22.14 - Analytical Method For Finding Absolute Maximum and Absolute Minimum

In the previous section, we we completed a discussion on critical points. In this section, we will see the analytical method to find absolute maximum and absolute minimum.

• We know how to find all critical points, with out the help of a graph.
Let x₁, x₂, x₃, . . .  be the critical points.
• The next step naturally, is to find the f values at those points.
So we get: f(x₁), f(x₂), f(x₃), . . .
• Then we compare those f values. After the comparison, we are inclined to believe that:
   ♦ The largest f value is the absolute maximum.
   ♦ The smallest f value is the absolute minimum.
• But before we make such a conclusion, there is another aspect which should be examined. It can be written in 5 steps:

1. Fig.22.38 below shows the graph of a function f.

Fig.22.38

2. x₁ and x₂ are the two critical points.
3. From the graph, it is clear that, f(x₁) will be larger than f(x₂). So we are inclined to conclude that, x₁ is the absolute maximum and x₂ is the absolute minimum.
4. But what if we are to find the extrema in the interval [a,b]?
In the fig., f(a) and f(b) are clearly marked.
• We see that:
   ♦ f(b) is larger than f(x₁).
   ♦ So the absolute maximum is f(b). Not f(x₁).
• We see that:
   ♦ f(a) is smaller than f(x₂).
   ♦ So the absolute minimum is f(a). Not f(x₂)
5. While finding the critical points, the end points a and b will not show up because, the derivatives at those points are not zero. Also, derivatives exist at those points.
• So it is clear that, before finalizing absolute maximum and absolute minimum, we need to check the endpoints also.

Now we will see some solved examples

Solved Example 22.49
Find the absolute maximum value and absolute minimum value of the following function in the given intervals:
(i) f(x) = x³, x ∈ [−2,2]
(ii) f(x) = sin x + cos x, x ∈ [0,π]
(iii) f(x) = 4x − (1/2)x², x ∈ [−2,9/2]
(iv) f(x) = (x−1)² + 3, x ∈ [−3,1]
Solution:
Part (i): f(x) = x³, x ∈ [−2,2]
Step I: Finding the critical points and evaluating f
1. We have: $\rm{f'(x)\,=\,3x^2}$
2. Equating f'(x) to zero, we get:
3x² = 0
⇒ x² = 0
⇒ x = 0
3. So the point in category I is: x = 0
4. We obtained f'(x) = 3x²
• This function is a polynomial function. It is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 0
6. Evaluating f at the critical point, we get:
f(0) = (0)³ = 0

Step II: Evaluating f at end points
1. f(−2) = (−2)³ = −8
2. f(2) = (2)³ = 8

Step III: Comparing the f values
1. Absolute maximum is 8, which occurs at x = 2
2. Absolute minimum is −8, which occurs at x = −2

• Fig.22.39 below shows the graph:

Fig.22.39

• The dashed magenta vertical lines represent the end points.
• We obtained the extrema because we checked the end points also. If rely solely on the critical points, we will not get the actual extrema.

Part (ii): f(x) = sin x + cos x, x ∈ [0,π]
Step I: Finding the critical points and evaluating f
1. We have: $\rm{f'(x)\,=\,\cos x - \sin x}$
2. This derivative must be equated to zero. For that, it should be rearranged as follows:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(x)}    & {~=~}    &{\cos x - \sin x}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\sin (\pi/2 -x) - \sin x}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{2 \cos \left(\frac{\pi/2 - x + x}{2} \right) \sin \left(\frac{\pi/2 - x - x}{2} \right)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{2 \cos \left(\frac{\pi}{4} \right) \sin \left(\frac{\pi}{4} - x \right)}    \\
\end{array}$                           

• Equating f'(x) to zero, we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2 \cos \left(\frac{\pi}{4} \right) \sin \left(\frac{\pi}{4} - x \right)}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{\sin \left(\frac{\pi}{4} - x \right)}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{\pi}{4} - x}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{x}    & {~=~}    &{\frac{\pi}{4}}    \\
\end{array}$

◼ Remarks:
(4) Magenta color: Here we do not require the general solution because, x lies within the given interval [0,π]

3. So the point in category I is: x = π/4
4. We obtained f'(x) = cos x − sin x
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = π/4
6. Evaluating f at the critical point, we get:
f(π/4) = sin(π/4) + cos(π/4)
= $\rm{\frac{1}{\sqrt2}+ \frac{1}{\sqrt2} ~=~\frac{2}{\sqrt2}~=~\sqrt2}$

