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Maxima and Minima of functions of one variable
Let c be a point in a domain D of a function f. Then f(c) is the
⇒ absolute maximum value of f on D if 𝑓(𝑐) ≥ 𝑓(𝑥) for all x in D.
⇒ absolute minimum value of f on D if 𝑓(𝑐) ≤ 𝑓(𝑥) for all x in D.
Definition:
Let c be a point in a domain D of a function f. Then f(c) is the
⇒ local maximum value of f if𝑓(𝑐) ≥ 𝑓(𝑥) when x is near c.
⇒ local minimum value of f if𝑓(𝑐) ≤ 𝑓(𝑥)when x is near c.
Critical Point:
A critical point of a function f is a point c in the domain of f such that either f ‘(c) = 0
or f ‘(c) does not exists.
If f has local maximum value or minimum value at c, then c is a critical point of f.
Example:
Find the critical points of the following functions
(i) 𝒇
𝟑
𝟐
(ii) 𝒇(𝒙) = 𝒙
𝟓
𝟒 − 𝟐𝒙
𝟏
𝟒
Solutions:
(i) 𝒇
𝟑
𝟐
′
2
′
2
1
3
Critical points are 𝑥 =
1
3
(ii) 𝒇
𝟓
𝟒
− 𝟐𝒙
𝟏
𝟒
′
5
4
1
4 −
1
4
−
3
4
′
1
4
1
4 ( 5 − 2 𝑥
− 1
1
4
1
4 = 0 , ( 5 − 2 𝑥
− 1
5 𝑥− 2
𝑥
2
5
Critical points are 𝑥 = 0 ,
2
5
Example:
Find the absolute maximum and absolute minimum of
(i) 𝑓
4
3
2
(ii) 𝑓
= 𝑥 − 2 𝑠𝑖𝑛𝑥 on [0 , 2𝜋]
(iii)
= 𝑥 − 𝑙𝑜𝑔𝑥 on [
1
2
, 2 ]
Solutions:
(i) 𝒇
𝟒
𝟑
𝟐
4
3
2
- 1 is continuous on [-2 , 3]
′
3
2
′
3
2
⇒ 𝑥 = 0 , − 1 , 2 are the critical points.
The values of 𝑓(𝑥) at critical points are
4
3
2
4
3
2
4
3
2
The value of 𝑓(𝑥) at the end points of the interval are
4
3
2
4
3
2
Absolute minimum value is 𝑓
Absolute maximum value is 𝑓
(ii) 𝒇 (iv) 𝒇
= 𝒙 − 𝟐𝒔𝒊𝒏𝒙 on [0 , 2 𝝅 ]
Solution:
= 𝑥 − 2 𝑠𝑖𝑛𝑥 is continuous on [0 , 2𝜋]
The values of f(x) at the end points of the intervals are
1
2
1
2
1
2
1
2
= 2 − log 2
Absolute maximum value is f(2) = 1.
Absolute minimum value is f(1) = 1
Rolle’s Theorem:
Let f be a function that satisfies the following three conditions:
𝑓 is continuous on the closed interval [𝑎, 𝑏]
𝑓 is differentiable on the open interval (𝑎, 𝑏)
Then there exists a number c in (𝑎, 𝑏) such that 𝑓 ‘(𝑐) = 0
Example:
Verify Rolle’s theorem for the following functions on the given interval
a) 𝒇
𝟑
𝟐
[
]
b) 𝒇
𝟏
𝟑
[
]
Solutions:
a) 𝒇
𝟑
𝟐
[
]
Solution:
is continuous on
[
]
𝑓(𝑥) is differentiable on [ 0 , 3 ]
Hence the Rolle’s theorem is not satisfied.
b) 𝒇
𝟏
𝟑
[
]
Solution:
is continuous on
[
]
is differentiable on
[
]
9
3
𝑥
3
′
1
2 √
𝑥
1
3
′
1
2 √𝑥
1
3
1
2 √
𝑥
1
3
3
2
Squaring, 𝑥 =
9
4
Hence Rolle’s theorem is verified.
Example:
Prove that equation 𝒙
𝟑
− 𝟏𝟓𝒙 + 𝒄 = 𝟎 has atmost one real root in the interval [-2 , 2]
Solution:
3
′
2
Now if there were two points x = a, b such that𝑓(𝑥) = 0
∴By Rolle’s theorem there exists a point 𝑥 = 𝑐 in between them, where 𝑓
′
Now 𝑓
′
2
2
Here both values lies outside [-2, 2]
∴ 𝑓 has no more than one zero.
⇒ 𝑓(𝑥) has exactly one real root.
Definition:
If the graph of 𝑓 lies above all of its tangents on an interval I, then it is called concave
upward on I. If the graph of 𝑓 lies below all of its tangents on an interval I, then it is called
concave downward on I.
Note:
Concave upward ≡ convex downward
Concave downward ≡ convex upward
Concavity Test
Definition:
(a) If 𝑓
′′
(𝑥) > 0 for all x in I, then the graph of 𝑓 is concave upward on I.
(b) If 𝑓
′′
(𝑥) < 0 for all x in I, then the graph of 𝑓 is concave downward on I.
