Chapter 6: Applications of Derivatives

Ans: Let the two numbers in $R$be ${{x}_{1}}$ and ${{x}_{2}}$.

$ {{x}_{1}}<{{x}_{_{2}}}\Rightarrow 3{{x}_{1}}<3{{x}_{2}} $ 

$ \Rightarrow 3{{x}_{1}}+17<3{{x}_{2}}+17 $ 

$ =f\left( {{x}_{1}} \right)<f\left( {{x}_{2}} \right)$ 

Thus, the function is strictly increasing on $R.$ 

Ans: Let the two numbers in $R$be ${{x}_{1}}$ and ${{x}_{2}}$.

Then, we have: 

$ {{x}_{1}}<{{x}_{2}}\Rightarrow 2{{x}_{1}}<2{{x}_{2}} $ 

$ \Rightarrow {{e}^{2{{x}_{1}}}}<{{e}^{2{{x}_{2}}}} $ 

$ \Rightarrow f\left( {{x}_{1}} \right)<f\left( {{x}_{2}} \right)$ 

Hence, $f$ is strictly increasing on $R.$ 

Ans: The given function is $f\left( x \right)=\sin x$.

$\therefore f'\left( x \right)=\cos x$ 

A) Since for each $x\in \left( 0,\frac{\pi }{2} \right)$, $\cos x>0$, we have $f'\left( x \right)>0$.

Hence, $f$ is strictly increasing in $\left( 0,\frac{\pi }{2} \right)$.

B) Since for each $x\in \left( \frac{\pi }{2},\pi  \right),\cos x<0,$ we have $f'\left( x \right)<0$.

Hence, $f$ is strictly increasing in $\left( \frac{\pi }{2},\pi  \right)$ 

C) It is clear from the results obtained in (A) and (B) that $f$ is neither increasing nor decreasing in $\left( 0,n \right)$.

Ans: The function provided is $f\left( x \right)=2{{x}^{2}}-3x$.

  $ f'\left( x \right)=4x-3 $ 

  $ \therefore f'\left( x \right)=0 $ 

  $ \Rightarrow x=\frac{3}{4} $ 

Now, the point $\frac{3}{4}$ divides the real line into two disjoint intervals i.e.,  

$\left( -\infty ,\frac{3}{4} \right)$  and $\left( \frac{3}{4},\infty  \right)$ .

In interval, $\left( -\infty ,\frac{3}{4} \right),f'\left( x \right)=4x-3<0$ 

Hence, the given function $\left( f \right)$ is strictly decreasing in interval $\left( -\infty ,\frac{3}{4} \right)$

In interval, $\left( \frac{3}{4},\infty  \right),f'\left( x \right)=4x-3>0$ 

Hence, the given function $\left( f \right)$ is strictly decreasing in interval $\left( \frac{3}{4},\infty  \right)$

 Ans: The given function is

 $ f\left( x \right)=2{{x}^{3}}-3{{x}^{2}}-36x+7 $ 

 $ f'\left( x \right)=6{{x}^{2}}-6x-36 $ 

 $ =6\left( {{x}^{2}}-x-6 \right) $ 

 $ =6\left( x+2 \right)\left( x-3 \right) $ 

 $ \therefore f'\left( x \right)=0 $ 

 $ \Rightarrow x=-2,3 $ 

The points $x=-2,3$ divide the real line into three disjoints intervals 

i.e., $\left( -\infty ,-2 \right),\left( -2,3 \right),$and $\left( 3,\infty  \right)$.

In intervals $\left( -\infty ,-2 \right)$ and $\left( 3,\infty  \right)$, ${f}'\left( x \right)$ is positive while in the interval $\left( -2,3 \right),f'\left( x \right)$ is negative.

Hence, the given function $\left( f \right)$is strictly increasing in intervals $\left( -\infty ,-2 \right)\cup \left( 3,\infty  \right)$ while the function $\left( f \right)$ is strictly decreasing in interval $\left( -2,3 \right)$.

Ans:  

a) Given that

$ f\left( x \right)={{x}^{2}}+2x-5 $ 

$ \therefore f'\left( x \right)=2x+2 $ 

Now, 

 $ f'\left( x \right)=0 $ 

 $ \Rightarrow x=-1 $

The real line is divided into two disjoint intervals $\left( -\infty ,-1 \right)and\left( -1,\infty  \right)$ by the point $x=-1$.

In interval $\left( -\infty ,-1 \right),$ 

$\therefore f'\left( x \right)=2x+2<0$ 

Therefore, $f$ is strictly decreasing in the interval $\left( -\infty ,-1 \right)$.

Thus $f$ is strictly decreasing for $x<-1$ 

In interval $\left( -1,\infty  \right),$ 

$\therefore f'\left( x \right)=2x+2>0$ 

Therefore, $f$ is strictly decreasing in interval $\left( -1,\infty  \right)$ 

Thus $f$ is strictly increasing for $x>-1$.

b) Given that,

$ f\left( x \right)=10-6x-2{{x}^{2}} $ 

$ \therefore f'\left( x \right)=-6-4x $ 

Now,

$ f'\left( x \right)=0 $ 

$ \Rightarrow x=\frac{-3}{2} $ 

The real line is divided into two disjoint intervals  $\left( -\infty ,-\frac{3}{2} \right)and\left( -\frac{3}{2},\infty  \right)$ by the point $x=-\frac{3}{2}$.

