Applications of Derivatives Set 1

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Example - 1

If \(g\left( x \right) = f\left( x \right) + f\left( {1 - x} \right)\) and \(\begin{align}f''\left( x \right) < 0\,{\text{for all}}\,x \in \left[ {0\,,1} \right],\end{align}\) prove that g(x) is increasing in [0, ½) and decreasing in (½, 1].

Solution: Our requirement is to somehow show that \(g'\left( x \right) > 0\,\,\forall x \in (0,\,\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} )\) and \(g'\left( x \right) < 0\forall x \in (\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} ,1]\).

From the given functional relation between (x) and g(x):

\[g'\left( x \right) = f'\left( x \right) - f'\left( {1 - x} \right)\]

Therefore, we must show that:

\[f'\left( x \right) > f'\left( {1 - x} \right)\,\,\,\,\,\,\,\forall x \in [0,\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} )\,\,\,\,\,...{\rm{ }}\left( i \right)\]

\[and\qquad  f'\left( x \right) < f'\left( {1 - x} \right)\,\,\,\,\,\,\,\,\,\forall x \in (\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} ,\,1]\,\,\,\,\,\,\,\,\,\,...{\rm{ }}\left( {ii} \right)\]

Since \(f''\left( x \right) < 0\,\,\,\forall x \in [0,\,1],\,\,f'\left( x \right)\) is decreasing on [0, 1]. This means that if we take any x value in [0, ½), (1– x) will be greater than x so that \(f'\left( {1 - x} \right)\) will be less than\(f'\left( x \right)\). In other words, (i) is satisfied by virtue of the fact that f'(x) is decreasing.

On similar lines, when we assume any x value in (½, 1], we will see that (ii) is also satisfied for the same reason (that f'(x) is decreasing).

\( \Rightarrow \quad  \) g(x) satisfies the stated assertion

 

Example - 2

Let \(\begin{align}f\left( x \right) = \frac{{\ln \left( {\pi  + x} \right)}}{{\ln \left( {e + x} \right)}}\end{align}\). Prove that f(x) is decreasing on  \(\left[ {0,\,\infty } \right)\)

Solution:

\(\begin{align}f'\left( x \right)=\frac{\begin{align}\frac{\ln \left( e+x \right)}{\pi +x}-\frac{\ln \left( \pi +x \right)}{e+x}\end{align}}{{{\left( \ln \left( e+x \right) \right)}^{2}}}\end{align}\)

\[\begin{align} &= \frac{{\left( {e + x} \right)\ln \left( {e + x} \right) - \left( {\pi  + x} \right)\ln \left( {\pi  + x} \right)}}{{\left( {e + x} \right)\left( {\pi  + x} \right){{\left( {\ln \left( {e + x} \right)} \right)}^2}}}\\& = \frac{{g\left( x \right)}}{{h\left( x \right)}}\qquad \text {{This substitution was done for convenience}}\end{align}\]

To determine the sign of \(f'\left( x \right)\) in \(\left[ {0,\,\infty } \right)\), we first note that \(h\left( x \right) > 0\,\,\forall \,x \in \left[ {0,\infty } \right)\), so that we need to only worry about the sign of g(x). The form of g(x) suggests that we can construct a new function \(G\left( x \right) = x\ln x\) to determine the sign of g(x) as follows:

\[\begin{align}&\qquad\;\;  G\left( x \right)=x\ln x \\\\& \Rightarrow \quad  G'\left( x \right)=1+\ln x \\\\
 & \Rightarrow\quad  G'\left( x \right)>0\,\,\forall x\in \left( \frac{1}{e},\infty  \right) \\ & and\,\,G'\left( x \right)<0\,\,\forall \,x\in \left( 0,\frac{1}{e} \right) \\\\ & \Rightarrow \quad  G\left( x \right)\,\,\text{is}\,\,\text{increasing}\,\text{on}\left( \frac{1}{e},\infty  \right) \\\\ & \Rightarrow \quad x\ln x\,\,\text{increases}\,\text{on}\left( \frac{1}{e},\infty  \right) \\ \\& \Rightarrow\quad  \,\,\left( \pi +x \right)\ln \left( \pi +x \right)>\left( e+x \right)\ln \left( e+x \right)\,\,\,\,\,\,\forall x\in \left[ 0,\infty  \right)\left\{ \begin{gathered}& \text{since}\left( \pi +x \right)>\left( e+x \right) \\ 
 & >\frac{1}{e}\forall x\in \left[ 0,\infty  \right) \\ \end{gathered} \right\} \\\\ & \Rightarrow \quad g\left( x \right)<0\ \,\,\,\,\,\,\forall \,x\in \left[ 0,\infty  \right) \\\\ & \Rightarrow \quad f'\left( x \right)<0\ \,\,\,\,\,\forall \,x\in \left[ 0,\infty  \right)\, \\\\ & \Rightarrow \quad  f\left( x \right)\text{is}\,\,\text{decreasing}\,\,\text{on}\left[ 0,\infty  \right) \\ 
\end{align}\]

 

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