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.
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We know how to find all critical points, with out the help of a graph.
Let x1, x2, x3, . . . be the critical points.
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The next step naturally, is to find the f values at those points.
So we get: f(x1), f(x2), f(x3), . . .
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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.
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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.
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| 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
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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.
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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:
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| Fig.22.39 |
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The dashed magenta vertical lines represent the end points.
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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:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{f'(x)} & {~=~} &{\cos x - \sin x} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\sin (\pi/2 -x) - \sin x} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{2 \cos \left(\frac{\pi/2 - x + x}{2} \right) \sin \left(\frac{\pi/2 - x - x}{2} \right)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{2 \cos \left(\frac{\pi}{4} \right) \sin \left(\frac{\pi}{4} - x \right)} \\
\end{array}$
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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.
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Absolute maximum occurs at the critical point.
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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 |
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The dashed magenta vertical lines represent the end points.
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Absolute minimum occurs at the critical point.
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Critical point is same as the right end point.
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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.
In the next section, we will see shape of graph.
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