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:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{(2-8x)^4 (x^2 - 9)^3}    \\ {~\color{magenta}    2    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^4 \left[3(x^2 - 9)^2 (2x)\right]~+~\left[4(2-8x)^3 (-8) \right] (x^2 - 9)^3}    \\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{(2-8x)^4 \left[(x^2 - 9)^2 (6x)\right]~-~\left[(32)(2-8x)^3 \right] (x^2 - 9)^3}    \\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{(2-8x)^3 (x^2 - 9)^2 \left[  (2-8x)  (6x)~-~(32)(x^2 - 9) \right]}    \\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{(2-8x)^3 (x^2 - 9)^2 \left[ 12x - 48 x^2 -32 x^2 + (32)(9) \right]}    \\ {~\color{magenta}    6    }    &{}    &{}    & {~=~}    &{(2-8x)^3 (x^2 - 9)^2 \left[ 12x - 80 x^2  + (32)(9) \right]}    \\ \end{array}$                           

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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