Chapter 6: Applications of Derivatives

Ans: 

i) The given function is $f\left( x \right)={{\left( 2x-1 \right)}^{2}}+3$.

It can be observed that ${{\left( 2x-1 \right)}^{2}}\ge 0$ for every $x\in R$.

Therefore, $f\left( x \right)={{\left( 2x-1 \right)}^{2}}+3\ge 3$ for every $x\in R$.

The minimum value of $f$ is attained when $2x-1=0$ 

$2x-1=0,x=\frac{1}{2}$ 

Minimum value of $f\left( \frac{1}{2} \right)={{\left( 2.\frac{1}{2}-1 \right)}^{2}}+3=3.$ 

Hence, function $f$ does not have a maximum value.

ii) The given function is $f\left( x \right)=9{{x}^{2}}+12x+2={{\left( 3{{x}^{2}}+2 \right)}^{2}}-2$.

It can be observed that ${{\left( 3{{x}^{2}}+2 \right)}^{2}}\ge 0$ for every $x\in R$.

Therefore, $f\left( x \right)={{\left( 3{{x}^{2}}+2 \right)}^{2}}-2\ge -2$ for every $x\in R$.

The minimum value of $f$ is attained when $3x+2=0$ 

$3x+2=0,x=-\frac{2}{3}$ 

Minimum value of $f\left( -\frac{2}{3} \right)={{\left( 3\left( \frac{-2}{3} \right)+2 \right)}^{2}}-2=-2.$ 

Hence, function $f$ does not have a maximum value.

iii) The given function is $f\left( x \right)=-{{\left( x-1 \right)}^{2}}+10.$.

It can be observed that ${{\left( x-1 \right)}^{2}}\ge 0$ for every $x\in R$.

Therefore, $f\left( x \right)=-{{\left( x-1 \right)}^{2}}+10\le 10$ for every $x\in R$.

The minimum value of $f$ is attained when $\left( x-1 \right)=0$ 

$x-1=0,x=0$ 

Minimum value of $f=f\left( 1 \right)=-{{\left( 1-1 \right)}^{2}}+10=10$ 

Hence, function $f$ does not have a maximum value.

vi) The given function is $g\left( x \right)={{x}^{3}}+1$.

Hence, function $g$ neither has a maximum value nor a minimum value.

Ans: We know that $\left| x+2 \right|\ge 0$ for every $x\in R$ 

Therefore, $f\left( x \right)=\left| x+2 \right|-1\ge -1$ for every $x\in R$

The minimum value of $f$ is attained when $\left| x+2 \right|=0$ 

 $\Rightarrow x=-2$ 

Minimum value of $f=f\left( -2 \right)=\left| -2+2 \right|-1=-1$ 

Hence, function $f$ does not have a maximum value.

Ans: We know that $-\left| x+1 \right|\le 0$ for every $x\in R$ 

Therefore, $g\left( x \right)=-\left| x+1 \right|+3\le 3$ for every $x\in R$

The minimum value of $g$ is attained when $\left| x+1 \right|=0$ 

 $\Rightarrow x=-1$ 

Minimum value of $g=g\left( -1 \right)=-\left| -1+1 \right|+3=3$ 

Hence, function $g$ does not have a maximum value.

Ans: We know that $-1\le \sin 2x\le 1$ 

$  -1+5\le \sin 2x+5\le 1+5 $  

$  4\le \sin 2x+5\le 6 $  

Hence, the maximum and minimum values of $h$ are $6\And 4$ respectively.

Ans: We know that $-1\le \sin 4x\le 1$ 

$  2\le \sin 4x+3\le 4 $  

$  2\le \left| \sin 4x+3 \right|\le 4 $  

Hence, the maximum and minimum values of $f$ are $4\And 2$ respectively.

Ans: Here, if a point ${{x}_{0}}$ is closest to $-1$ , then we find $\frac{{{x}_{0}}}{2}+1<{{x}_{0}}+1$ for all ${{x}_{0}}\in \left( -1,1 \right)$.

Also, if ${{x}_{1}}$ is closet to $-1$ , then we find${{x}_{1}}+1<\frac{{{x}_{1}}+1}{2}+1$ for all ${{x}_{0}}\in \left( -1,1 \right)$.

Hence, function $h\left( x \right)$ has neither maximum nor minimum value at $\left( -1,1 \right).$

Ans: $f\left( x \right)={{x}^{2}}$ 

$\therefore f'\left( x \right)=2x$ 

Now, $f'\left( x \right)=0\Rightarrow x=0$ 

Thus, $x=0$ is the only critical point that could be the point of local maxima or local minima of $f.$ 

We have ${f}''\left( 0 \right)=2$, which is positive.

