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Maximum And Minimum Values

Learn graphical & formal definition of absolute maximum and minimum values of a function. Sketch the graph of a function with desired local and absolute extrema values.

Graphical Definition of Maximum and Minimum values of a Function

Consider the graph of a function \('f'\).

  • The highest point on the graph of \('f'\) is (7, 5). We say that,

\(f(7)=5\) is the largest value the function can take.

  • The lowest point on the graph of \(f\) is (4, –2). We say that,

\(f(4)=–2\) is the smallest value of function.

  • The largest value of the function in an interval is called the absolute maximum value of the function and similarly smallest value of the function in the interval is called absolute minimum value of the function.
  • 5 is the absolute maximum value of \('f'\) and –2 is the absolute minimum value of \(f\).
  • height of highest point = absolute maximum of \(f\).

Height of lowest point = absolute minimum value of \(f\).

Illustration Questions

Given is the graph of \(f\) the absolute maximum value of \('f'\) is

A 5

B 6

C –4

D 10

×

Highest point of graph is (3, 6)

\(\therefore\) Absolute maximum value = 6

Given is the graph of \(f\) the absolute maximum value of \('f'\) is

image
A

5

.

B

6

C

–4

D

10

Option B is Correct

Illustration Questions

Given is the graph of \(f\) , the absolute minimum value of \('f'\) is

A 4

B –1

C –2

D 3

×

Lowest point of graph is (–2, –1)

\(\therefore\) Absolute minimum value of \(f\) is  –1.

Given is the graph of \(f\) , the absolute minimum value of \('f'\) is

image
A

4

.

B

–1

C

–2

D

3

Option B is Correct

Formal Definition of Absolute Maximum and Minimum

  • Let  \(c\) be \(a\) number in the domain \(D\) of \('f'\). Then \(f(c)\) is called absolute maximum or global maximum value of \(f\) in \(D\) if

\(f(c)\geq f(x)\) for all \(x\) in \(D\).

  • Let \(c\) be \(a\) number in the domain \(D\) of \('f'\). Then \(f(c)\) is called absolute minimum or global minimum value of \(f\) in \(D\) if

\(f(c)\leq f(x)\) for all \(x\) in \(D\).

  • The maximum and minimum values of \(f\) are together called extreme values of \(f\).

The absolute maximum value of \(f\) is \(f(\beta)\) and absolute minimum value is \(f(\alpha)\).

Illustration Questions

Consider the graph of a function \(f\) given, choose the correct option.

A The absolute maximum value of \(f\) is 5

B The absolute minimum value of \(f\) is 2

C The absolute minimum value of \(f\) is –3

D The absolute maximum value of \(f\) is 0

×

The highest and lowest point on the graph are (6, 8) and (–2, –3) respectively.

\(\therefore\) Absolute maximum value = 8

Absolute minimum value = –3

\(\therefore\) Option 'c' is correct.

Consider the graph of a function \(f\) given, choose the correct option.

image
A

The absolute maximum value of \(f\) is 5

.

B

The absolute minimum value of \(f\) is 2

C

The absolute minimum value of \(f\) is –3

D

The absolute maximum value of \(f\) is 0

Option C is Correct

Local Maxima and Local Minima

  • The number \(f(c)\) is called the local maximum value of a function \('f'\) if

\(f(c)\geq f(x)\) for all \(x\) which are near \('c'\) or in some interval containing \('c'\).

\(A,\,B,\,C,\,D\) are all points of local maxima or \(f(a),\,f(b),\,f(c),\,f(d)\) are all local maximum values of function.

  • The point or the value of \(x\) at which local maximum value occurs is called local maxima.
  • Note that local word is used to convey that there can be more than one such maximas for the same function.

Illustration Questions

The graph of the function \('f'\) is as given. The local maximum values of \('f'\) are

A 1, –1 and 3

B –2, 4 and 3

C \(\dfrac{1}{4}\), 2 and 6

D 4, –3 and 1

×

Local maxima occurs at \(c\) when \(f(c)\geq f(x)\) for all \(x\) near \(c\).

Such a situation occurs at \(x=-1,\;x=1,\;x=5\).

Absolute maximum value is 4, it occur at \(x=-4\) but it is not local maxima.

\(\therefore\) Local maximum values of \('f'\) are \(f(-1),\,f(1)\) and \(f(5)\)  i.e. 1, –1 and 3.

The graph of the function \('f'\) is as given. The local maximum values of \('f'\) are

image
A

1, –1 and 3

.

B

–2, 4 and 3

C

\(\dfrac{1}{4}\), 2 and 6

D

4, –3 and 1

Option A is Correct

Local Minima of a Function

  • The number \(f(c)\) is called the local minimum value of a function \('f'\) if

\(f(c)\leq f(x)\) for all \(x\) near \(c\) or in some interval contain \('c'\).