In interval $\left( -\infty ,-\frac{3}{2} \right)$ i.e., when $x<-\frac{3}{2},f'\left( x \right)=-6-4x<0.$ 

$\therefore f$ is strictly increasing for $x<-\frac{3}{2}$ 

In interval $\left( -\frac{3}{2},\infty  \right)$ i.e., when $x>\frac{3}{2},f'\left( x \right)=-6-4x<0$.

$\therefore f$ is strictly increasing for $x<-\frac{3}{2}$ 

c) Given that,

 $ f\left( x \right)=-2{{x}^{3}}-9{{x}^{2}}-12x+1 $ 

 $ \therefore f'\left( x \right)=-6{{x}^{2}}-18x-12 $ 

 $ =-6\left( {{x}^{2}}+3x+2 \right) $ 

 $ =-6\left( x+1 \right)\left( x+2 \right) $ 

Now,

 $ f'\left( x \right)=0 $ 

 $ \Rightarrow x=-1 $ 

and $x=-2$ 

The real line is divided into three disjoint interval $\left( -\infty ,-2 \right),\left( -2,-1 \right)and\left( -1,\infty  \right)$ by the points $x=-1,-2$  when $x<-2$ and $x>-1$,

$f'\left( x \right)=-6\left( x+1 \right)\left( x+2 \right)<0$ 

$\therefore f$ is strictly increasing for $x<-2<x>-1$ 

Now, in interval $\left( -2,-1 \right)$ i.e., when $-2<x<-1$,

$f'\left( x \right)=-6\left( x+1 \right)\left( x+2 \right)>0$ 

$\therefore f$ is strictly increasing for $-2<x<-1$

d) Given that,

$ f\left( x \right)=6-9x-{{x}^{2}} $ 

$ \therefore f'\left( x \right)=-9-2x $ 

Now, $f'\left( x \right)=0$ gives $x=-\frac{9}{2}$ 

The real line is divided into two disjoint intervals $\left( -\infty ,-\frac{9}{2} \right)and\left( -\frac{9}{2},\infty  \right)$ by the point $x=-\frac{9}{2}$.

In interval $\left( -\infty ,-\frac{9}{2} \right)$ i.e., for $x<\frac{9}{2},$ 

$\therefore f$  is strictly increasing for $x<\frac{9}{2}$

In interval $\left( \frac{9}{2},\infty  \right)$ i.e., for $x<-\frac{9}{2},f'\left( x \right)=-9-2x>0$ 

$\therefore f$  is strictly increasing for $x<-\frac{9}{2}$

In interval $\left( -\frac{9}{2},\infty  \right)$ i.e., for $x>-\frac{9}{2},f'\left( x \right)=-9-2x<0$ 

$\therefore f$  is strictly increasing for $x>-\frac{9}{2}$

e) Given that,

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

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

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

$ =6{{\left( x+1 \right)}^{2}}{{\left( x-3 \right)}^{2}}\left( x-1 \right) $ 

Now, 

$ f'\left( x \right)=0 $ 

$ \Rightarrow x=-1,3,1 $  

The points $x=-1,1,3$ divided the real line into four disjoint intervals. i.e., $\left( -\infty ,-1 \right),\left( -1,1 \right),\left( 1,3 \right)and\left( 3,\infty  \right)$ 

In intervals $\left( -\infty ,-1 \right)and\left( -1,1 \right),f'\left( x \right)=6{{\left( x+1 \right)}^{2}}{{\left( x-3 \right)}^{2}}\left( x-1 \right)<0$ 

$\therefore f$ is strictly decreasing in interval $\left( -\infty ,-1 \right)and\left( -1,1 \right)$ 

In intervals $\left( 1,3 \right)and\left( 3,\infty  \right),f'\left( x \right)=6{{\left( x+1 \right)}^{2}}{{\left( x-3 \right)}^{2}}\left( x-1 \right)>0$ 

$\therefore f$ is strictly decreasing in the interval $\left( 1,3 \right)and\left( 3,\infty  \right)$.

