critical points/ stationary points , turning points,Increasing, decreasing functions, absolute maxima & Minima, Local Maxima & Minima , convex upward & convex downward - first & second derivative tests.

1. Differential Calculus-
Increasing & Decreasing
Functions

2. Increasing function
Let 𝑦 = 𝑓(𝑥) be a differentiable function on an interval (𝑎, 𝑏). If for any two
points 𝑥1, 𝑥2 ∈(a,b) such that 𝑥1 < 𝑥2, there holds the inequality 𝑓(𝑥1) ≤ 𝑓(𝑥2), the
function is called increasing (or non-decreasing) in this interval.
If this inequality is strict, i.e. 𝑓(𝑥1) < 𝑓(𝑥2), then the function 𝑦 = 𝑓(𝑥) is said to
be strictly increasing on the interval ( 𝑎 , 𝑏).
when x1 < x2 then
f(x1) ≤ f(x2)
Increasing
when x1 < x2 then
f(x1) < f(x2)
Strictly Increasing

4. Decreasing Function
Let 𝑦 = 𝑓(𝑥) be a differentiable function on an interval (𝑎, 𝑏). If for any two
points 𝑥1, 𝑥2 ∈(a,b) such that 𝑥1 < 𝑥2, there holds the inequality 𝑓(𝑥1) ≥
𝑓(𝑥2), the function is called decreasing (or non-increasing) in this interval.
If this inequality is strict, i.e. 𝑓 𝑥1 > 𝑓(𝑥2), then the function 𝑦 = 𝑓(𝑥) is
said to be strictly decreasing on the interval ( 𝑎 , 𝑏).
when x1 < x2 then
f(x1) ≥ f(x2)
Decreasing
when x1 < x2 then
f(x1) > f(x2)
Strictly Decreasing

6. Examples
1. Consider 𝑓(𝑥) = 𝑥
Clearly 2 < 3 ,
𝑓(2) = 2 & 𝑓(3) = 3
⇒ 𝑓 2 < 𝑓(3)
⟹ 𝑓 𝑖𝑠 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
2. Consider 𝑓(𝑥) = −𝑥
Clearly 2 < 3 ,
𝑓 2 = −2 & 𝑓(3) = −3
⇒ 𝑓 3 < 𝑓(2)
⟹ 𝑓 𝑖𝑠 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
https://www.desmos.com/calculator/uy361axm7t

7. Interval Type of function
(-5,-2) Increasing
(-2,1) Constant
(1,3) Increasing
(3,5) decreasing

9. Method to check whether given function y = f(x) is increasing and
decreasing functions
• Step 1. Find f’(x)
• Step 2. Let f’(x) = 0 and find the values of x
let it be 𝑥1 and 𝑥2 (critical points)
• Step 3. Split the real line (−∞, ∞ ) using the points 𝑥1 𝑎𝑛𝑑 𝑥2
−∞, 𝑥1 , 𝑥1, 𝑥2 , [𝑥2, ∞)
• Step 4. Check the value of f’(x) in each interval at any point in that interval
• If f’(x) > 0 , then the function is increasing in that particular interval
• If f’(x) < 0 , then the function is decreasing in that particular interval
• If f’(x) = 0 , then the function is constant in that particular interval

10. Problem 1
𝐈𝐟 𝒇(𝒙) = 𝒙 𝟑
+ 𝒙 𝟐
− 𝒙 + 𝟏 , then find intervals on which 𝒇 is increasing or
decreasing.
Solution:
Given 𝑓(𝑥) = 𝑥3 + 𝑥2 − 𝑥 +1 ------- (1)
Differ. w.r.to x
𝑓′
𝑥 = 3𝑥2
+ 2𝑥 − 1 ------- (2)
Put 𝒇′ 𝒙 = 𝟎 & solve
⟹ 3𝑥2
+ 2𝑥 − 1 = 0
⟹ 3𝑥2
+ 3𝑥 − 𝑥 − 1 = 0
⟹ 3𝑥(𝑥 + 1) − (𝑥 + 1) = 0
⟹ (3𝑥 − 1)(𝑥 + 1) = 0
⟹ 3𝑥 − 1 = 0 or 𝑥 + 1 = 0
⟹ 𝑥 =
1
3
or 𝑥 = −1
Now we have to check the sign of first derivative in every interval to find the
function is increasing or decreasing.

