1. Increasing and decreasing function

2. Strictly increasing function:
A function f(x) is said to be a strictly increasing function on (a, b) if
x1<x2 => f(x1)<f(x2) for all x1, x2∈(a, b)
Thus, f(x) is strictly increasing on (a, b) if the values of f(x) increase with increase in the value of x.
Graphically, f(x) is increasing on (a, b) if the graph y=f(x) moves up as x moves to the right. The graph in fig is the graph of strictly increasing on (a, b).

3. Ex: Show that the function f(x) =2x+3 strictly increasing on R.
Solution: let x1, x2∈R and let x1<x2 then
x1<x2 =>3 x1<3x2
3x1+5<3x2+5=>f(x1)<f(x2) for all x1, x2∈(a, b).
So, f(x) is increasing function on R.
The graph in fig is also evidence of strictly increasing on (a, b). 

4. Strictly decreasing function
A function f(x) is said to be a strictly increasing function on (a, b) if
x1<x2 => f(x1)>f(x2) for all x1, x2∈(a, b)
Thus, f(x) is strictly increasing on (a, b) if the values of f(x) decrease with increase in the value of x.Graphically, f(x) is decreasing on (a, b) if the graph y=f(x) moves down as x moves to the right. The graph in fig is the graph of strictly decreasing on (a, b).

5. Ex: Show that the function f(x) =-3x+7 strictly decreasing on R.
Solution: let x1, x2 ∈ R and let x1<x2 then
x1<x2 =>3 x1<3x2
-3x1+7>-3x2+5=>f(x1)>f(x2) for all x1, x2∈(a, b).
So, f(x) is increasing function on R.
The graph in fig is also evidence of strictly increasing on (a, b).

6. Necessary and sufficient condition for monotonicity of function
In this section we intend to see how we can use derivative of a function to determine where it is increasing and where it is decreasing.
Necessary condition: if f(x) is an increasing function on(a, b) then the tangent at every point on the curve y=f(x) makes an acute angle 𝜃 with the positive direction of x- axis.

7. 𝐻𝑒𝑟𝑒, tan 𝜃 = 𝑑𝑦 𝑑𝑥 >0 𝑜𝑟 𝑓 ′ 𝑥 >0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥𝜖 (𝑎, 𝑏)
It is evident from fig that if f(x) is a decreasing function on (a, b), then tangent at every point on the curve y=f(x) makes and obtuse angle 𝜃 with the positive direction of x- axis.
∴ tan 𝜃 = 𝑑𝑦 𝑑𝑥 <0 𝑜𝑟 𝑓 ′ 𝑥 <0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥𝜖 (𝑎, 𝑏)
Thus, f’(x) > 0 for all 𝑥𝜖 (𝑎, 𝑏) is the necessary condition for a function f(x) to be increasing on a given interval (a, b).
Thus, f’(x) < 0 for all 𝑥𝜖 (𝑎, 𝑏) is the necessary condition for a function f(x) to be decreasing on a given interval (a, b).

8. Sufficient condition
Theorem Let f be a differentiable real function defined on an open interval (a, b).
If f’(x) > 0 for all 𝑥𝜖 (𝑎, 𝑏), then f(x) increasing on (a, b)
If f’(x) < 0 for all 𝑥𝜖 (𝑎, 𝑏), then f(x) decreasing on (a, b)
Let x1, x2 𝜖 (𝑎, 𝑏) such that x1<x2. Consider the sub interval [x1, x2­].since f(x) is differentiable in (a, b) and [x1, x2­]⊂ (a, b). therefore f(x) is continuous on [x1, x2] and differentiable on (x1, x2) such that
𝑓 ′ 𝑐 = 𝑓 𝑥 2 −𝑓( 𝑥 1 ) 𝑥 2 − 𝑥 (1)

