# 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 $$g'\left( x \right) > 0\,\,\forall x \in (0,\,\raise.5ex\hbox{1}\kern-.1em/\kern-.15em\lower.25ex\hbox{2} )$$ and $$g'\left( x \right) < 0\forall x \in (\raise.5ex\hbox{1}\kern-.1em/\kern-.15em\lower.25ex\hbox{2} ,1]$$.

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:

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

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

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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