\(A,\,B,\,C\) are the point of local minima. \(f(a),\;f(b),\;f(c)\) are the local minimum values.

  • The point or the value of x at which local minimum value occur is called local minima.
  • Note that 'local' word is used to convey that there can be more than one such minima for a function.

Illustration Questions

The graph of the function \('f'\) is as given. The local minimum values of \('f'\) are

A 0 and –2

B \(\dfrac{5}{4}\) and –1

C –1 and 2

D 1 and –2

×

Local minimum value occurs when \(f(c)\leq f(x)\) for all \(x\) near \(c\).

Such a situation occurs at \(x=\dfrac{1}{4},\;x=3,\;x=-2\)

Absolute minimum value of the function \('f'\) is –2 and if occur at \(x=\dfrac{1}{4}\) and \(x=3\). It is also the local minimum value.

\(\therefore\) Local minimum value of \('f'\) are \(f(-2),\;f\left(\dfrac{1}{4}\right)\) and \(f(3)\)

i.e.  0, –2 and –2

\(\rightarrow\) 0 and –2

Absolute minimum value of a function may not be one of its local minimum value.

The graph of the function \('f'\) is as given. The local minimum values of \('f'\) are

image
A

0 and –2

.

B

\(\dfrac{5}{4}\) and –1

C

–1 and 2

D

1 and –2

Option A is Correct

Sketching the graph of a function with desired local and absolute extrema values

  • If some particular absolute maximum and minimum values of a function are desired then  mark the highest and lowest points on the graph according to the value given.
  • Suppose a function has absolute maximum at \(x=5\) and absolute minimum at \(x=-2\) then there can be many possible graphs, one of which can be.

  • Similarly the intermediate top and bottom points can be added to the graph according to desired local extremas (i.e. local maximas and minimas)
  • If a function \(f\) has absolute maximum at \(x=3\), absolute minimum at \(x=-4\), local maximum at \(x=1\) and local minimum at \(x=2\) then its graph can be as follows.

Illustration Questions

Sketch the graph of a function \('f'\) which is continuous in [–2, 7] and has an absolute maxima at \(x=-1\), absolute minima at \(x=7\) and local minima at \(x=3\) local maxima at \(x=6\).

A

B

C

D

×

Absolute maxima at \(x=-1\) 

\(\Rightarrow\) highest point should be above (–1, 0)

Absolute minima at \(x=7\) 

\(\Rightarrow\) lowest point should be above or below (7, 0)

Local maxima at \(x=6\)

\(\Rightarrow\,f(6)>f(x)\,\forall \,x\) near 6.

Local minima at \(x=3\)

\(\Rightarrow\,f(3)<f(c)\,\forall \,x\) near 3.

Such a situation is in the graph of option 'a'.

Sketch the graph of a function \('f'\) which is continuous in [–2, 7] and has an absolute maxima at \(x=-1\), absolute minima at \(x=7\) and local minima at \(x=3\) local maxima at \(x=6\).

A image
B image
C image
D image

Option A is Correct

Finding Absolute Maximum and Minimum Values (Trigonometric Functions)

To find the absolute maximum and absolute minimum value of a function in closed interval [a, b]:

  1. Find all critical numbers of \(f\) in [a, b] and then value of \(f\) at these critical numbers.
  2. Find value of \(f\) at the end point i.e. \(f(a)\) and \(f(b)\).
  3. The largest of the values from step (1) and (2)  is the absolute maximum and smallest of the values is absolute minimum value of the function. 

Illustration Questions

Find the absolute maximum and absolute minimum value of the function. \(f(x)=x\,sin\,x+cos\,x-\dfrac{1}{4}x^2\) in \(\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right]\)

A Absolute maximum value=5, Absolute minimum value =1

B Absolute maximum value= \(\dfrac{6\pi\sqrt3\,-\pi^2+18}{36}\), Absolute minimum value =1

C Absolute maximum value=1, Absolute minimum value = –1

D Absolute maximum value= \(\dfrac{6\pi+\pi^2}{3}\), Absolute minimum value =2

×

Step -1 \(\to\) Find the critical numbers of \(x\) in \(\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right]\). It is differentiable for all values of \(x\).

