Applications of Derivatives Set 10

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Example - 22

If $$\alpha ,\beta$$ and $$\gamma$$ are the roots of $${x^3} + {x^2} - 5x - 1 = 0,$$find the value of $$\left[ \alpha \right] + \left[ \beta \right] + \left[ \gamma \right].$$

Solution: It is obvious that the required value can be found out if for each root we can find out the two (successive) integers between which that root lies. For example, if $$\alpha$$ lies between $$I$$ and  $$I+1$$ , then [$$\alpha$$] = $$I$$ and so on.

Let

$$f\left( x \right) = {x^3} + {x^2} - 5x - 1$$

$$\Rightarrow \qquad f'\left( x \right) = 3{x^2} + 2x - 5$$

$$= \left( {x - 1} \right)\left( {3x + 5} \right)$$

$$\Rightarrow \qquad f'\left( x \right) > 0\,\,{\rm{for}}\,\,x \in \left( { - \infty , - 5/3} \right) \cup \left( {1,\infty } \right)$$ $${\rm{and}}\,f'\left( x \right) < 0\,\,{\rm{for}}\,\,x \in \left( { - 5/3,1} \right)$$

\begin{align}\Rightarrow \qquad f\left( x \right){\text{increases}}\,{\text{in}}\left( { - \infty ,\frac{{ - 5}}{3}} \right),\,{\text{decreases}}\,\,{\text{in}}\,\left( {\frac{{ - 5}}{3},1} \right) \text{and again increases in} \left( {1,\infty } \right)\end{align}

Also \begin{align}f\left( {\frac{{ - 5}}{3}} \right) \equiv \frac{{128}}{{27}}{\rm{and}}\,f\left( 1 \right) = - 4.\end{align} Additionally, $$f(0)=-1$$

The approximate graph for f(x) is drawn above.

We see that the three roots of f(x) lie not far from the origin and their approximate location can be found out by evaluating f(x) for different integers close to 0.

\begin{align}&\quad \;f\left( 0 \right)=0+0-0-1=-1<0 \\\\ & \left. \begin{gathered}& f\left( 1 \right)=1+1-5-1=-4<0 \\\\ & f\left( 2 \right)=8+4-10-1=1>0 \\\\ \end{gathered} \right\}\begin{gathered}\text{One}\,\text{root}\,\,\alpha \,\,\text{lies}\,\,\text{between}\,1\,\,\text{and}\,\,2 \\\\ \Rightarrow\quad [\alpha ]=1\end{gathered} \\ \\ & \quad\left. f\left( -1 \right)=-1+1+5-1=4>0 \right\}\begin{gathered}\text{One}\,\text{root}\,\,\beta \,\,\text{lies}\,\,\text{between}\,0\,\,\text{and}\,\,-1 \\\\ \Rightarrow\quad [\beta ]=-1 \end{gathered}\\ \\ & \left. \begin{gathered}\\\\& f\left( -2 \right)=-8+4+10-1\,\,=\,5>0 \\\\ & f\left( -3 \right)=-27+9+15-1=-4<0 \end{gathered} \right\}\begin{gathered}\text{One}\,\text{root}\,\,\gamma \,\,\text{lies}\,\,\text{between}\,-2\,\,\text{and}\,\,-3 \\\\ \Rightarrow\quad [\gamma ]=-3 \end{gathered}\\ \\ &\quad \Rightarrow\quad [\alpha ]+[\beta ]+[\gamma ]=-3 \\ \end{align}

Example - 23

The function $$y = f\left( x \right)$$ is represented parametrically as:

\begin{align}&{x = g\left( t \right) = {t^5} - 5{t^3} - 20t + 7}\\\\&{y = h\left( t \right) = 4{t^3} - 3{t^2} - 18t + 3}\end{align}\,\,\,\,\left( { - 2 < t < 2} \right)

Find the extrema points of this function.