Step II: Evaluating f at end points
1. f(0) = sin 0 + cos 0 = (0+1) = 1
2. f(π) = sin π + cos π (0−1) = −1

Step III: Comparing the f values
1. Absolute maximum is √2, which occurs at x = π/4
2. Absolute minimum is −1, which occurs at x = π

• Fig.22.40 below shows the graph:

Fig.22.40

• The y-axis and the dashed magenta vertical line represent the end points.
• Absolute maximum occurs at the critical point.
• Absolute minimum occurs at the right end point.

Part (iii): f(x) = 4x − (1/2)x², x ∈ [−2,9/2]
Step I: Finding the critical points and evaluating f
1. We have: $\rm{f'(x)\,=\,4 - x}$
2. This derivative must be equated to zero. We get:
4 − x = 0
⇒ x = 4

3. So the point in category I is: x = 4
4. We obtained f'(x) = 4 − x
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 4
6. Evaluating f at the critical point, we get:
f(4) = 4(4) − (1/2) (4)² = (16 − 8) = 8

Step II: Evaluating f at end points
1. f(−2) = 4(−2) − (1/2) (−2)² = (−8 − 2) = −10
2. f(9/2) = 4(9/2) − (1/2) (9/2)² = (18 − 81/8) =$\rm{7 \frac{7}{8}}$

Step III: Comparing the f values
1. Absolute maximum is 8, which occurs at x = 4
2. Absolute minimum is −10, which occurs at x = −2

• Fig.22.41 below shows the graph:

Fig.22.41

• The dashed magenta vertical lines represent the end points.
• Absolute maximum occurs at the critical point.
• Absolute minimum occurs at the left end point.

Part (iv): f(x) = (x−1)² + 3, x ∈ [−3,1]
Step I: Finding the critical points and evaluating f
1. We have: $\rm{f'(x)\,=\,2(x-1)\,=\,2x - 2}$
2. This derivative must be equated to zero. We get:
2x − 2 = 0
⇒ 2x = 2
⇒ x = 1

3. So the point in category I is: x = 1
4. We obtained f'(x) = 2x − 2
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1
6. Evaluating f at the critical point, we get:
f(1) = (1−1)² + 3 = (0+3) = 3

Step II: Evaluating f at end points
1. f(−3) = (−3 −1)² + 3 = (16+3) = 19
2. f(1) = (1 −1)² + 3 = (0+3) = 3

Step III: Comparing the f values
1. Absolute maximum is 19, which occurs at x = −3
2. Absolute minimum is 3, which occurs at x = 1

• Fig.22.42 below shows the graph:

Fig.22.42

• The dashed magenta vertical lines represent the end points.
• Absolute minimum occurs at the critical point.
• Critical point is same as the right end point.
• Absolute maximum occurs at the left end point.

Solved Example 22.50
Find the absolute maximum value and absolute minimum value of the function:
$\rm{f(x)\,=\,\sqrt{x}\,-\,\sqrt{x^3}}$, x ∈ [0,4]
Solution:
Step I: Finding the critical points and evaluating f
1. First we find the derivative:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(x)}    & {~=~}    &{\sqrt{x}\,-\,\sqrt{x^3}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{x^{1/2}\,-\,x^{3/2}}    \\
{~\color{magenta}    3    }    &{\implies}    &{f'(x)}    & {~=~}    &{(1/2)x^{-1/2}\,-\,(3/2)x^{1/2}}    \\
\end{array}$                           

2. This derivative must be equated to zero.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(1/2)x^{-1/2}\,-\,(3/2)x^{1/2}}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{1}{2 \sqrt x} \,-\, \frac{3 \sqrt x}{2}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{1}{2} \,-\, \frac{3 x}{2}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{1 - 3x}{2}}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{1 - 3x}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{\implies}    &{3x}    & {~=~}    &{1}    \\
{~\color{magenta}    7    }    &{\implies}    &{x}    & {~=~}    &{\frac{1}{3}}    \\
\end{array}$                           

◼ Remarks:
(3) Magenta color: Here we multiply the whole equation by √x.