11. Consider the real line (−∞, ∞)
Split the line using the points 𝑥 = −1, 𝑥 =
1
3
The intervals are −∞, −1 , −1,
1
3
,
1
3
, ∞
Consider the interval −∞, −1
Choose any point in −∞, −1 , say −2
Put 𝑥 = −2 𝑖𝑛 (2)
2 ⟹ 𝑓′
2 = 3 −2 2
+ 2 −2 − 1
= 3 4 − 4 − 1 = 12 − 5 = 7 > 0 (+ value)
Hence f(x) is increasing in the interval −∞, −1

12. Consider the interval −1,
1
3
Choose any point in −1,
1
3
, say 0
Put 𝑥 = 0 𝑖𝑛 (2)
2 ⟹ 𝑓′
0 = 3 0 2
+ 2 0 − 1
= −1 < 0 (- value)
Hence f(x) is decreasing in the interval −1,
1
3
Consider the interval
1
3
, ∞
Choose any point in
1
3
, ∞ , say 2
Put 𝑥 = 2 𝑖𝑛 (2)
2 ⟹ 𝑓′
2 = 3 2 2
+ 2 2 − 1
= 3 4 + 4 − 1 = 12 + 3 = 15 > 0 (+ value)
Hence f(x) is increasing in the interval
1
3
, ∞

13. 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑆𝑖𝑔𝑛 𝑜𝑓 𝑓 ’(𝑥) 𝑓(𝑥)
−∞, −1 + Increasing
−1,
1
3
− decreasing
1
3
, ∞ + increasing

14. Note :
To find the critical points/stationary points:
When 𝑥 =
1
3
, 1 ⟹ 𝑓 𝑥 =
1
3
3
+
1
3
2
−
1
3
+ 1
=
1
27
+
1
9
−
1
3
+ 1 =
1+3−9+27
27
=
22
27
When 𝑥 = −1, 1 ⟹ 𝑓(𝑥) = −1 3
+ −1 2
– (−1) + 1
= -1+1+1+1 =2
The critical points are
1
3
,
22
27
, −1,2
points (𝑥, 𝑦) = (𝑥, 𝑓(𝑥))
Or
Simply we can write 𝑥 =
1
3
, 𝑥 = −1 are critical points

15. Problem 2:
Find the intervals of increase and decrease of the function 𝒇(𝒙) = 𝒙 𝟑
− 𝟑𝒙 + 𝟐.
Solution:
Given 𝑓 𝑥 = 𝑥3
− 3𝑥 + 2 −−−− −(1)
Diff . (1) w.r.to x
𝑓′(𝑥) = 3𝑥2
− 3 ------------- (2)
Put 𝒇′
𝒙 = 𝟎 & solve
3𝑥2
− 3 = 0
3𝑥2
= 3
𝑥2
= 1
⟹ 𝑥 = ± 1 = −1 𝑜𝑟 1
𝒙 = −𝟏 𝒂𝒏𝒅 𝒙 = 𝟏
Now we have to check the sign of first derivative in every interval to find the function is
increasing or decreasing.

16. Consider the real line (−∞, ∞)
Split the line using the points 𝒙 = −𝟏 𝒂𝒏𝒅 𝒙 = 𝟏
The intervals are −∞, −1 , −1,1 , (1, ∞)
Consider the interval (−∞, −1),
Choose any point in (−∞, −1), say -2
Put x = −2 in (2)
2 ⟹ 𝑓’(−2) = 3(−2)2 − 3
𝑓’(−2) = 12 − 3
𝑓’(−2) = 9 > 0
Hence f(x) is increasing in the interval −∞, −1

17. Consider the interval (−1, 1),
Choose any point in (−1, 1), say 𝑥 = 0
𝑃𝑢𝑡 𝑥 = 0 𝑖𝑛 (2)
𝑓′(0) = 3 0 2
− 3
𝑓′(0) = −3 < 0
Hence f(x) is decreasing in the interval (−1, 1)
Consider the interval (1, ∞),
Choose any point in (1, ∞), say 𝑥 = 2
𝑃𝑢𝑡 𝑥 = 2 𝑖𝑛 (2)
𝑓′(2) = 3 2 2
− 3
𝑓′(2) = 12 − 3
𝑓′(2) = 9 > 0
Hence, the function f(x) is increasing in the interval (1, ∞),

18. 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑆𝑖𝑔𝑛 𝑜𝑓 𝑓 ’(𝑥) 𝑓(𝑥)
−∞, −1 + Increasing
−1,1 − decreasing
(1, ∞) + increasing

19. Problem 3:
Find the intervals in which the function f(x) = 2x3 – 3x2 -36x +7 is Strictly Increasing &
Strictly Decreasing.
Solution:
Given 𝑓(𝑥) = 2𝑥3
– 3𝑥2
− 36𝑥 + 7 --------- (1)
Diff (1) w.r.to x,
𝑓ꞌ(𝑥) = 6𝑥2
– 6𝑥 − 36 -------- (2)
Put 𝒇′
𝒙 = 𝟎 & solve
6 𝑥2
– 𝑥 − 6 = 0
6 𝑥2
– 3𝑥 + 2𝑥 – 6 = 0
6[𝑥(𝑥 – 3) + 2(𝑥 – 3)]=0
6 𝑥 + 2 𝑥 – 3 = 0
⟹ 𝑥 = −2, 3
Now we have to check the sign of first derivative in every interval to find the function is
increasing or decreasing.

