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