Definition:
A point P on a curve 𝑦 = 𝑓(𝑥)is called an inflection point iff is continuous there and the
curve changes from concave upward to concave downward or from concave downward to
concave upward at P.
The Second Derivative Test
Definition:
Suppose 𝑓′′ is continuous near c,
(a) If 𝑓
′
= 0 and 𝑓
′′
(𝑐) > 0 , then f has a local minimum at c.
(b) If 𝑓
′
= 0 and 𝑓
′′
(𝑐) < 0 , then f has a local maximum at c.
Example:
Find where the function 𝒇
𝟒
𝟑
𝟐
- 𝟓 is increasing and where it is
decreasing.
Solution:
Given 𝑓
4
3
2
′
3
2
2
′
⇒ 𝑥 = 0 , 2 , − 1 are the critical values.
We divide the real line into intervals whose end points are the critical points. 𝑥 = 0 , 2 , − 1 and
list them in a table
Interval 12 𝑥 𝑥 − 2 𝑥 + 1 𝑓 ‘(𝑥) 𝑓(𝑥)
𝑥 < − 1 - - - - decreasing
− 1 < 𝑥 < 0 - - + + increasing
𝑥 > 2 + + + + increasing
∴The function is increasing in − 1 < 𝑥 < 0 and 𝑥 > 2 and it is decreasing in 𝑥 < − 1 and
Example:
Find the local maximum and minimum values of 𝒚 = 𝒙
𝟓
− 𝟓𝒙 + 𝟑 using both the first
and second derivative tests.
Solution:
Given 𝑦 = 𝑓
5
′
4
′
4
4
4
2
⇒ 𝑥 = 1 , − 1 are the critical points.
Interval Sign of 𝑓′
Behaviour of f
1 < x < ∞ + increasing
First derivative test tells us that
(i) Local maximum at 𝑥 = − 1
Second derivative test tells us that
𝑓(𝑥)changes from increasing to decreasing at 𝑥 = − 3. Thus the function has a local maximum
𝑥 = − 3 and local maximum value is 𝑓
3
2
𝑓(𝑥) changes from decreasing to increasing at 𝑥 = 2. Thus the function has a local minimum
𝑥 = 2 and local minimum value is 𝑓
3
2
For concavity test ,𝑓
′′
1
2
We divide the real line into intervals whose end points are the critical points 𝑥 = −
1
2
and list
them in a table.
Interval
𝑓’’(𝑥) concavity
downward
upward
Since the curve changes from concave downward to concave upward at 𝑥 = −
1
2
The point of inflection is
[
1
2
1
2
]
1
2
1
2
3
1
2
2
1
2
1
8
1
4
1
4
3
4
− 1 + 3 + 72
4
74
4
37
2
Hence the point of inflection are (−
1
2
37
2
Example:
Find the interval of concavity and the inflection points. Also find the extreme values
on what interval is f increasing or decreasing.
a) (𝒙) = 𝒔𝒊𝒏𝒙 + 𝒄𝒐𝒔𝒙 , 𝟎 ≤ 𝒙 ≤ 𝟐𝝅
b) 𝒇
𝟐𝒙
−𝒙
c) 𝒇
Solution:
a) (𝒙) = 𝒔𝒊𝒏𝒙 + 𝒄𝒐𝒔𝒙 , 𝟎 ≤ 𝒙 ≤ 𝟐𝝅
′
′
𝜋
4
5 𝜋
4
are the critical points.
Interval Sign of f ’ Behaviour of f
(i) Maximum at
𝜋
4
′
𝜋
4
𝜋
4
𝜋
4
1
√ 2
1
√ 2
2
√ 2
(ii) Minimum at
5 𝜋
4
′
5 𝜋
4
5 𝜋
4
5 𝜋
4
′′
′′
3 𝜋
4
7 𝜋
4
Interval Sign of 𝑓’’ Behaviour of f
′′
2 𝑥
−𝑥
′′
2 𝑥
−𝑥
2 𝑥
−𝑥
3 𝑥
1
4
1
4
1
3
1
4
1
3
1
3
Interval Sign of 𝑓′′
Behaviour of f
Concave up
Concave up
No inflection points.
c) 𝑓
′
′
1
2
2 𝜋
3
4 𝜋
3
are the critical points.
Interval Sign of 𝑓’ Behaviour of f
− decreasing
The first derivatives test tells us that there is a
(i) Local maximum at
2 𝜋
3
𝑓 (
2 𝜋
3
2 𝜋
3
2 𝜋
3
) = 3. 83
(ii)Local minimum at
4 𝜋
3
𝑓 (
4 𝜋
3
4 𝜋
3
4 𝜋
3
) = 2. 46
𝑓
′′
( 𝑥
) = − 2 𝑠𝑖𝑛𝑥
𝑓
′′
( 𝑥
) = 0 ⇒ − 2 𝑠𝑖𝑛𝑥 = 0
⇒ 𝑠𝑖𝑛𝑥 = 0 ⇒ 𝑥 = 0 , 𝜋, 2 𝜋
Inflection points are
Interval Sign of 𝑓′′
Behaviour of f
0 < 𝑥 < 𝜋 + Concave up
Concave down