20. Consider the real line (−∞, ∞)
Split the line using the points 𝒙 = −𝟐 𝒂𝒏𝒅 𝒙 = 𝟑
The intervals are (−∞, −2), (−2, 3), (3, ∞)
Consider the interval (−∞, −2),
Choose any point in (−∞, −2), 𝑠𝑎𝑦 𝑥 = −3
Put 𝑥 = −3 in (2)
𝑓’(−3) = 6[(𝑥 + 2) (𝑥 – 3)]
𝑓’(−3) = 6[(− 3 + 2) (− 3 – 3)]
𝑓’(−3) = 6[(−1) (− 6)]
= 6(6)
= 36 > 0
Hence the function f(x) is strictly increasing in the interval (−∞, −2)

21. Consider the interval (−2, 3),
Choose any point in (−2, 3), 𝑠𝑎𝑦 𝑥 = 0
𝑃𝑢𝑡 𝑥 = 0 𝑖𝑛 2
𝑓′(0) = 6[(0 + 2) (0 – 3)]
𝑓′(0) = 6( – 6)
= – 36 < 0
Hence the function f(x) strictly decreasing in the interval (−2, 3)
Consider the interval (3, ∞),
Choose any point in (3, ∞), 𝑠𝑎𝑦 𝑥 = 4
𝑃𝑢𝑡 𝑥 = 4 𝑖𝑛 (2)
𝑓′(4) = 6[(4 + 2) (4 – 3)]
𝑓′(4) = 6[6.1]
= 36 > 0
Hence, the function is strictly Increasing in the interval (3, ∞)

22. 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑆𝑖𝑔𝑛 𝑜𝑓 𝑓 ’(𝑥) 𝑓(𝑥)
(−∞, −2), + Increasing
(−2, 3) − decreasing
(3, ∞)
+ increasing

23. Differential Calculus-
Maximum &
Minimum(Extreme values)
of Functions

24. Note :
There are two types of maximum & Minimum
One is Absolute / Global Maximum or Minimum.
Another one is Local/relative Maximum or Minimum
The maximum or minimum over the entire function is called an "Absolute" or
"Global" maximum or minimum.
The maximum or minimum over a particular interval of the function is called
“Local” or “Relative” maximum or minimum.
There is only one global maximum (and one global minimum) but there can be
more than one local maximum or minimum.

25. A function 𝑓(𝑥) has an absolute (or global) maximum at 𝑥 = 𝑎
if 𝑓(𝑥) ≤ 𝑓(𝑎) for every 𝑥 in the domain we are working on.
A function 𝑓(𝑥) has an absolute (or global) minimum at 𝑥 = 𝑎
if 𝑓(𝑥) ≥ 𝑓(𝑎) for every 𝑥 in the domain we are working on.
A function 𝑓(𝑥) has a relative (or local) maximum at 𝑥 = 𝑎
if 𝑓(𝑥) ≤ 𝑓(𝑎) for every 𝑥 in some open interval around 𝑥 = 𝑎.
A function 𝑓(𝑥) has a relative (or local) minimum at 𝑥 = 𝑎
if 𝑓(𝑥) ≥ 𝑓(𝑎) for every 𝑥 in some open interval around 𝑥 = 𝑎.
Note :
A function f has a local extremum at a if f has a local maximum at a or f has a
local minimum at a.
A function f has a absolute extremum at a if f has a absolute maximum at a or f
has a absolute minimum at a.

28. We know that the gradient of a graph is given by
𝑑𝑦
𝑑𝑥
or 𝑓’(𝑥)
Consequently,
𝑑𝑦
𝑑𝑥
= 0 at points A, B and C. All of these points are known as stationary
points

29. Stationary Points/ Critical points
Any point at which the tangent to the graph is horizontal is called a stationary
point.
We can locate stationary points by looking for points at which
𝑑𝑦
𝑑𝑥
= 0 or
𝑓’(𝑥) = 0
Turning Points
Note that at points A and B the curve has turns. These two stationary points
are referred to as turning points. Point C is not a turning point because,
although the graph is flat for a short time, the curve continues to go down as
we look from left to right.
So, all turning points are stationary points.
But not all stationary points are turning points (e.g. point C).
In other words, there are points for which
𝑑𝑦
𝑑𝑥
= 0 which are not turning points.
At a turning point
𝑑𝑦
𝑑𝑥
= 0.
Not all points where
𝑑𝑦
𝑑𝑥
= 0, are turning points, i.e. not all stationary points
are turning points