Therefore, by second derivative test,$x=0$ is a point of local minima and local minimum value of $f$ at $x=0$ is $f\left( 0 \right)=0.$ 

Ans: $g\left( x \right)={{x}^{3}}-3x$ 

$\therefore {g}'\left( x \right)=3{{x}^{2}}-3$ 

Now,

$  {g}'\left( x \right)=0\Rightarrow 3{{x}^{2}}=3\Rightarrow x=\pm 1 $  

$  {g}''\left( x \right)=6x $  

$  {g}''\left( 1 \right)=6>0 $  

$  {g}''\left( -1 \right)=-6<0 $  

By second derivative test, $x=1$ is a point of local minima and local minimum value of $g$ at $x=1$ is $g\left( 1 \right)={{1}^{3}}-3=-2.$

Ans: $h\left( x \right)=\sin x+\cos .0<x<\frac{\pi }{2}$ 

$  \therefore {h}'\left( x \right)=\cos x+\sin x $  

$  {h}'\left( x \right)=0\Rightarrow \sin x=\cos x\Rightarrow \tan x=1 $  

$  \Rightarrow x=\frac{\pi }{4}\in \left( 0,\frac{\pi }{2} \right) $  

$  {h}''\left( x \right)=-\sin x-\cos x $  

$  {h}''\left( \frac{\pi }{4} \right)=-\left( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right) $  

$  =-\frac{1}{\sqrt{2}}=-\sqrt{2}<0 $  

Therefore, by second derivative test, $x=\frac{\pi }{4}$ is a point of local maxima and the local maximum value of $h$ at $x=\frac{\pi }{4}$   

$  h\left( \frac{\pi }{4} \right)=\sin \frac{\pi }{4}+\cos \frac{\pi }{4} $  

$  =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}. $

Ans: $f\left( x \right)=\sin x-\cos x,0<x<2\pi $ 

$  \therefore {f}'\left( x \right)=\cos x+\sin x $  

$  {f}'\left( x \right)=0\Rightarrow \cos x=-\sin x $  

$  \tan x=-1\Rightarrow x=\frac{3\pi }{4},\frac{7\pi }{4}\in \left( 0,2\pi  \right) $  

$  {f}''\left( x \right)=-\sin x+\cos x $  

$  {f}''\left( \frac{3\pi }{4} \right)=-\sin \frac{3\pi }{4}+\cos \frac{3\pi }{4}=-\sqrt{2}<0 $  

$  {f}''\left( \frac{7\pi }{4} \right)=-\sin \frac{7\pi }{4}+\cos \frac{7\pi }{4}=\sqrt{2}>0 $  

Therefore, by second derivative test, $x=\frac{3\pi }{4}$ is appoint of local maxima and the local maximum value of $f$ at $x=\frac{3\pi }{4}$ is 

$f\left( \frac{3\pi }{4} \right)=\sin \frac{3\pi }{4}\cos \frac{3\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}.$ However, $x=\frac{7\pi }{4}$ is a point of local minima and the local minimum value of $f$ at $x=\frac{7\pi }{4}$

is $f\left( \frac{7\pi }{4} \right)=\sin \frac{7\pi }{4}-\cos \frac{7\pi }{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2.}$

Ans: $f\left( x \right)={{x}^{3}}-6x+9x+15$ 

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

$  f\left( x \right)=0\Rightarrow 3\left( {{x}^{2}}-4x+3 \right)=0 $  

$  \Rightarrow 3\left( x-1 \right)\left( x-3 \right)=0 $  

$  \Rightarrow x=1,3 $  

Now,

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

$  {f}''\left( 1 \right)=6\left( 1-2 \right)=-6<0 $  

$  {f}''\left( 3 \right)=6\left( 3-2 \right)=-6>0 $  

Therefore, by second derivative test, $x=1$ is a point of local maxima and the local maximum value of $f$ at $x=1$ is $f\left( 1 \right)=1-6+9+15=19$. However, $x=3$ is a point of local minima and the local minimum value of $f$ at $x=3$ is $f\left( 3 \right)=27-54+27+15=15.$

Ans: $g\left( x \right)=\frac{x}{2}+\frac{2}{x},x>0$ 

$  \therefore {g}'\left( x \right)=\frac{1}{2}-\frac{2}{{{x}^{2}}} $  

$  Now, $  

$  {g}'\left( x \right)=0gives\frac{2}{{{x}^{2}}}=\frac{1}{2} $  

$  \Rightarrow {{x}^{2}}=4\Rightarrow x=\pm 2 $  

Since $x>0$ we take $x=2$.

