Examples on Monotonicity Set 3

Go back to  'Applications of Derivatives'

Example - 15

Let \(\begin{align}f\left( x \right) = \left\{ {\begin{gathered}{x{e^{ax}},}\qquad\qquad\quad{x \le 0}\\{x + a{x^2} - {x^3},}{\;\;\;\;\;x > 0}\end{gathered}} \right\}\end{align},\) where a is a positive constant. Find the intervals in which f '(x) is increasing.

Solution: Notice that we are required to find the intervals of increase of f '(x) and not f (x). Therefore, we need to first determine f '(x) from f (x), and then check the sign of the derivative of f '(x) in different intervals, i.e, the sign of f ''(x).

Observe that f (x) is continuous and differentiable at x = 0 so that f '(x) is defined at x = 0.

Therefore,

\[f'\left( x \right) = \left\{ {\begin{align}{\left( {1 + ax} \right){e^{ax}},} \qquad{x \le 0}\\{1 + 2ax - 3{x^2},} \qquad{x > 0}\end{align}} \right\}\]

Notice again that f '(x) is also continuous and differentiable at x = 0 so that f ''(x) is also defined at x = 0.

\[f''\left( x \right) = \left\{ {\begin{align}{\left( {2 + ax} \right)a{e^{ax}},} \qquad{x \le 0}\\{2a - 6x,}\qquad{x > 0}\end{align}} \right\}\]

\(\begin{align}{\rm\bf{Interval(s)\,of \;strict\;increase\;for \;f '(x):}}\quad &f''\left( x \right) > 0\\\\ \Rightarrow \qquad & 2 + ax > 0\left( {{\text{if }}\;x \le 0} \right){\text{ and }}\;2a - 6x > 0\left( {{\text{if }}\;x > 0} \right)\\\\\Rightarrow \qquad & x > \frac{{ - 2}}{a}\;\left( {{\text{if }}\;x \le 0} \right){\text{ and}}\;x < \frac{a}{3}\;\left( {{\text{if }}\;x > 0} \right)\\\\ \Rightarrow \qquad &\frac{{ - 2}}{a} < x \le 0\,\,\,{\rm{and}}\,\,\,0 < x < \frac{a}{3}\\\\ \Rightarrow \qquad & \frac{{ - 2}}{a} < x < \frac{a}{3}\end{align}\)

Therefore, f '(x) is strictly increasing on the interval \(\begin{align}\left( {\frac{{ - 2}}{a},\frac{a}{3}} \right)\end{align}.\)

Example - 16

For what values of \(\lambda \) does the function\(f\left( x \right) = \left( {\lambda  + 2} \right){x^3} - 3\lambda {x^2} + 9\lambda x - 1\) decrease for all x?

\(\begin{align}{\rm\bf{Solution:}} \quad &f '(x)\;{\text {must be negative for all x if f (x) is to decrease for all x.}}\\\\&f'\left( x \right) = 3\left( {\lambda + 2} \right){x^2} - 6\lambda x + 9\lambda\\\\ &f'\left( x \right) < 0 \qquad \forall\;\; x \in \mathbb{R}\end{align}\)

\[\begin{align}&\Rightarrow \qquad \qquad D\,\,{\rm{of }}\,f'\left( x \right) < 0 \qquad \qquad \qquad  \qquad \rm{and}  \qquad \lambda  + 2 < 0\\\\&\Rightarrow  \qquad \qquad 36{\lambda ^2} - 108\lambda \left( {\lambda  + 2} \right) < 0 \qquad  \qquad   \rm{and} \qquad  \lambda  + 2 < 0\\\\&\Rightarrow \qquad \qquad {\lambda ^2} - 3\lambda \left( {\lambda  + 2} \right) < 0 \qquad  \qquad  \qquad \rm{and} \qquad \lambda  <  - 2\\\\&\Rightarrow \qquad \qquad  2{\lambda ^2} + 6\lambda  > 0     \qquad  \qquad  \qquad  \qquad \;\;\;          \rm{and} \qquad  \lambda  <  - 2\\\\&\Rightarrow \qquad \qquad \,\lambda  <  - 3\,\,{\rm{or  }}\lambda  > 0 \qquad  \qquad \qquad \quad \;\;   \rm{and} \qquad \lambda  <  - 2\\\\&\Rightarrow \qquad \qquad  \lambda  <  - 3\end{align}\]

Therefore, if  \(\lambda  \in \left( { - \infty , - 3} \right),f\left( x \right)\) will decrease for all x.