In the previous section, we saw some miscellaneous examples. In this section, we will see a few more miscellaneous examples.
Solved example 22.77
Find the intervals in which the function given by
$\rm{f(x)\,=\,\frac{3}{10} x^4 - \frac{4}{5} x^3 - 3 x^2 + \frac{36}{5} x + 11}$
is (a) strictly increasing (b) strictly decreasing.
Solution:
1. First we write the derivative:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{f(x)} & {~=~} &{\frac{3}{10} x^4 - \frac{4}{5} x^3 - 3 x^2 + \frac{36}{5} x + 11} \\
{~\color{magenta} 2 } &{\implies} &{f'(x)} & {~=~} &{\frac{3(4)}{10} x^3 - \frac{4(3)}{5} x^2 - 3(2) x + \frac{36}{5} + 0} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{6}{5} x^3 - \frac{12}{5} x^2 - 6 x + \frac{36}{5}} \\
\end{array}$
2. Equating the above result to zero, we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{f'(x)} & {~=~} &{0} \\
{~\color{magenta} 2 } &{\implies} &{\frac{6}{5} x^3 - \frac{12}{5} x^2 - 6 x + \frac{36}{5}} & {~=~} &{0} \\
{~\color{magenta} 3 } &{\implies} &{6 x^3 - 12 x^2 - 30 x + 36} & {~=~} &{0} \\
{~\color{magenta} 4 } &{\implies} &{x^3 - 2 x^2 - 5 x + 6} & {~=~} &{0} \\
\end{array}$
3. Solving the above third degree equation, we get:
x = −2, x = 1 and x = 3
•
So the domain R can be divided into four intervals:
(−∞,−2), (−2,1), (1,3) and (3,∞)
4. Consider the first interval (−∞,−2)
−3 is a convenient point in this interval.
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{f'(-3)} & {~=~} &{\frac{6}{5} (-3)^3 - \frac{12}{5} (-3)^2 - 6 (-3) + \frac{36}{5}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{-28.8} \\
\end{array}$
• So f '(x) is −ve in this interval. That means, f(x) is decreasing in this interval.
• Note that, f '(−2) will be zero. But −2 is not an input value. So we can write:
The given f is strictly decreasing in (−∞,−2)
5. Consider the second interval (−2,1)
0 is a convenient point in this interval.
$\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{f'(0)} &
{~=~} &{\frac{6}{5} (0)^3 - \frac{12}{5} (0)^2 - 6 (0) +
\frac{36}{5}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{7.2} \\
\end{array}$
• So f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.
• Note that, f '(−2) and f '(0) will be zero. But −2 and 1 are not an input values. So we can write:
The given f is strictly increasing in (−2,1)
6. Consider the third interval (1,3)
2 is a convenient point in this interval.
$\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{f'(2)} &
{~=~} &{\frac{6}{5} (2)^3 - \frac{12}{5} (2)^2 - 6 (2) +
\frac{36}{5}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{-4.8} \\
\end{array}$
• So f '(x) is −ve in this interval. That means, f(x) is decreasing in this interval.
• Note that, f '(1) and f '(3) will be zero. But 1 and 3 are not an input values. So we can write:
The given f is strictly decreasing in (1,3)
7. Consider the fourth interval (3,∞)
4 is a convenient point in this interval.
$\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{f'(4)} &
{~=~} &{\frac{6}{5} (4)^3 - \frac{12}{5} (4)^2 - 6 (4) +
\frac{36}{5}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{21.6} \\
\end{array}$
• So f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.
• Note that, f '(3) will be zero. But 3 is not an input value. So we can write:
The given f is strictly increasing in (3,∞)
Solved example 22.78
Show that the function f given by
f(x) = tan−1(sin x + cos x), x > 0 is always a strictly increasing function in (0, π/4).
Solution:
1. First we write the derivative:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{f(x)} & {~=~} &{\tan^{-1}(\sin x + \cos x)} \\
{~\color{magenta} 2 } &{\implies} &{f'(x)} & {~=~} &{\left(\frac{1}{1 + (\sin x + \cos x)^2} \right) (\cos x - \sin x)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\left(\frac{1}{1 + \sin^2 x + 2 \sin x \cos x + \cos^2 x} \right) (\cos x - \sin x)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\left(\frac{1}{2 + \sin(2 x)} \right) (\cos x - \sin x)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{\cos x - \sin x}{2 + \sin(2 x)}} \\
\end{array}$
2. In this problem, there is no need to equate f'(x) to zero and find the intervals. This is because, we are already given an interval. We are asked to show that, the function is always strictly increasing in that interval.
•
The given interval is (0, π/4). The boundaries of this interval may not be consecutive critical points. So the usual method of checking using a 'convenient number' cannot be used. We must show the general case.
3. We obtained the derivative as: $\frac{\cos x - \sin x}{2 + \sin(2 x)}$
•
The largest input possible from the given interval is a value just below π/4.
•
Even if the input is π/4, sin(2x) will be in the first quadrant. Sine is always +ve in the first quadrant. So (2 + sin(2x)) is +ve in the given interval.
•
That means, denominator of the derivative is +ve in the given interval.
4. Now we can write, the derivative will be +ve if the numerator is also +ve.
That means: Derivative will be +ve if cos x − sin x > 0.
⇒ cos x / sin x − sin x / sin x > 0
⇒ cot x − 1 > 0
⇒ cot x > 1
•
In the given interval (0, π/4), tan x is always less than 1. So cot x is always greater than 1.
•
So we can write: the numerator is also +ve.