Solution: We need to employ parametric differentiation here to determine the points where \begin{align}\frac{{dy}}{{dx}} = 0\end{align}:

\begin{align}&g'\left( t \right) = \frac{{dx}}{{dt}} = 5{t^4} - 15{t^2} - 20\\\\&\qquad\qquad \quad= 5\left( {{t^4} - 3{t^2} - 4} \right)\\\\& \qquad\qquad \quad= 5\left( {{t^2} - 4} \right)\left( {{t^2} + 1} \right)\end{align}

For$$\;t \in \left( { - 2,2} \right),\,\,\,{t^2} - 4 < 0\,\,\,{\rm{so}}\,\,{\rm{that}}\,\,\,g'\left( t \right) < 0\forall t \in \left( { - 2,\,\,2} \right),\,\,g'\left( t \right) \ne 0\,\,{\rm{for}}\,{\rm{any}}\,t\,{\rm{in}}\left( { - 2,2} \right)$$

\begin{align}&h'\left( t \right) = \frac{{dy}}{{dt}} = 12{t^2} - 6t - 18\\\\&\qquad = 6\left( {2{t^2} - t - 3} \right)\\\\& \qquad = 6\left( {2t - 3} \right)\left( {t + 1} \right)\\\\&h'\left( t \right) = 0\,\,\,{\rm{when}}\,\,t = - 1,\,\,3/2\\\\ &\Rightarrow \quad \frac{{dy}}{{dx}} = \frac{{h'\left( t \right)}}{{g'\left( t \right)}} = 0\,\,{\rm{when}}\,\,t = - 1,\,\,3/2\,\end{align}

Now,

\begin{align}\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{g'\left( t \right)h''\left( t \right)-h'\left( t \right)g''\left( t \right)}{{{\left( g'\left( t \right) \right)}^{3}}}\end{align}

\begin{align} &\quad \quad \text{ }\!\!~\!\!\text{ }\quad \text{ }\!\!~\!\!\text{ }\quad ={{\left. \frac{{h}''\left( t \right)}{{{\left( {g}'\left( t \right) \right)}^{2}}} \right|}_{t=-1,3/2}}\left( {h}'\left( t \right)=0\;\text{for}\;t=-1,3/2 \right) \\ \\ & {{\left. {h}''\left( t \right) \right|}_{t=-1}}={{\left. 24t-6 \right|}_{t=-1}}=-30<0 \\ \\ & {{\left. {h}''\left( t \right) \right|}_{t=3/2}}={{\left. 24t-6 \right|}_{t=3/2}}=30>0 \\\\ & \Rightarrow \text{ }\!\!~\!\!\text{ }\quad {{\left. \frac{{{d}^{2}}y}{d{{x}^{2}}} \right|}_{t=-1}}<0\;\text{and}\;{{\left. \frac{{{d}^{2}}y}{d{{x}^{2}}} \right|}_{t=3/2}}>0 \\\\ & \Rightarrow \text{ }\!\!~\!\!\text{ }\quad t=\text{ }\!\!~\!\!\text{ }-1\text{ is a maximum and }t=3/2\text{ is a minimum for }y=f(x) \\ \end{align}

Example - 24

Prove the following inequality:

\begin{align}x - \frac{{{x^3}}}{3} < {\tan ^{ - 1}}x < x - \frac{{{x^3}}}{6}\,\,\,\,\forall x \in \left( {0,\,\,1} \right]\end{align}

Solution: Consider

$f\left( x \right) = {\tan ^{ - 1}}x - \left( {x - \frac{{{x^3}}}{3}} \right)$

\begin{align}& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= {\tan ^{ - 1}}x - x + \frac{{{x^3}}}{3}\\\\& \Rightarrow f'\left( x \right) = \frac{1}{{1 + {x^2}}} - 1 + {x^2}\\\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^4}}}{{1 + {x^2}}} > 0\,\,\,\,\,\forall x \in \left( {0,\,\,1} \right]\end{align}

\begin{align} \Rightarrow \qquad & \text{ }\!\!~\!\!\text{ }f\left( x \right)\text{is increasing on }(0,1] \\ \Rightarrow \qquad &\text{ }\!\!~\!\!\text{ }f\left( x \right)>f\left( 0 \right)\forall x\in (0,1] \\ \end{align}

Since (0) = 0

$\Rightarrow {\tan ^{ - 1}}x > x - \frac{{{x^3}}}{3}\,\,\,\,\,\forall x \in (0,\,\,1]$

The right hand side inequality can analogously be proved.

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