30. Absolute Maximum
(Global Maximum)
and
Absolute Minimum
(Global Minimum)

31. Method to find Absolute Maximum(Global Maximum) and Absolute
Minimum (Global Minimum) of given function y = f(x) in [a, b]
(First derivative test)
Step 1. Find f’(x)
Step 2. Let f’(x) = 0 and find the values of x
let it be 𝑥1 and 𝑥2 (critical points)
Step 3. Find the value of f(x) at the points a, b, 𝑥1 and 𝑥2
say 𝑓(𝑎) , 𝑓(𝑏) , 𝑓(𝑥1) , 𝑓(𝑥2)
The highest value of these 𝒇(𝒂) , 𝒇(𝒃) , 𝒇(𝒙 𝟏) , 𝒇(𝒙 𝟐)
is the absolute / global maximum of the function
The lowest value of these 𝒇(𝒂) , 𝒇(𝒃) , 𝒇(𝒙 𝟏) , 𝒇(𝒙 𝟐)
is the absolute / global minimum of the function

32. Problem 1:
Find the absolute maximum and absolute minimum over the specified interval and state where those
values occur. 𝒇 𝒙 = 𝒙 𝟐
+ 𝟑𝒙 − 𝟐 𝒐𝒗𝒆𝒓 [𝟏, 𝟑]
Solution:
Given 𝑓 𝑥 = 𝑥2
+ 3𝑥 − 2 ------ (1)
Diff . w.r.t 𝑥
𝑓’(𝑥) = 2𝑥 + 3
Put 𝑓’(𝑥) = 0
⟹ 2𝑥 + 3 = 0
⟹ 𝑥 = −
3
2
Put 𝑥 = −
3
2
in (1)
𝑓 −
3
2
= −
3
2
2
+ 3 −
3
2
− 2
=
9
4
−
9
2
− 2 =
9−18−36
4
= −
45
4
Given interval is [1,3]
Put 𝑥 = 1 in (1)
𝑓(1) = 1 2
+ 3(1) − 2
= 1 + 3 − 2 = 2
Put 𝑥 = 3 in (1)
𝑓(3) = 2 2
+ 3(2) − 2
= 4 + 6 − 2 = 8

33. 𝑥 𝑓(𝑥) 𝑐𝑜𝑛𝑐𝑙𝑢𝑠𝑖𝑜𝑛
𝑥 = −
3
2
−
45
4
Absolute
Minimum
𝑥 = 1 2
𝑥 = 3
8
Absolute
Maximum

34. Problem 2:
Find the absolute maximum and absolute minimum of 𝒇(𝒙) = 𝒙 − 𝟐 𝟐 in
[1,4]
Solution:
Given 𝑓(𝑥) = 𝑥 − 2 2
------- (1)
Diff. w.r.to x, 𝑓’(𝑥) = 2(𝑥 − 2)
Put 𝑓’(𝑥) = 0
2(𝑥 − 2) = 0
𝑥 − 2 = 0
𝑥 = 2
Put 𝑥 = 2 in (1)
𝑓 2 = 2 − 2 2
= 0
Given interval [1,4]
Put 𝑥 = 1 in (1)
𝑓 1 = 1 − 2 2 = −1 2 = 1
Put 𝑥 = 4 in (1)
𝑓 4 = 4 − 2 2
= 2 2
= 4

35. 𝑥 𝑓(𝑥) 𝑐𝑜𝑛𝑐𝑙𝑢𝑠𝑖𝑜𝑛
𝑥 = 2 0
Absolute
Minimum
𝑥 = 1 1
𝑥 = 4
4
Absolute
Maximum

36. Problem 3:
Find the absolute maxima and minima of the function 𝒇(𝒙) = 𝒙 𝒙 − 𝒙 𝟐 on
the interval [𝟎, 𝟏]
Solution:
Given 𝑓(𝑥) = 𝑥 𝑥 − 𝑥2 = 𝑥 𝑥 − 𝑥2
1
2 --------- (1)
Diff (1) w.r.t x
𝑓’ 𝑥 = 𝑥
1
2
𝑥 − 𝑥2
1
2
−1
1 − 2𝑥 + 𝑥 − 𝑥2
1
2 .1
= 𝑥
1
2
𝑥 − 𝑥2 −
1
2 1 − 2𝑥 + 𝑥 − 𝑥2
1
2
= 𝑥
1
2 𝑥−𝑥2
1
2
1 − 2𝑥 + 𝑥 − 𝑥2
1
2
=
𝑥 1−2𝑥
2 𝑥−𝑥2
1
2
+ 𝑥 − 𝑥2
1
2
=
𝑥 1−2𝑥 + 𝑥−𝑥2
1
2 2 𝑥−𝑥2
1
2
2 𝑥−𝑥2
1
2