Ans: We are given that,

$ y=\log \left( 1+x \right)-\frac{2x}{2+x} $ 

$ \therefore \frac{dy}{dx}=\frac{1}{1+x}-\frac{\left( 2+x \right)\left( 2 \right)-2x\left( 1 \right)}{{{\left( 2+x \right)}^{2}}} $ 

 $ =\frac{{{x}^{2}}}{{{\left( 2+x \right)}^{2}}} $ 

Now,

$ \frac{dy}{dx}=0 $ 

 $ \Rightarrow \frac{{{x}^{2}}}{{{\left( 2+x \right)}^{2}}}=0 $ 

 $ \Rightarrow {{x}^{2}}=0 $ 

$ \Rightarrow x=0 $  

Since $x>-1$ point $x=0$ divides the domain $\left( -1,\infty  \right)$ in two disjoint 

intervals i.e., $-1<x<0$ and $x>0.$ 

 When $-1<x<0$we have:

  $ x<0\Rightarrow {{x}^{2}}>0 $ 

 $ x>-1\Rightarrow \left( 2+x \right)>0 $ 

 $ \Rightarrow {{\left( 2+x \right)}^{2}}>0 $ 

 \[\therefore y=\frac{{{x}^{2}}}{{{\left( 2+x \right)}^{2}}}>0\]

Also, when $x>0$ 

 $ x<0\Rightarrow {{x}^{2}}>0, $ 

 $ {{\left( 2+x \right)}^{2}}>0 $ 

 $ \therefore y=\frac{{{x}^{2}}}{{{\left( 2+x \right)}^{2}}}>0 $ 

 Hence, the function $f$ is increasing throughout this domain.

Ans: We have,         

  $ y={{\left[ x\left( x-2 \right) \right]}^{2}} $ 

 $ ={{\left[ {{x}^{2}}-2x \right]}^{2}} $ 

 $ \therefore \frac{dy}{dx}=2\left( {{x}^{2}}-2x \right)\left( 2x-2 \right) $ 

 $ =4x\left( x-2 \right)\left( x-1 \right) $ 

 $ \therefore \frac{dy}{dx}=0 $ 

 $ \Rightarrow x=0,1,2 $ 

The points $x=0,1,2$ divide the real line into four disjoint intervals i.e., 

$\left( -\infty ,0 \right),\left( 0,1 \right),\left( 1,2 \right)and\left( 2,\infty  \right).$ 

In intervals $\left( -\infty ,0 \right)and\left( 1,2 \right),\frac{dy}{dx}<0$ 

$\therefore y$ is strictly decreasing in intervals $\left( -\infty ,0 \right)and\left( 1,2 \right)$ 

However, in intervals $\left( 0,1 \right)and\left( 2,\infty  \right)$, $\frac{dy}{dx}>0$ 

$\therefore y$ is strictly decreasing in intervals $\left( 0,1 \right)and\left( 2,\infty  \right)$

$\therefore y$ is strictly decreasing in intervals $0<x<1$  and $x>2$.

Ans: We have,

 $ y=\frac{4\sin \theta }{\left( 2+\cos \theta  \right)}-\theta  $ 

 $ \therefore \frac{dy}{d\theta }=\frac{\left( 2+\cos \theta  \right)\left( 4\cos \theta  \right)-4\sin \theta \left( -\sin \theta  \right)}{{{\left( 2+\cos \theta  \right)}^{2}}}-1 $ 

 $ =\frac{8\cos \theta +4{{\cos }^{2}}\theta +4{{\sin }^{2}}\theta }{{{\left( 2+\cos \theta  \right)}^{2}}}-1 $ 

 $ =\frac{8\cos \theta +4}{{{\left( 2+\cos \theta  \right)}^{2}}}-1 $ 

Now,           

$ \frac{dy}{d\theta }=0 $ 

$ \Rightarrow \frac{8\cos \theta +4}{{{\left( 2+\cos \theta  \right)}^{2}}}=1 $ 

$ \Rightarrow 8\cos \theta +4=4+{{\cos }^{2}}\theta +4\cos \theta  $ 

$ \Rightarrow {{\cos }^{2}}\theta -4\cos \theta =0 $ 

$ \Rightarrow \cos \theta =0,4 $ 

Since, 

$ \cos \theta \ne 4, $ 

$ \cos \theta =0 $ 

$ \Rightarrow \theta =\frac{\pi }{2} $ 

Now, $\frac{dy}{d\theta }=\frac{\cos \left( 4-\cos \theta  \right)}{{{\left( 2+\cos  \right)}^{2}}}$ 

In the interval $\left[ 0,\frac{\pi }{2} \right]$, we have $\cos \theta >0$,

Also, $4>\cos \theta \Rightarrow 4-\cos \theta >0$ 

$\therefore \cos \theta \left( 4-\cos \theta  \right)>0$ and also ${{\left( 2+\cos \theta  \right)}^{2}}>0$ 

$ \Rightarrow \frac{\cos \theta \left( 4-\cos \theta  \right)}{{{\left( 2+\cos \theta  \right)}^{2}}}>0 $ 

$ \Rightarrow \frac{dy}{dx}>0 $  

Therefore $y$ is strictly increasing in the interval $\left( 0,\frac{\pi }{2} \right)$ 

Hence, $y$ is increasing in the interval $\left( 0,\frac{\pi }{2} \right)$.

Ans: The given function is $f\left( x \right)\log x$.

\[f'\left( x \right)=\frac{1}{x}\] 

It is clear that for $x>0,f'\left( x \right)=\frac{1}{x}>0$ 

Hence, $f\left( x \right)\log x$ is strictly increasing in the interval $\left( 0,\infty  \right)$.