Examples on Monotonicity Set 1

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Example – 11

Determine the intervals in which the following functions are increasing or decreasing:

\(\qquad \qquad{\rm{\bf(a)}}\;\;{f( x) = {x^3} - x} \qquad \qquad {\rm{\bf(b)}}\;\;f\left( x \right) = {x^3} - 6{x^2} + 11x - 6\)

 Solution: In this and subsequent questions where we are required to find out the intervals of increase/decrease, we first determine f '(x).  f(x) increases in all intervals where f '(x) > 0 and decreases in all intervals where f '(x) < 0.

               (a)      \(f'\left( x \right) = 3{x^2} - 1\)

\(\begin{align}\rm{{\bf Interval(s)\;of\;strict\;increase:}}\;\;&f'\left( x \right) > 0\\\\
 &\Rightarrow \,\,\,3{x^2} - 1 > 0\\\\ &\Rightarrow \,\,\,x < \frac{{ - 1}}{{\sqrt 3 }}{\rm{\, or \,}}x > \frac{1}{{\sqrt 3 }}\end{align}\)

\(\begin{align}\rm{\bf{Interval(s)\;of\;strict\;decrease:}}\;\;&f'\left( x \right) < 0\\\\&
\Rightarrow \,\,\,3{x^2} - 1 < 0\\\\&\Rightarrow \,\,\,\frac{{ - 1}}{{\sqrt 3 }} < x < \frac{1}{{\sqrt 3 }}\end{align}\)

Therefore, f (x) increases in \(\begin{align}\left( { - \infty ,\frac{{ - 1}}{{\sqrt 3 }}} \right) \cup \left( {\frac{1}{{\sqrt 3 }},\infty } \right)\end{align}\) and decreases in\(\begin{align}\left( {\frac{{ - 1}}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)\end{align}\). The graph for f(x) confirms this: (to plot the graph, the knowledge of roots of f(x) helps, which is easy  to obtain for this example; \({x^3} - x = 0 \Rightarrow x = 0, \pm 1)\)

(b)  \(f'\left( x \right) = 3{x^2} - 12x + 11\)

The roots of f '(x) are \(\begin{align}\frac{{12 \pm \sqrt {12} }}{6} = \frac{{6 \pm \sqrt 3 }}{3} = 2 \pm \frac{1}{{\sqrt 3 }}\end{align}\)

\(\begin{align}{\rm\bf{Interval(s)\;of\;strict\;increase:}}\;\;&f'\left( x \right) > 0\\\\&\Rightarrow \,\,\,3{x^2} - 12x + 11 > 0\\\\&\Rightarrow \,\,\,x < 2 - \frac{1}{{\sqrt 3 }}\;{\rm{  or }}\;x > 2 + \frac{1}{{\sqrt 3 }}\end{align}\)

\(\begin{align}{\rm\bf{Interval(s)\;of\;strict\;decrease:}}\;\;&f'\left( x \right) < 0\\\\&\Rightarrow \,\,\,3{x^2} - 12x + 11 < 0\\\\&\Rightarrow \,\,\,2 - \frac{1}{{\sqrt 3 }} < x < 2 + \frac{1}{{\sqrt 3 }}\end{align}\)

f(x) can be factorised as (x – 1)(x – 2)(x – 3) so that the roots of f(x) are x = 1, 2, 3. The graph for f(x) is approximately sketched below:

Example - 12

Determine the values of x for which \(f(x) = {x^x}\), x > 0 is increasing or decreasing.

Solution:   To find f '(x), we first take the logarithm of both sides of the given equation:

    \[\ln \left( {f\left( x \right)} \right) = x\ln x\]

Differentiating both sides, we get:

\(\begin{align}&\frac{1}{{f\left( x \right)}} \cdot f'\left( x \right) = x \cdot \frac{1}{x} + \ln x \cdot 1\\\\ &\qquad \qquad \quad \;\;= 1 + \ln x\\\\\Rightarrow \qquad &f'\left( x \right) = {x^x}\left( {1 + \ln x} \right)\end{align}\)

\(\begin{align}{\rm\bf{Interval(s)\;of \;strict\;increase}:}\;\; &f'\left( x \right) > 0\\\\ \Rightarrow \quad& {x^x}\left( {1 + \ln x} \right) > 0\\\\\Rightarrow \quad &1 + \ln x > 0\\\\\Rightarrow \quad &x > {e^{ - 1}} = \frac{1}{e}\\\\
{\rm\bf{Interval(s)\;of\;strict\;decrease:}}\;\;&f'\left( x \right) < 0\\\\\Rightarrow \quad &{x^x}\left( {1 + \ln x} \right) < 0\\\\\Rightarrow \quad &x < \frac{1}{e}\end{align}\)

 \(\begin{align}\text{To plot the graph of f(x), notice that} \mathop {\lim }\limits_{x \to 0} f\left( x \right)&= \mathop {\lim }\limits_{x \to 0} {x^x}\\\\&= \mathop {\lim }\limits_{x \to 0} {e^{x\ln x}} = {e^{\mathop {\lim }\limits_{x \to 0} .x\ln x}} = {e^0} = 1.\\\\\rm{Also,} \qquad \qquad &\quad\mathop {\lim }\limits_{x \to \infty } \left( {f\left( x \right)} \right) = \infty .\end{align}\)

 f(x) decreases in (0, 1/e) and increases in \(\begin{align}\left( {\frac{1}{e},\infty } \right)\end{align}.\)  

Download practice questions along with solutions for FREE:
Applications of Derivatives
grade 11 | Questions Set 1
Applications of Derivatives
grade 11 | Answers Set 1
Applications of Derivatives
grade 11 | Questions Set 2
Applications of Derivatives
grade 11 | Answers Set 2