9. Since f’(x) > 0 for all 𝑥𝜖 (𝑎, 𝑏). So in particular, f’(c)>0
𝑓 ′ 𝑐 >0=> 𝑓 𝑥 2 −𝑓( 𝑥 1 ) 𝑥 2 − 𝑥 1 >0 [Using (i)]
=> 𝑓 𝑥 2 −𝑓( 𝑥 1 )>0 [∴ 𝑥 2 − 𝑥 1 >0 𝑤ℎ𝑒𝑛 𝑥1,<𝑥2­ ]
=> 𝑓 𝑥 2 >𝑓 𝑥 1 => 𝑓( 𝑥 1 )< 𝑓 𝑥 2
Since x1, x2 are arbitrary points in (a, b). Therefore
𝑥1,<𝑥2­=> 𝑓( 𝑥 1 )< 𝑓 𝑥 2 for all 𝑥1, 𝑥2­ 𝜖 (𝑎, 𝑏)
Hence, f(x) increasing on (a, b).
Since f’(x) < 0 for all 𝑥𝜖 (𝑎, 𝑏). So in particular, f’(c)<0
𝑓 ′ 𝑐 <0=> 𝑓 𝑥 2 −𝑓( 𝑥 1 ) 𝑥 2 − 𝑥 1 <0 [Using (i)]
=> 𝑓 𝑥 2 −𝑓 𝑥 1 <0 [∴ 𝑥 2 − 𝑥 1 >0 𝑤ℎ𝑒𝑛 𝑥1,<𝑥2­ ]
=> 𝑓 𝑥 2 <𝑓 𝑥 1 => 𝑓 𝑥 1 > 𝑓 𝑥 2
𝑥1,<𝑥2­=> 𝑓( 𝑥 1 )> 𝑓 𝑥 2 for all 𝑥1, 𝑥2­ (𝑎, 𝑏)
Hence, f(x) decreasing on (a, b). 

10. Algorithm for finding the interval in which a function is increasing or decreasing
Calculate f’(x) =0 for critical points
Let c1, c2, c3 are the roots of f’(x) = 0.
Cut the no line at c1, c2, c3
Write the interval in table −∞, 𝑐 1 , 𝑐 1 , 𝑐 2 𝑐 1 , 𝑐 2 , 𝑐 2 , 𝑐 3 , 𝑐 3 , ∞
Consider any point in the interval
See the sign of f’(x) in that interval and accordingly determine the function is increasing or decreasing in that interval.

11. Solution f’(x) = 6x2+18x+12 =6(x2-3x+2) =6(x-2) (x-1) = f’(x)=0
Ex 1: Find the interval in which the function f(x) =2x3-9x2+12x+15 is (a) increasing (b) decreasing
Solution f’(x) = 6x2+18x+12
=6(x2-3x+2)
=6(x-2) (x-1)
= f’(x)=0
=>x=2 and x=1
−∞ ∞
Clearly from the table f(x) is increasing on −∞, 1 ∪ 2, ∞ and f(x) is decreasing on (1, 2).
Interval
−∞, 𝟏
(1, 2)
𝟐, ∞
f’(x)
+ve
-ve
f(x)
Increasing
Decreasing
increasing

12. Ex 2: Find the interval in which the function f(x) =2x3-9x2+12x+20 is (a) increasing (b) decreasing
Solution f’(x) = 6x2+18x+12
=6(x2+3x+2)
=6(x+2) (x+1)
f’(x)=0
=>x=-2 and x=-1
Clearly from the table f(x) is increasing on −∞, 1 ∪ −1, ∞ and f(x) is decreasing on (1, 2). −∞ ∞
Interval
−∞, 1
(-2, -1)
−1, ∞
f’(x)
+ve
-ve
f(x)
Increasing
Decreasing

13. Exercise  
Ex 1: Find the interval in which the function f(x) =(x-1)3(x-2)2 is (a) increasing (b) decreasing.
Ex 2: Find the interval in which the function f(x) = 𝑥 4 − 𝑥 3 3 is (a) increasing (b) decreasing.
Ex 3: Find the interval in which the function f(x) = log 1+𝑥 − 2𝑥 2+𝑥 is (a) increasing (b) decreasing.
Ex 4: Determine the interval in which the function f(x) =x4 - 8x3 +22x2 +24x +21 is (a) increasing (b) decreasing.
Ex 5: Find the interval in which the function f(x) = x4 - 2x2 is (a) increasing (b) decreasing.
Ex 6: Find the interval in which the function f(x) = x4 - 2x2 is (a) increasing (b) decreasing.
Ex 7: Find the interval in which the function f(x) = 𝑥 𝑥 , 𝑥≠0is (a) increasing (b) decreasing.
Ex8: Separate [0,𝜋/2] into subintervals in which f(x) =sin3x is increasing or decreasing.