37. =
𝑥 1−2𝑥 +2(𝑥−𝑥2)
2 𝑥−𝑥2
1
2
=
𝑥−2𝑥2+2𝑥−2𝑥2
2 𝑥−𝑥2
1
2
=
3𝑥−4𝑥2
2 𝑥−𝑥2
1
2
Put 𝑓’(𝑥) = 0
3𝑥−4𝑥2
2 𝑥−𝑥2
1
2
= 0
⟹ 3𝑥 − 4𝑥2
= 0
⟹ 𝑥 3 − 4𝑥 = 0
⟹ 𝑥 = 0 𝑜𝑟 3 − 4𝑥 = 0
⟹ 𝑥 = 0 , 𝑥 =
3
4
Put 𝑥 =
3
4
in (1)
𝑓
3
4
=
3
4
3
4
−
3
4
2
1
2
=
3
4
3 4 − 3 2
42
1
2
=
3
4
12−9
42
1
2
=
3
4
3
1
2
4
=
3
1+
1
2
16
=
3
3
2
16
Put 𝑥 = 0 in (1)
𝑓 0 = 0 0 − 0 2
1
2 = 0
Given interval [0,1]
Put 𝑥 = 1 in (1)
𝑓 1 = 1 1 − 1 2
1
2 = 0

38. 𝑥 𝑓(𝑥) 𝑐𝑜𝑛𝑐𝑙𝑢𝑠𝑖𝑜𝑛
𝑥 =
3
4
3
3
2
16
Absolute Maximum
𝑥 = 0 0 Absolute Minimum
𝑥 = 1
0 Absolute Minimum

39. Local Maximum
(Relative Maximum)
and
Local Minimum
(Relative Minimum)

40. Condition for Minimum (Second Derivative Test)
Notice that to the left of the minimum point,
𝑑𝑦
𝑑𝑥
is negative because the tangent has negative gradient. At the
minimum point,
𝑑𝑦
𝑑𝑥
= 0. To the right of the minimum point
𝑑𝑦
𝑑𝑥
is positive, because here the tangent has a
positive gradient. So,
𝑑𝑦
𝑑𝑥
goes from negative, to zero, to positive as x increases. In other words,
𝑑𝑦
𝑑𝑥
must be
increasing as x increases.
In fact, we can use this observation, once we have found a stationary point, to check if the point is a
minimum. If
𝑑𝑦
𝑑𝑥
is increasing near the stationary point then that point must be minimum.
Now, if the derivative of
𝑑𝑦
𝑑𝑥
is positive then we will know that
𝑑𝑦
𝑑𝑥
is increasing; so we will know that the
stationary point is a minimum. Now the derivative of
𝑑𝑦
𝑑𝑥
, called the second derivative, is written
𝑑2 𝑦
𝑑𝑥2 . We
conclude that if
𝑑2 𝑦
𝑑𝑥2 is positive at a stationary point, then that point must be a minimum turning point.
If
𝑑𝑦
𝑑𝑥
= 0 at a point, and if
𝑑2 𝑦
𝑑𝑥2 > 0 there, then that point must be a minimum.

41. Notice that to the left of the maximum point,
𝑑𝑦
𝑑𝑥
is positive because the tangent has positive
gradient. At the maximum point,
𝑑𝑦
𝑑𝑥
= 0 . To the right of the maximum point
𝑑𝑦
𝑑𝑥
is negative,
because here the tangent has a negative gradient. So,
𝑑𝑦
𝑑𝑥
goes from positive, to zero, to negative
as x increases.
In fact, we can use this observation to check if a stationary point is a maximum. If
𝑑𝑦
𝑑𝑥
is
decreasing near a stationary point then that point must be maximum.
Now, if the derivative of
𝑑𝑦
𝑑𝑥
is negative then we will know that
𝑑𝑦
𝑑𝑥
is decreasing; so we will
know that the stationary point is a maximum. As before, the derivative of
𝑑𝑦
𝑑𝑥
, the second
derivative is
𝑑2 𝑦
𝑑𝑥2 . We conclude that if
𝑑2 𝑦
𝑑𝑥2 is negative at a stationary point, then that point must be
a maximum turning point.
If
𝑑𝑦
= 0 at a point, and if
𝑑2𝑦
2 < 0 there, then that point must be a maximum.
Condition for Maximum (Second Derivative Test)

