# Examples on Monotonicity Set 2

Go back to  'Applications of Derivatives'

Example – 13

Separate the interval $$\left[ {0,{\rm{ }}\pi /2} \right]{\rm{ }}$$ into sub-intervals in which $$f\left( x \right) = {\sin ^4}x + {\cos ^4}x$$  is increasing or decreasing.

\begin{align}{\rm\bf{Solution:}} f'\left( x \right) &= 4{\sin ^3}x\cos x - 4{\cos ^3}x\sin x\\\\& = 4\sin x\cos x\left( {{{\sin }^2}x - {{\cos }^2}x} \right)\\\\&= 2\sin 2x\left( { - \cos 2x} \right)\\\\ &= - \sin 4x\end{align}

We now need to consider the sign of f '(x) in the interval $$\left[ {0,{\rm{ }}\pi /2} \right]{\rm{ }}$$.

\begin{align}{\rm\bf{Interval (s)\; of \;strict\; increase:}}\;\;&f'\left( x \right) > 0\\\\\Rightarrow \qquad &- \sin 4x > 0\\\\\Rightarrow \qquad &\sin 4x < 0\\\\\Rightarrow\qquad &\pi < 4x < 2\pi \qquad\qquad \left( \begin{gathered}{\rm{This\,\, range\,\, of \,4}}x{\rm{\,\, will \,\,ensure }}\\{\rm{that\,\, }}x\;{\rm{itself \,\,lies\,\, in }}\left[ {0,\frac{\pi }{2}} \right]\end{gathered} \right)\\\Rightarrow \qquad &\frac{\pi }{4} < x < \frac{\pi }{2}\\\\{\rm\bf{Interval(s)\; strict \;decrease:}} \;\;&f'\left( x \right) < 0\\\\\Rightarrow \qquad &- \sin 4x < 0\\\\\Rightarrow \qquad &0 < 4x < \pi\\\\\Rightarrow \qquad &0 < x < \pi /4\end{align}

Therefore, f(x) decreases in  $$\left[ {0,{\rm{ }}\pi /4} \right]$$and increases in \begin{align}\left[ {\frac{\pi }{4},\frac{\pi }{2}} \right]\end{align}. The minimum value in \begin{align}\left[ {0,\pi /2} \right]\end{align} is at \begin{align}x = \pi /4\end{align} equal to \begin{align}f\left( x \right) = \frac{1}{2}\end{align}   and the maximum value is at x = 0 or  \begin{align}x = \frac{\pi }{2}\end{align} equal to f(x) = 1. The graph is approximately sketched below:

Example – 14

Determine the intervals of monotonicity of the function \begin{align}f\left( x \right) = \frac{{{x^2} + x + 1}}{{{x^2} - x + 1}}\end{align} .

\begin{align}{\rm\bf{Solution:}} \quad f'\left( x \right)&= \frac{{\left( {{x^2} - x + 1} \right)\left( {2x + 1} \right) - \left( {{x^2} + x + 1} \right)\left( {2x - 1} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\\\&= \frac{{\left( {2{x^3} - {x^2} + x + 1} \right) - \left( {2{x^3} + {x^2} + x - 1} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\\\ &= \frac{{ - 2\left( {{x^2} - 1} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\end{align}

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

Therefore, f(x) strictly increases in (–1, 1) and strictly decreases in $$\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)$$ .

We will be able to sketch the graphs of such functions accurately after going through the section on Maxima / Minima. However, you are still urged to give it a try for this example using the knowledge you’ve gained upto this point.