
+
increasing on [1,∞)

The given is increasing on (-∞,-1] ∪ [1,∞) and decreasing on [-1,1]

Example 3:

Find the intervals in which f (x) = x - 2 sin x is increasing or decreasing

Solution:

f (x) = x - 2 sin x

f '(x) = 1 - 2 cos x

f '(x) = 0

1 - 2 cos x = 0

- 2 cos x = -1

cos x = 1/2

x = cos ⁻ ¹ (1/2)

x = Π/3,5Π/3

x = Π/3                  x = 5Π/3

increasing and decreasing intervals  increasing and decreasing intervals

We can split this as three intervals (0,Π/3) (Π/3,5Π/3) (5Π/3,2Π). Now let us see the given function is increasing or decreasing in which intervals.

Interval 1 - 2 cos x f ' (x) Intervals of increasing/decreasing
0 < x < Π/3
-
-
Decreasing on (0,Π/3]
Π/3 < x < 5Π/3
+
+
Increasing on [Π/3,5,Π/3]
5Π/3 < x < 2Π
-
-
Decreasing on [5Π/3,2Π)

The given is increasing on [Π/3,5,Π/3] and decreasing on (0,Π/3]
∪ [5Π/3,2Π).

These are the examples in the topic increasing and decreasing intervals. By practicing these kinds of problems you can understand this topic clearly.

Related Topics

Quote on Mathematics

“Mathematics, without this we can do nothing in our life. Each and everything around us is math.

Math is not only solving problems and finding solutions and it is also doing many things in our day to day life.  They are:

It divides sorrow and multiplies forgiveness and love.

Some people would not be able accept that the subject Math is easy to understand. That is because; they are unable to realize how the life is complicated. The problems in the subject Math are easier to solve than the problems in our real life. When we people are able to solve all the problems in the complicated life, why can we not solve the simple math problems?

Many people think that the subject math is always complicated and it exists to make things from simple to complicate. But the real existence of the subject math is to make things from complicate to simple.”

increasing and decreasing intervals to Monotonic Function