42. Method to find Local Maximum(Relative Maximum) and Local
Minimum (Relative Minimum)of the given function y = f(x)
(Second derivative test)
Step 1. Find 𝑓’(𝑥)
Step 2. Let 𝑓’(𝑥) = 0 and find the values of 𝑥
let it be 𝑥1 and 𝑥2 (critical points)
Step 3. Find 𝑓’’(𝑥)
Step 4. Find the value of 𝑓’’(𝑥) at 𝑥1 and 𝑥2
If 𝑓’’(𝑥) < 0, then at that particular point 𝑓(𝑥) has maximum value
The value 𝑓(𝑥) at this point is local/ relative maximum
If 𝑓’’(𝑥) > 0, then at that point 𝑓(𝑥) has minimum value
The value 𝑓(𝑥) at this point is local/ relative minimum

43. Problem 1:
Find the maxima and minima of the function 𝟐𝒙 𝟑
− 𝟑𝒙 𝟐
− 𝟑𝟔𝒙 + 𝟏𝟎
Solution:
Given 𝑓 𝑥 = 2𝑥3
− 3𝑥2
− 36𝑥 + 10 -------- (1)
Diff. w.r.to x,
𝑓′
𝑥 = 6𝑥2
− 6𝑥 − 36 ---------- (2)
Put 𝑓′
𝑥 = 0
⟹ 6𝑥2
− 6𝑥 − 36 = 0
⟹ 6 𝑥2
− 𝑥 − 6 = 0
⟹ 𝑥2
− 𝑥 − 6 = 0
⟹ 𝑥 − 3 𝑥 + 2 = 0
⟹ 𝑥 − 3 = 0 , 𝑥 + 2 = 0 ⇒ 𝑥 = 3 , 𝑥 = −2 (These are called critical points)

44. Diff. (2) w.r.to x,
𝑓′′
𝑥 = 12𝑥 − 6
𝑓′′
𝑥 𝑎𝑡 𝑥=3 = 12 3 − 6 = 36 − 6 = 30 > 0 ( + 𝑖𝑣𝑒)
Hence at 𝑥 = 3 , 𝑓 𝑥 has minimum
𝑓′′
𝑥 𝑎𝑡 𝑥=−2 = 12 −2 − 6 = −24 − 6 = −30 < 0 (− 𝑖𝑣𝑒)
Hence at 𝑥 = −2 , 𝑓 𝑥 has maximum
To find the maximum & minimum values:
Maximum value :
Put 𝑥 = −2 in (1)
𝑓 −2 = 2(−2)3
−3 −2 2
− 36(−2) + 10

45. = 2(−8) −3 4 + 72 + 10 = −16 − 12 + 82 = −28 + 82 = 54
Hence the maximum value of the function 𝑓(𝑥) is 54 at the point 𝑥 = −2
Minimum Value:
Put 𝑥 = 3 in (1)
𝑓 3 = 2(3)3
−3 3 2
− 36(3) + 10
= 2(27) −3 9 − 36 3 + 10 = 54 − 27 − 108 + 10 = 26 − 84 = −71
Hence the minimum value of the function 𝑓(𝑥) is -71 at the point 𝑥 = 3

46. 𝑥 𝑓(𝑥)
Maximum
Value
-2 54
Minimum
value
3 -71

47. Problem 2:
Find the extreme values of the function 𝒚 = 𝒙 𝟒
− 𝟖𝒙 𝟑
+ 𝟐𝟐𝒙 𝟐
− 𝟐𝟒𝒙 + 𝟏𝟐
Solution:
Given f(x) = 𝑥4
− 8𝑥3
+ 22𝑥2
− 24𝑥 + 12 -------- (1)
Diff. w.r.to 𝑥,
𝑓′(𝑥) = 4 𝑥3 − 24𝑥2 + 44𝑥 − 24 -------(2)
Put 𝑓′(𝑥) = 0
⟹ 4 𝑥3
− 24𝑥2
+ 44𝑥 − 24 = 0
⟹ 4 𝑥3
− 6𝑥2
+ 11𝑥 − 6 = 0
⟹ 𝑥3
− 6𝑥2
+ 11𝑥 − 6 = 0

48. So 𝑥 = 1 is a root and also we will get 𝑥2
− 5𝑥 + 6 = 0
⟹ 𝑥 − 3 𝑥 − 2 = 0
⟹ 𝑥 − 3 = 0 , 𝑥 − 2 = 0
⟹ 𝑥 = 3 , 𝑥 = 2
Hence the roots are 𝑥 = 1, 𝑥 = 2 , 𝑥 = 3 (critical points)
Diff. (2) w.r.to 𝑥,
𝑓’’ 𝑥 = 12𝑥2 − 48𝑥 + 44
𝑓’’ 𝑥 𝑎𝑡 𝑥 =1 = 12 1 2
− 48 1 + 44
= 12 − 48 + 44 = 8 > 0
Hence at 𝑥 = 1, the function 𝑓(𝑥) has minimum.