3. So the point in category I is: x = 1/3
4. We obtained $\rm{f'(x) \,=\, (1/2)x^{-1/2}\,-\,(3/2)x^{1/2}}$
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1/3 = 0.3333
6. Evaluating f at the critical point, we get:
$\rm{f(1/3)\,=\,\sqrt{1/3}\,-\,\sqrt{(1/3)^3}~=~0.3849}$

Step II: Evaluating f at end points
1. $\rm{f(0)\,=\,\sqrt{0}\,-\,\sqrt{(0)^3}~=~0}$
2. $\rm{f(4)\,=\,\sqrt{4}\,-\,\sqrt{(4)^3}~=~-6}$

Step III: Comparing the f values
1. Absolute maximum is 0.3849, which occurs at x = 1/3
2. Absolute minimum is −6, which occurs at x = 4

• Fig.22.43 below shows the graph:

Fig.22.43

• The y-axis and the dashed magenta vertical line represent the end points.
• Absolute maximum occurs at the critical point.
• Absolute minimum occurs at the right end point.

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In the next section, we will see shape of graph.

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22.13 - Solved Examples on Critical Points

In the previous section, we saw critical points. We saw a solved example also. In this section, we will see a few more solved examples on critical points.

Solved Example 22.43
Find all critical points for the function:
$\rm{f(x)\,=\,2x^3 - 6x^2 + 6x + 5}$
Solution:
1. We have: $\rm{f'(x)\,=\,6x^2 - 12x + 6}$
2. Equating f'(x) to zero, we get:
6(x² − 2x + 1) = 0
⇒ 6(x² − 2x + 1) = 0
⇒ (x² − 2x + 1) = 0
⇒ (x − 1)² = 0
⇒ (x − 1) = 0
⇒ x = +1
3. So the point in category I is: x = 1
4. We obtained f'(x) = 6(x² − 2x + 1)
• This function is a polynomial function. It is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1
6. Fig.22.32 shows the graph:

Fig.22.32

• From the graph, it is clear that, x=1 is a critical point but it is not an extremum.

Solved Example 22.44
Find all critical points for the function:
$\rm{f(x)\,=\,(2-8x)^4 (x^2 - 9)^3}$
Solution:
1. First we will write the derivative:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{(2-8x)^4 (x^2 - 9)^3}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^4 \left[3(x^2 - 9)^2 (2x)\right]~+~\left[4(2-8x)^3 (-8) \right] (x^2 - 9)^3}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^4 \left[(x^2 - 9)^2 (6x)\right]~-~\left[(32)(2-8x)^3 \right] (x^2 - 9)^3}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^3 (x^2 - 9)^2 \left[  (2-8x)  (6x)~-~(32)(x^2 - 9) \right]}    \\
{~\color{magenta}    5    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^3 (x^2 - 9)^2 \left[ 12x - 48 x^2 -32 x^2 + (32)(9) \right]}    \\
{~\color{magenta}    6    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^3 (x^2 - 9)^2 \left[ 12x - 80 x^2  + (32)(9) \right]}    \\
\end{array}$                           

2. Equating the derivative to zero, we get three equations:
(i) (2 − 8x)³ = 0
(ii) (x² − 9)² = 0
(iii) 12x − 80x² + (32)(9) = 0

• Solving the first equation, we get:
$\rm{x = \color{yellow} {\frac{1}{4}}}$
• Solving the second equation, we get:
$\rm{x = \color{yellow} {\pm 3}}$
• The third equation can be rearranged as:
3x − 20x² + 72 = 0
Solving this, we get:
$\rm{x = \color{yellow} {\frac{3 \pm \sqrt{5769}}{40}}}$
3. So there are five points in category I. They are written in yellow color.
4. Consider the derivative written in (1)
• This function is a polynomial function. It is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only five critical points are the ones we wrote in yellow color in (2)
6. Fig.22.33 shows the graph:

Fig.22.33

• From the graph, it is clear that:
   ♦ Points A, B and C are critical points but they are not extrema.
   ♦ Point D is a local minimum.
   ♦ Point E is the absolute minimum.
   ♦ This function does not have absolute maximum.