14. Type II on proving THE MONOTONICITY of a function on given interval(s)
Ex 1: Prove that the function f(x) =x3-3x2+3x-100 is increasing on R.
Solution :
we have
f(x) =x3-3x2+3x-100
f’(x)= 3x2 -6x +3
=3(x-1)2
Now, 𝑥𝜖𝑅=> (x-1)2 ≥0 =>f’(x) ≥0 thus, f’(x) ≥0 for all x∈R.

15. We have f(x)= x3 – 8=> f’(x) =3x2
Ex2: State when a function is said to be increasing function on [a, b]. Test whether the function f(x) = x3 – 8 in increasing on [1, 2].
We have
f(x)= x3 – 8=>
f’(x) =3x2
Clearly f’(x) >0 for all x ∈[1, 2]. So f(x) is increasing on [1, 2].

16. Ex3:Find the least value of a such that the function f given by f(x)=x2+ax+1 is strictly increasing on (1, 2).
Given, f(x)=x2+ax+1
f’(x)=2x+a
For f to be strictly increasing on(1, 2), f’(x)≥0 𝑖𝑛 (1, 2)
Now 2x+a ≥0 for all x∈(1, 2)
Least value of 2x+a ≥0 for all x∈(1, 2)
2×1+a ≥0
a ≥−2
Hence required value if x are given by -2≤𝑎<∞
∴𝑙𝑒𝑎𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑎=−2.

17. Ex4: Prove that 𝑓 𝑥 1+𝑥 < log 1+𝑥 <𝑥 𝑜𝑟 𝑥>0.
To prove: 𝑥 1+𝑥 < log 1+𝑥 𝑓𝑜𝑟 𝑥>0
Let, f(x) = 𝑥 1+𝑥 − log 1+𝑥 (1)
Then, f’(x)= 1+𝑥 .1.−𝑥 𝑥 2 − 𝑥 1+𝑥 = 𝑥 2 − 1 1+𝑥
= 1−1−𝑥 1+𝑥 2 =− 𝑥 1+𝑥 2
In (0, ∞), 𝑓 ′ 𝑥 =− 𝑥 1+𝑥 2 <0∴𝑓 𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑖𝑛 0, ∞
Also, f(X) is continuous at x=0
Hence f is decreasing function in (0, ∞)
∴𝑥>0⇒𝑓 𝑥 <𝑓 0
=> 𝑥 1+𝑥 − log 1+𝑥 <0⇒ 𝑥 1+𝑥 < log 1+𝑥 (2)

18. To prove: log 1+𝑥 <𝑥 𝑓𝑜𝑟 𝑥>0
Let g(x)=log(1+x)-x (3)
Then g’(x)= 1 1+𝑥 −1=− 𝑥 1+𝑥 <0 𝑖𝑛 0, ∞
∴𝑔 𝑖𝑠 𝑎 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 0, ∞
Also g(x) is continuous at x=0.
∴𝑔 𝑥 𝑖𝑠 𝑎 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 0, ∞
∴ 𝑥>0⇒𝑔 𝑥 <𝑔 0
⇒ log 1+𝑥 −𝑥<0
⇒ log 1+𝑥 <𝑥 (4)
From (2) and (4), we have 𝑥 1+𝑥 < log 1+𝑥 <𝑥 𝑓𝑜𝑟 𝑥>0.

19. Ex 5: The interval in which y=x2e-x is increasing with respect to x is , (a) (−∞, ∞) (b) (-2, 0) (c) (2, ∞) (d) (0, 2).
Given y=x2e-x
∴ 𝑑𝑦 𝑑𝑥 =x2e−x − 𝑥 2 𝑒 −𝑥 =𝑥 𝑒 −𝑥 2−𝑥
Since 𝑒 −𝑥 >0,
therefore sign scheme for f ′ x will be same as that of x 2−x 2−x
Sign scheme for f’(x) i.e. for x(2-x) is
Since f’(x)>0 in (0, ∞), therefore f is an increasing function in (0, 2).
Ans. (d)

20. Exercise
Find the intervals in which the following function are increasing or decreasing.
f(x) = 2x3 -15x2 + 36x +1
f(x) = 2x3 -9x2 + 12x - 5