5. We see that, f '(x) is +ve in the interval (0, π/4).
Therefore, f(x) is increasing in this interval.
6. Note that,
♦ f '(0) is 1/2. It is +ve.
♦ f '(π/4) is zero.
•
But π/4 is not an input value. So we can write:
The given f is strictly increasing in (0, π/4)
7. Fig.22.75 below shows the graph.
Fig.22.75 |
Solved example 22.79
Find the intervals in which the function f given by
$f(x)\,=\,\frac{4 \sin x - 2x - x \cos x}{2 + \cos x}$
is (i) increasing (ii) decreasing.
Solution:
1. First we write the derivative:
Using quotient rule, we get:
• $f'(x)\,=\,-{\frac{\cos x(\cos x - 4)}{(2 + \cos x)^2}}$
2. Next, we equate the above result to zero:
•
The denominator cannot be zero. So we get:
cos x (cos x − 4) = 0
•
cos x cannot be 4. So we get: cos x = 0
3. Solving the equation cos x = 0, we get:
x = π/2 and x = 3π/2 as the principal solutions
•
So the angle of the unit circle can be divided into three intervals:
(0, π/2), (π/2, 3π/2) and (3π/2, 2π)
4. Consider the first interval (0, π/2)
•
π/4 is a convenient point in this interval.
$f'(\pi/4)\,=\,-{\frac{\cos (\pi/4)(\cos (\pi/4) - 4)}{(2 + \cos (\pi/4))^2}}$
•
Here, the only −ve term is (cos(π/4) − 4).
• So f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.
5. Consider the second interval (π/2, 3π/2)
•
π is a convenient point in this interval.
$f'(\pi)\,=\,\frac{\cos (\pi)(\cos (\pi) - 4)}{(2 + \cos (\pi))^2}$
•
Here, both terms in the numerator are −ve. The denominator is +ve.
• So f '(x) is −ve in this interval. That means, f(x) is decreasing in this interval.
6. Consider the third interval (3π/2, 2π)
•
7π/4 is a convenient point in this interval.
$f'(7 \pi/4)\,=\,\frac{\cos (7\pi/4)(\cos (7\pi/4) - 4)}{(2 + \cos (7\pi/4))^2}$
•
cos(7π/4) = cos(π + 3π/4) = −cos(3π/4)
= −[cos(π/2 + π/4)] = −[−sin(π/4)] = sin(π/4)
•
So here, the only −ve term is (cos(7π/4) − 4).
• Thus f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.
7. Now we can write about the intervals.
•
f is:
♦ Increasing in (0, π/2) and (3π/2, 2π)
♦ Decreasing in (π/2, 3π/2)
8. Fig.22.76 below shows the graph.
Fig.22.76 |
•
Note that:
♦ π/2 = 1.57
♦ 3π/2 = 4.71
Solved example 22.80
Find the intervals in which the function f given by
$f(x)\,=\,x^3 + \frac{1}{x^3}$
is (i) increasing (ii) decreasing.
Solution:
1. First we write the derivative:
$f'(x)\,=\,3x^2 \,-\, \frac{3}{x^4}$
2. Next, we equate the above result to zero:
$3x^2 \,-\, \frac{3}{x^4}~=~0$
⇒ $x^2 \,-\, \frac{1}{x^4}~=~0$
⇒ $x^2 ~=~\frac{1}{x^4}$
⇒ $x^6 ~=~1$
⇒ $x\,=\,1~\text{and}~x\,=\,-1$
3. So the domain can be divided into three intervals:
(−∞, −1), (−1, 1) and (1, ∞)
4. Consider the first interval (−∞, −1)
•
−2 is a convenient point in this interval.
$f'(-2)\,=\,3(-2)^2 \,-\, \frac{3}{(-2)^4}$
= $12\,-\, \frac{3}{16}$
• So f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.
5. Consider the second interval (−1, 1)
•
0.5 is a convenient point in this interval.
$f'(0.5)\,=\,3(0.5)^2 \,-\, \frac{3}{(0.5)^4}$
= $\frac{3}{4}\,-\, \frac{30000}{625}$
• So f '(x) is −ve in this interval. That means, f(x) is decreasing in this interval.
6. Consider the third interval (1, ∞)
• 2 is a convenient point in this interval.
$f'(2)\,=\,3(2)^2 \,-\, \frac{3}{(2)^4}$
= $12\,-\, \frac{3}{16}$
• So f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.
7. Now we can write about the intervals.
•
f is:
♦ Increasing in (−∞, −1) (1, ∞)
♦ Decreasing in (−1, 1)
Solved example 22.81
Let f be a function defined on [a,b] such that f '(x) > 0 for all x ∈ (a,b). Then prove that f is an increasing function on (a,b).
Solution:
1. Let x1, x2 ∈ (a,b) such that x1 < x2.
2. f is defined on [a,b]. Also, [x1,x2] is a subset of [a,b].
So f is continuous and differentiable in [x1,x2]
3. By mean value theorem,
There exists a number c in (x1,x2) such that:
$f'(c)\,=\,\frac{f(x_2) - f(x_1)}{x_2 - x_1}$
4. Given that, f '(x) > 0 for all x ∈ (a,b).
•
So f '(c) > 0
⇒ $\frac{f(x_2) - f(x_1)}{x_2 - x_1}~>~0$
⇒ f(x2) − f(x1) > 0
⇒ f(x2) > f(x1)
5. So we can write:
When x2 > x1, we will obtain: f(x2) > f(x1)
•
That means, f is an increasing function on (a,b).
In the next section, we will see a few more miscellaneous examples.
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