Solved Example 22.45
Find all critical points for the function:
$\rm{f(x)\,=\,4 \sqrt{x}\,-\, x^2}$
Solution:
1. First we will write the derivative:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{4 \sqrt{x}\,-\, x^2}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{4 (1/2)(x)^{-1/2}\,-\, 2x}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{2(x)^{-1/2}\,-\, 2x}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{2 \left[(x)^{-1/2}\,-\, x \right]}    \\
\end{array}$                           

2. Equating the derivative to zero, we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2 \left[(x)^{-1/2}\,-\, x \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{(x)^{-1/2}\,-\, x}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{(x)^{-1/2}}    & {~=~}    &{x}    \\
{~\color{magenta}    4    }    &{\implies}    &{(x)^{-1}}    & {~=~}    &{x^2}    \\
{~\color{magenta}    5    }    &{\implies}    &{x^3}    & {~=~}    &{1}    \\
{~\color{magenta}    6    }    &{\implies}    &{x}    & {~=~}    &{1}    \\
\end{array}$

3. So the point in category I is: x = 1
4. We obtained $\rm{f'(x) = 2 \left[(x)^{-1/2}\,-\, x \right]}$
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1
6. Fig.22.34 shows the graph:

Fig.22.34

• From the graph, it is clear that:
   ♦ x=1 is a critical point.
   ♦ Also, x=1 is an extremum. It is the absolute maximum.

Solved Example 22.46
Find all critical points for the function:
$\rm{f(x)\,=\,\frac{1}{x-1}}$
Solution:
1. First we will write the derivative:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\frac{1}{x-1}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{(x-1)^{-1}}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(-1)(x-1)^{-2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{(x-1)^2}}    \\
\end{array}$                           

2. Equating the derivative to zero, we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{-1}{(x-1)^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{-1}    & {~=~}    &{0}    \\
\end{array}$                           
• This equation does not have any solution.

3. So there are no points in category I.
4. We obtained $\rm{f'(x) = \frac{-1}{(x-1)^2}}$
• This function is not defined at x = 1. But the f(x) is also not defined at x = 1. It is meaningless to say that f'(x) is not available at a point where f(x) itself is not available. So we need not consider the point x=1.
• Therefore, there will be no points in category II.
5. So there are no critical points for the given function
6. Fig.22.35 shows the graph:

Fig.22.35

• From the graph, it is clear that:
   ♦ When x decreases, f(x) approaches zero.
   ♦ When x increases, then also f(x) approaches zero.
   ♦ When x approaches 1 from the left, f(x) approaches −∞.
         ✰ So there is no absolute minimum.
   ♦ When x approaches 1 from the right, f(x) approaches +∞.
         ✰ So there is no absolute maximum.

Solved Example 22.47
Find all critical points for the function:
$\rm{f(x)\,=\,\tan x}$
Solution:
1. First we will write the derivative:
f'(x) = sec²x

2. Equating the derivative to zero, we get:
sec²x = 0                  
• This equation does not have any solution.

3. So there are no points in category I.
4. We obtained f'(x) = sec²x
• This function is not defined at x = π/2. But the f(x) is also not defined at x = π/2. It is meaningless to say that f'(x) is not available at a point where f(x) itself is not available. So we need not consider the point x = π/2.
• Therefore, there will be no points in category II.
5. So there are no critical points for the given function
6. Fig.22.36 shows the graph:

Fig.22.36

• From the graph, it is clear that:
   ♦ There is no absolute maximum.
   ♦ There is no absolute minimum.
   ♦ No tangent can be drawn in the horizontal direction.
         ✰ That means, there is no point where the derivative is zero.
   ♦ The function does not exist at $\rm{x = \frac{n \pi}{2}}$
         ✰ Where n is any integer.

Solved Example 22.48
Find all critical points for the function:
$\rm{f(x)\,=\,\sin^2 x}$
Solution:
1. First we will write the derivative:
f'(x) = 2 sin x cos x

2. Equating the derivative to zero, we get two equations:
(i) sin x = 0                  
(ii) cos x = 0
• Solving the first equation, we get:
x = 0, π, 2π, 3π, . . . so on
• Solving the first equation, we get:
x = π/2, 3π/2, 5π/2, . . . so on
• The above two results can be combined as:
$\rm{x \,=\,\frac{n \pi}{2}}$
   ♦ Where n is any integer.

3. So there are infinite points in category I.
4. We obtained f'(x) = 2 sin x cos x
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.

5. So the critical points for the given function are given by:
$\rm{x \,=\,\frac{n \pi}{2}}$
   ♦ Where n is an integer.
6. Fig.22.37 shows the graph:

Fig.22.37

• From the graph, it is clear that:
   ♦ Absolute maximum occur at x = π/2, 3π/2, 5π/2, . . . so on.
   ♦ Absolute minimum occur at x = 0, 2π/2, 4π/2, 6π/2, . . . so on.