49. 𝑓’’ 𝑥 𝑎𝑡 𝑥 =2 = 12 2 2
− 48 2 + 44
= 12 4 − 96 + 44 = 48 − 96 + 44 = −4 < 0
Hence at 𝑥 = 2, the function 𝑓(𝑥) has maximum.
𝑓’’ 𝑥 𝑎𝑡 𝑥 =3 = 12 3 2
− 48 3 + 44
= 12 9 − 144 + 44 = 108 − 144 + 44 = 8 > 0
Hence at 𝑥 = 3, the function 𝑓(𝑥) has minimum.
To find the maximum & minimum values:
Maximum value :
Put 𝑥 = 2 in (1)
𝑓 2 = (2)4
−8 2 3
+ 22 2 2
− 24(2) + 12
= 16 − 8 8 + 22 4 − 24 2 + 12
= 16 − 64 + 88 − 48 + 12 = 4

50. Minimum Value:
Put 𝑥 = 1 in (1)
𝑓 1 = (1)4−8 1 3 + 22 1 2 − 24(1) + 12
= 1 − 8 + 22 − 24 + 12 = 3
(OR)
Put 𝑥 = 3 in (1)
𝑓 1 = (3)4−8 3 3 + 22 3 2 − 24(3) + 12
= 81 − 8(27) + 22(9) − 72 + 12
= 81 − 216 + 198 − 72 + 12 = 3

51. 𝐱 𝐟(𝐱)
Maximum Value 2 4
Minimum value 3,1 3

52. Concave upward
and
Concave downward

53. Concave upward
Concave upward is when the slope increases
Concave Upward is also called Convex or Convex Downward
A piece of the graph of 𝑓 is concave upward if the curve ‘bends’ upward.
𝑓(𝑥) is concave up on an interval I if all of the tangents to the curve on I are below the
graph of 𝑓(𝑥).
https://www.desmos.com/calculator/m9gglm2r8x
For example, the popular parabola 𝑦 = 𝑥2
is concave upward in its entirety.

54. Concave downward
Concave downward is when the slope decreases
Concave Upward is also called Convex or Convex Downward
A piece of the graph of 𝑓 is concave downward if the curve ‘bends’ downward.
𝑓(𝑥) is concave down on an interval II if all of the tangents to the curve on II are
above the graph of 𝑓(𝑥).
For example, a ‘flipped’ version 𝑦 = −𝑥2
of the popular parabola is concave
downward in its entirety
https://www.desmos.com/calculator/ozas4xltl4

55. Point of inflection
A point where the concavity changes (from up to down or down to up) is called
a point of inflection (POI).
A point of inflection of the graph of a function 𝑓 is a point where the second
derivative 𝑓′′ is 0.
If the graph of 𝑦 = 𝑓 (𝑥) has a point of inflection then 𝑦 = 𝑓 "(x) = 0.

58. Method to find concave / Convex upward/downward of given
function y = f(x)
Step 1. Find 𝑓’(𝑥)
Step 2 Find 𝑓’’(𝑥)
Step 3. Find the value of 𝑥 when 𝑓’’(𝑥) = 0
Let it be 𝑥1 and 𝑥2
Step 4. Split the real line (−∞, ∞ ) using the points 𝑥1 𝑎𝑛𝑑 𝑥2
−∞, 𝑥1 , 𝑥1, 𝑥2 , [𝑥2, ∞)
Step 5. Check the value of 𝑓’’(𝑥) in each interval at any point in that interval
If 𝒇’’(𝒙) > 𝟎 , then the function is concave upwards in that particular interval
If 𝒇’’(𝒙) < 𝟎 , then the function is concave downwards in that particular interval

59. Problem 1
Find the interval in which the function 𝒇(𝒙) = 𝟓𝒙 𝟑 + 𝟑𝟎𝒙 𝟐 + 𝒙 + 𝟏
Concave down/up.
Solution:
Given 𝑓(𝑥) = 5𝑥3 + 30𝑥2 + 𝑥 + 1 ------ (1)
Diff (1) w.r.to x,
𝑓′
𝑥 = 15𝑥2
+ 60𝑥 + 1 -------- (2)
Diff (2) w.r.to x,
𝑓′′ 𝑥 = 30𝑥 + 60 --------- (3)
Put 𝑓′′ 𝑥 = 0
⟹ 30𝑥 + 60 = 0
⟹ 30(𝑥 + 2) = 0
⟹ (𝑥 + 2) = 0
⟹ 𝑥 = −2 (point of inflection)
Consider the real line (−∞, ∞)

60. Split the real line by 𝑥 = −2
−∞ −2 +∞
The intervals are −∞, −2 , (−2, ∞)
Consider the interval −∞, −2
Choose any point in −∞, −2 , say −3
Put 𝑥 = −3 in (3)
𝑓′′
−3 = 30 −3 + 60 = −90 + 60 = −30 < 0
Hence in this interval , the function 𝑓(𝑥) is concave down.
Consider the interval (−2, ∞)
Choose any point in −2, ∞ , say 0
Put 𝑥 = 0 in (3)
𝑓′′ 0 = 30 0 + 60 = 60 > 0
Hence in this interval, the function 𝑓(𝑥) is concave up.