\(f'(x)\)\(=0\)

\(\Rightarrow\,\dfrac{d}{dx}\left(xsin\,x+cos\,x-\dfrac{x^2}{4}\right)=0\)

\(\Rightarrow\,xcos\,x+sin\,x-sin\,x-\dfrac{x}{2}=0\)

\(\Rightarrow\,x\left(cos\,x-\dfrac{1}{2}\right)=0\)

\(\Rightarrow\,x=0,\;or\;cos\;x=\dfrac{1}{2}\)

\(\Rightarrow\,x=0\;or\;x=\dfrac{\pi}{3},\;x=-\dfrac{\pi}{3}\)

Find \(f(0)=1\)

\(f\left(\dfrac{\pi}{3}\right)=\dfrac{\pi}{3}\,sin(\dfrac{\pi}{3})+cos(\dfrac{\pi}{3})-\dfrac{\pi^2}{36}\)

\(=\dfrac{\pi}{2\sqrt3}+\dfrac{1}{2}-\dfrac{\pi^2}{36}\simeq\,1.13\)

\(f\left(-\dfrac{\pi}{3}\right)=\dfrac{\pi}{3}\,sin(\dfrac{\pi}{3})+cos(\dfrac{\pi}{3})-\dfrac{\pi^2}{36}\)

\(=\dfrac{\pi}{2\sqrt3}+\dfrac{1}{2}-\dfrac{\pi^2}{36}\)  \(\simeq\,1.13\)

Step-2 \(\to\) Find \(f(a)\;\&\;f(b)\) 

\(f\left(-\dfrac{\pi}{2}\right)=-\dfrac{\pi}{2}\,sin\left(-\dfrac{\pi}{2}\right)+cos\left(-\dfrac{\pi}{2}\right)-\left(\dfrac{\left(-\dfrac{\pi}{2}\right)^2}{{4}}\right)\)

\(=\dfrac{\pi}{2}-\dfrac{\pi^2}{16}\)  \(\simeq\,0.95\)

\(f\left(\dfrac{\pi}{2}\right)=\dfrac{\pi}{2}\,sin\dfrac{\pi}{2}+cos\dfrac{\pi}{2}-\left(\dfrac{\left(\dfrac{\pi}{2}\right)^2}{{4}}\right)\)

\(=\dfrac{\pi}{2}-\dfrac{\pi^2}{16}\)

Step-3 \(\to\) Compare \(f\left(\dfrac{\pi}{2}\right),\,f\left(-\dfrac{\pi}{2}\right),\,f(0),\,f\left(\dfrac{\pi}{3}\,\right),f\left(-\dfrac{\pi}{3}\right)\)

Largest \(=f\left(\dfrac{\pi}{3}\right)=f\left(-\dfrac{\pi}{3}\right)=\dfrac{6\pi\sqrt3-\pi^2+18}{36}=\) Absolute maximum value

Smallest \(=f(0)=1=\) Absolute minimum value

Find the absolute maximum and absolute minimum value of the function. \(f(x)=x\,sin\,x+cos\,x-\dfrac{1}{4}x^2\) in \(\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right]\)

A

Absolute maximum value=5, Absolute minimum value =1

.

B

Absolute maximum value= \(\dfrac{6\pi\sqrt3\,-\pi^2+18}{36}\), Absolute minimum value =1

C

Absolute maximum value=1, Absolute minimum value = –1

D

Absolute maximum value= \(\dfrac{6\pi+\pi^2}{3}\), Absolute minimum value =2

Option B is Correct

Finding absolute maximum and minimum values of a function by sketching the graph of function

  • If we have to find the absolute maximum and minimum value of a piecewise function i.e. a function which is defined by different expression in different intervals, we first sketch the graph of the function.
  • If the compact functions of piecewise functions are linear in nature, then graph is easy to sketch.
  • The highest and lowest point on the graph will give the absolute maximum and absolute minimum value of the function.

e.g.  Look at the graph below. \(f\) has absolute minimum value at \(x=2\).

Illustration Questions

The absolute maximum and minimum value of a function \(f(x)= \begin{cases} 2-x&if&0\leq x\leq3\\ 2x+1&if&3< x\leq5\\ \end{cases}\)  is

A Absolute maximum value = 5, Absolute minimum value =11

B Absolute maximum value = 11, Absolute minimum value = –1

C Absolute maximum value = 5, Absolute minimum value = –1

D Absolute maximum value = 2, Absolute minimum value = –3

×

Sketch the graph of \(f\).

For \(x>3\) and \(x\leq5\to f(x)=\) segment of straight line \(y=2x+1\)

For \(x\leq3\) and \(x>0\to f(x)=\) segment of straight line \(y=2-x\)

image

From the graph we observe that 

Absolute maximum value \(=f(5)=11\)

Absolute minimum value \(=f(3)=-1\)

The absolute maximum and minimum value of a function \(f(x)= \begin{cases} 2-x&if&0\leq x\leq3\\ 2x+1&if&3< x\leq5\\ \end{cases}\)  is

A

Absolute maximum value = 5, Absolute minimum value =11

.

B

Absolute maximum value = 11, Absolute minimum value = –1

C

Absolute maximum value = 5, Absolute minimum value = –1

D

Absolute maximum value = 2, Absolute minimum value = –3

Option B is Correct

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