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In the next section, we will see the analytical method for calculating Maxima and Minima.

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22.12 - Maxima And Minima

In the previous section, we saw change in quantity. In this section, we will see maxima and minima.

Some basics can be written in 4 steps:
1. Consider the function f(x) = sin x.
• We can give any real number as the input. The maximum possible output is 1.
    ♦ We will not get an output greater than 1
• We can give any real number as the input. The minimum possible output is −1.
    ♦ We will not get an output lesser than −1
• We have seen this fact in our trigonometry classes.
• Also this fact can be visually seen in the graph of the sine function.
2. Consider another function f(x) = 3x² + 5x + 7.
• It’s graph is shown in fig.22.26 below:

Fig.22.26

• We see that:
    ♦ when x increases, the graph goes up to +∞.
    ♦ when x decreases, then also the graph goes up to +∞.
• So there is no maximum value for this function.
3. Let us try to find the minimum value.
• From the graph, we get a feeling that, −4 is the minimum value.
• But when a horizontal line (shown in yellow color) is drawn through (0,−4), it cuts the graph at two points A and B.
• A portion of the graph is below the yellow line. That means, −4 is not the minimum value.
4. Often in science, engineering and business problems, we will want to find the exact maximum and minimum values of various functions. Derivatives can help us to achieve this goal. In this section, we will first become familiar with absolute extrema.

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Some basics about absolute extrema can be written in 5 steps:
1. Consider the function f(x) = x² + 2.
The graph is shown in fig.22.27 below:

Fig.22.27

2. We see that:
    ♦ when x increases, the graph goes up to +∞.
    ♦ when x decreases, then also the graph goes up to +∞.
• So there is no maximum value for this function.
3. Let us try to find the minimum value.
From the graph, we get a feeling that, 2 is the minimum value. When a horizontal line (shown in yellow color) is drawn through (0,2), it just touches the graph at a point. No portion of the graph is below the yellow line. That means, 2 is indeed the minimum value.
4. Analytically also, we can prove that 2 is the minimum value. It can be done in 3 steps:
(i) In the given function, there are two terms: x² and 2
    ♦ x² will be always +ve
    ♦ 2 is +ve
    ♦ The two terms are being added together
(ii) So the value of the function will be always greater than or equal to 2.
(iii) Therefore, the minimum value is 2, which occurs when x = 0
5. In this situation we say two points:
(i) The function f(x) = x² + 2 has an absolute minimum.
    ♦ The absolute minimum is 2
    ♦ The absolute minimum occurs at x = 0
(ii) The function f(x) = x² + 2 does not have an absolute maximum.

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Now we can define absolute extrema. It can be written in 7 steps:
1. f is a function defined over an interval I.
2. c is an element in I. That is., c∈I
3. We take each element of I and use it as the input for f.
And we find that each output is less than f(c).
Symbolically, we write this as: f(c) ≥ f(x)  for all x∈I
4. We take each element of I and use it as the input for f.
And we find that each output is greater than f(c).
Symbolically, we write this as: f(c) ≤ f(x)  for all x∈I
5. If the situation in (3) occur, we say that:
f has an absolute maximum on I at c.
6. If the situation in (4) occur, we say that:
f has an absolute minimum on I at c.
7. If either of (3) or (4) occur, we say that:
f has an absolute extremum on I at c.
(extremum is the singular of extrema)

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Extreme Value Theorem
This can be explained in 3 steps:
1. When we are given a function f, it may fall into any one of the following four categories:
(i) f has an absolute maximum but no absolute minimum.
(ii) f has an absolute minimum but no absolute maximum.
(iii) f has an absolute maximum and an absolute minimum.
(iv) f has neither absolute maximum nor absolute minimum.
2. But if two conditions are satisfied, we can guarantee that, there will be an absolute maximum and there will be an absolute minimum.
Condition 1:
• The interval under consideration, must be closed and bounded.
Explanation for "closed and bounded: We know that, a closed interval has square brackets on both ends. That is, []. For an interval to be bounded, neither of the end points should be infinity. So for condition 1, the interval must be in the form [a,b].
Condition 2:
• f must be continuous over [a,b]
3. The guarantee given by the two conditions is known as extreme value theorem.