61. Interval
𝐱
value
Sign of
𝐟’’(𝐱)
Concave
down/up
−∞, −𝟐 -3 -
Concave
down
(−𝟐, ∞) 0 + Concave up

62. Mean value Theorem
If f(x) is continuous in [a, b] and differentiable in (a, b) then there exist a point ‘c’
in (a, b) such that 𝑓’ 𝑐 =
𝑓 𝑏 – 𝑓 𝑎
𝑏−𝑎
Problem 1:
Check whether the function 𝒇 𝒙 = 𝒙 𝟑
− 𝟓𝒙 𝟐
− 𝟑𝒙 in [𝟏, 𝟑] satisfied the Mean
Value Theorem.
Solution
Given 𝑓 𝑥 = 𝑥3
− 5𝑥2
− 3𝑥 -------- (1)
Diff. (1) w.r.to x,
𝑓′ 𝑥 = 3𝑥2 − 10𝑥 − 3 --------- (2)
Given interval is [1, 3]
Here 𝑎 = 1, 𝑏 = 3
W.k.t, by Mean Value Theorem , there exist a point 𝑐 ∈ (𝑎, 𝑏) such that
𝑓’ 𝑐 =
𝑓 𝑏 – 𝑓 𝑎
𝑏−𝑎
-------- (3)
Now 𝑓 𝑏 = 𝑓 3 = 3 3
− 5 3 2
− 3 3 , ( In (1), put 𝑥 = 3)
= 27 − 45 − 9 = −27

63. Now 𝑓 𝑎 = 𝑓 1 = 1 3
− 5 1 2
− 3(1), ( In (1), put 𝑥 = 1)
= 1 − 5 − 3 = −7
LHS of (3) is
𝑓 𝑏 – 𝑓 𝑎
𝑏−𝑎
=
−27− −7
3−1
=
−27+7
2
= −
20
2
= −10 ---(4)
RHS of (3) is 𝑓′(𝑐) ,
Put 𝑥 = 𝑐 in (2)
𝑓′
𝑐 = 3𝑐2
− 10𝑐 − 3 ----- (5)
Hence from (3), (4) and (5),
−10 = 3𝑐2
− 10𝑐 − 3
3𝑐2
− 10𝑐 − 3 + 10 = 0
3𝑐2
− 10𝑐 + 7 = 0
3𝑐2
− 3𝑐 − 7𝑐 + 7 = 0
3𝑐(𝑐 − 1) − 7(𝑐 − 1) = 0
3𝑐 − 7 𝑐 − 1 = 0
3𝑐 − 7 = 0, 𝑐 − 1 = 0
𝑐 =
7
3
, 1
Clearly
7
3
∈ 1,3 , Hence Mean Value Theorem satisfied.

64. Roll’s Theorem
If 𝑓(𝑥) is continuous in [a, b] and differentiable in (𝑎, 𝑏), also 𝑓(𝑏) = 𝑓(𝑎) then
there exist a point ‘c’ in (𝑎, 𝑏) such that 𝑓’ 𝑐 = 0
Problem 1:
Verify the Rolle’s theorem for 𝒇 𝒙 = 𝒙 𝟑
+ 𝟓𝒙 𝟐
− 𝟔𝒙 in (0,1)
Solution:
Given 𝑓 𝑥 = 𝑥3
+ 5𝑥2
− 6𝑥 ------ (1)
Diff. (1) w.r.to x,
𝑓′ 𝑥 = 3𝑥2 + 10𝑥 − 6 -------- (2)
Given interval is (0,1)
Here a = 0 , b = 1
𝑓 𝑎 = 𝑓 0 = 0 3
+ 5 0 2
− 6 0 = 0
𝑓 𝑏 = 𝑓 1 = 1 3
+ 5 1 2
− 6 1 = 1 + 5 − 6 = 0
Clearly 𝑓(0) = 𝑓(1)
To find a ‘𝑐’ in (0,1) such that 𝑓′ 𝑐 = 0
Put 𝑥 = 𝑐 in (2) & use 𝑓′ 𝑐 = 0, 3𝑐2 + 10𝑐 − 6 = 0

65. 3𝑐2 + 10𝑐 − 6 = 0
𝑐 = −
5
3
±
43
3
( use the formula
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
, a =3, b=10, c=-6 )
𝑐 = 0.519 , −3.852
Clearly 𝑐 = 0.519 ∈ (0,1)
Hence Rolle’s theorem is verified