Applications of Derivatives Set 12

Example - 26

Find the minimum value of

\[f(x) = |\sin x + \cos x + \tan x + \cot x + \sec x + {\rm{cosec}}\;x\;|\]

Solution:         \(\begin{align}f(x) = \left| {\;\sin x + \cos x + \frac{1}{{\sin x\cos x}} + \frac{{\sin x + \cos x}}{{\sin x\cos x}}\;} \right|\end{align}\)

Using \(\sin x + \cos x = t,\) and therefore \(\sin x\cos x = \frac{{{t^2} - 1}}{2}\), we have

\[\begin{align}&f(t) = \left| {\;t + \frac{{2(1 + t)}}{{{t^2} - 1}}\;} \right| = \left| {\;t + \frac{2}{{t - 1}}\;} \right|\\\\&\,\,\,\,\,\,\,\,\,\,\, = \left| {\;(t - 1) + \frac{2}{{t - 1}}\; + 1\;} \right|\\\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \le 2\sqrt 2  + 1\end{align}\]

The maximum value is \(2\sqrt 2  + 1\)

Example - 27

Let \(f(x) = \left( {{x^2} + bx + c} \right){e^x}\)

(a) Does \(f > 0\;{\rm{imply}}\;f' > 0\)             (b) Does \(f' > 0\;{\rm{imply}}\;f > 0\)

Solution:         We have \(\begin{align}f'(x) = \left( {{x^2} + (b + 2)x + b + c} \right){e^x}\end{align}\). Thus,

\[f > 0   \Leftrightarrow  {b^2} - 4c < 0\;\;{\rm{and}}\;\;f' > 0\;\;\; \Leftrightarrow \;\;\;{b^2} - 4c + 4 < 0\]

(a) Let\(f > 0\). Thus,\({b^2} - 4c < 0\), but nothing can be said about the sign of \({b^2} - 4c + 4\), i.e., it may be positive \(f > 0\) Thus,  does not necessarily imply  \(f' > 0\).

(b) Let  \(f > 0.\) Thus,\({b^2} - 4c + 4 < 0\), which necessarily implies that \({b^2} - 4c < 0,\;{\rm{i}}{\rm{.e}}.,\;f > 0\).

Example - 28

Find the minimum value of \(|f(x)|\) given by

\[f(x) = \frac{{{{\left( {x + \frac{1}{x}} \right)}^6} - \left( {{x^6} + \frac{1}{{{x^6}}}} \right) - 2}}{{{{\left( {x + \frac{1}{x}} \right)}^3} + \left( {{x^3} + \frac{1}{{{x^3}}}} \right)}}\]

Solution:         Substitute \(\begin{align}{\left( {x + \frac{1}{x}} \right)^3} = a\;\;{\rm{and}}\;\;{x^3} + \frac{1}{{{x^3}}} = b\end{align}\). Thus,

\[\begin{align}&f(x) = \left| {\;\frac{{{a^2} - {b^2}}}{{a + b}}\;} \right| = \;|a - b|\; = \left| {\;{{\left( {x + \frac{1}{x}} \right)}^3} - {x^3} - \frac{1}{{{x^3}}}\;} \right|\\\\&\,\,\,\,\,\,\,\,\,\,\, = 3\left| {\;x + \frac{1}{x}\;} \right| \ge 6\end{align}\]

The minimum value of \(|f(x)|\)is 6.