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Local extrema
We have seen absolute extrema. Now we will see local extrema. It can be explained in 7 steps:
1. In fig.22.28 below, the function f is defined on (−∞,∞).

Fig.22.28

• The function f has:
   ♦ A peak point at x = a
   ♦ A valley point at x = b
   ♦ A peak point at x = c
2. It is clear that, f does not have an absolute minimum. This is because:
   ♦ As x decreases, the value of f approaches −∞
   ♦ As x increases, then also, the value of f approaches −∞
3. It is clear that, the peak at x = c, represents the absolute maximum.
4. The peak at x = a, also has significance.
• The value f(a) is larger than the other values in the immediate vicinity.
• This fact can be expressed mathematically in the form of three conditions:
(i) There is a positive number 'h' such that, we can form the interval (a−h,a+h) on the x-axis.
(ii) Take each input x value from this interval. Write the corresponding f(x) value for each of those inputs.
(iii) All those f(x) values will be less than f(a).
5. If we can find a number 'h' which satisfies the three conditions above, then we say that, x =a is a local maximum.
• So the function has an absolute maximum and a local maximum.
• But the absolute maximum at x = c satisfies the three conditions. So it is a local maximum also.
• We can write:
The function has two local maxima. The local maxima at x = c happens to be the absolute maximum also.
6. Similarly, x = b is a local minimum.
• So the function does not have an absolute minimum. But it has a local minimum.
7. Local maximum is also called relative maximum.
• Similarly, local minimum is also called relative minimum.

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Critical points
This can be explained in 6 steps:
1. On many occasions, graph of the function may not be available. Even if we have the graph, it may not be easy to point out the absolute and local extrema. For example, in the fig.22.26 that we saw at the beginning of this section, there are several points below the horizontal yellow line. We will be able to point out the absolute minimum, only if we zoom in.
2. So it is clear that, we need an analytical method to find the extrema. In this regard, we must first become familiar with critical points.
3. Consider the graph in fig.22.28 above.
• Local extrema occur at x=a, x=b and x=c. We can write two important facts about these three points.
(i) At x=a and x=b, the derivative f'(x) is zero. (Note that, the tangents at these points are horizontal. So slope of tangents at these points is zero)
(ii) At x=c, the derivative does not exist. (At corner points, derivative does not exist)
4. Consider all possible input values of a function (ie., all elements of the domain). Based on the two facts written in (3), we can form two categories:
Category I:
The input values at which derivative is zero, will fall in this category.
Category II:
The input values at which derivative does not exist, will fall in this category.
5. For a given function, if an input x value, falls in category I OR category II, then that x value is a critical point.
• So for the graph in fig.22.28 above, x=a, a=b and x=c are critical points.
6. The above step (5) gives us an effective way to find critical points. But we must keep in mind that, all critical points do not represent extrema. Let us see an example:
• Fig.22.29 below shows the graph of f(x) = x³.
The derivative is given by: f'(x) = 3x²
So at x = 0, the derivative is zero. (tangent is horizontal)
Since the derivative is zero, it is a critical point. But it is not an extremum.

Fig.22.29

• Let us see another example.
Fig.22.30 below shows the graph of $\rm{f(x) \,=\,2(x)^{5/3} \,-\,x^{2/3}}$.

Fig.22.30

At x = 0, we cannot draw the tangent because it is a cusp point. Remember that, we cannot draw tangents at corner points and cusp points.
So at x = 0, the derivative does not exist.
Since the derivative does not exist, it is a critical point. But it is not an extremum.

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Now we will see a solved example on critical points
Solved Example 22.42
Find all critical points for the function f(x) = x³ − 3x + 3
Solution:
1. We have f'(x) = 3x² − 3 = 3(x² − 1)
2. Equating f'(x) to zero, we get:
3(x² − 1) = 0
⇒ (x² − 1) = 0
⇒ x = +1 or x = −1
4. So the points in category I are: x = 1 and x = −1
5. We obtained f'(x) = 3(x² − 1)
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
6. So the critical points are: x = 1 and x = −1
7. Fig.22.31 shows the graph:
 

Fig.22.31

• From the graph, it is clear that:
   ♦ x=1 is a critical point. And it is the absolute minimum.
   ♦ x=−1 is a critical point. And it is the absolute maximum.

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In the next section, we will see a few more solved examples on critical points.

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