Now, 

$  {g}''\left( x \right)=\frac{4}{{{x}^{3}}} $  

$  {g}''\left( 2 \right)=\frac{4}{{{2}^{3}}}=\frac{1}{2}>0 $  

Therefore, by second derivative test,  $x=2$ is a point of local minima and the local minimum value of $g$ at $x=2$ is $g\left( 2 \right)=\frac{2}{2}+\frac{2}{2}=1+1=2.$

Ans: $g\left( x \right)=\frac{1}{{{x}^{2}}+2}$ 

$  \therefore {g}'\left( x \right)=\frac{-\left( 2x \right)}{{{\left( {{x}^{3}}+2 \right)}^{2}}} $  

$  {g}'\left( x \right)=0\Rightarrow \frac{-\left( 2x \right)}{{{\left( {{x}^{3}}+2 \right)}^{2}}}=0\Rightarrow x=0 $  

Now, for values close to $x=0$ and to the left of $0,{g}'\left( x \right)>0$. Also, for values close to $x=0$ and to the right of ${g}'\left( x \right)<0.$ 

Therefore, by first derivative test $x=0$ is a point of local maxima and the local maximum value of $g\left( 0 \right)is\frac{1}{0+2}=\frac{1}{2}.$

Ans: $f\left( x \right)=x\sqrt{1-x},x>0$ 

  $  \therefore {f}'\left( x \right)=x\sqrt{1-x}+x.\frac{1}{2\sqrt{1-x}}\left( -1 \right)=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}} $  

 $  =\frac{2\left( 1-x \right)-x}{2\sqrt{1-x}}=\frac{2-3x}{2\sqrt{1-x}} $  

 $  {f}'\left( x \right)=0\Rightarrow \frac{2-3x}{2\sqrt{1-x}}=0 $  

 $  \Rightarrow 2-3x=0\Rightarrow x=\frac{2}{3} $  

 $  {f}''\left( x \right)=\frac{1}{2}\left[ \frac{\sqrt{1-x}\left( -3 \right)-\left( 2-3x \right)\left( \frac{-1}{2\sqrt{1-x}} \right)}{1-x} \right] $  

 $  =\frac{\sqrt{1-x}\left( -3 \right)+2\left( 2-3x \right)\left( \frac{1}{2\sqrt{1-x}} \right)}{2\left( 1-x \right)} $  

 $  =\frac{3x-4}{4{{\left( 1-x \right)}^{\frac{3}{2}}}} $  

 $  {f}'\left( \frac{2}{3} \right)=\frac{2-4}{4{{\left( \frac{1}{3} \right)}^{\frac{3}{2}}}}=\frac{-1}{2{{\left( \frac{1}{3} \right)}^{\frac{3}{2}}}}<0 $   

Therefore, by second derivative test, $x=\frac{2}{3}$ is a point of local 

maxima and the local maximum value of $f$ at $x=\frac{2}{3}$ is $f\left( \frac{2}{3} \right)=\frac{2-4}{4{{\left( \frac{1}{3} \right)}^{\frac{3}{2}}}}=\frac{-1}{2{{\left( \frac{1}{3} \right)}^{\frac{3}{2}}}}<0$.

Ans: $f\left( x \right)={{e}^{x}}$ 

$\therefore {f}'\left( x \right)={{e}^{x}}$ 

Now, if  ${f}'\left( x \right)=0$ then ${{e}^{x}}=0$. But the exponential function can never assume $0$ for any value of $x$.

Therefore, there does not exist $c\in R$ such that ${f}'\left( c \right)=0$.

Hence, function $f$ does not have maxima or minima.

Ans: We have,

$  g\left( x \right)=\log x $  

$  \therefore {g}'\left( x \right)=\frac{1}{x} $  

Since $\log x$ is defined for a positive number $x,{g}'\left( x \right)>0$ for any $x$

Therefore, there does not exist  $c\in R$ such that ${g}'\left( c \right)=0$.

Hence, function $g$ does not have maxima or minima.

Ans: We have,

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

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

Now,

Therefore, there does not exist $c\in R$ such that ${h}'\left( c \right)=0$.

Hence, function $h$ does not have maxima or minima.

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

$\therefore {f}'\left( x \right)=3{{x}^{2}}$ 

Now,

${f}'\left( x \right)=0\Rightarrow x=0$ 

Then, we evaluate the value of $f$ at critical point $x=0$ and at endpoints of the interval $\left[ -2,2 \right]$.

$  f\left( 0 \right)=0 $  

 $  f\left( -2 \right)={{\left( -2 \right)}^{3}}=-8 $  

 $  f\left( 2 \right)={{\left( 2 \right)}^{3}}=8 $  

Hence, we can conclude that the absolute maximum value of $f$ on $\left[ -2,2 \right]$ is $8$ occurring at $x=-2$. Also, the absolute minimum value of $f$  on $\left[ -2,2 \right]$ is $-8$ occurring at $x=-2$.

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

$\therefore {f}'\left( x \right)=\cos x-\sin x$ 

Now,

$  {f}'\left( x \right)=0\Rightarrow \sin x=\cos x $  

$  \Rightarrow \tan x=1\Rightarrow x=\frac{\pi }{4} $  

Then, we evaluate the value of $f$ at critical point $x=\frac{\pi }{4}$ and at the endpoints of the interval $\left[ 0,\pi  \right]$.

  $  f\left( \frac{\pi }{4} \right)=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2} $  

 $  f\left( 0 \right)=\sin 0+\cos 0=1 $  

 $  f\left( \pi  \right)=\sin \pi +\cos \pi =1 $  

Hence, we can conclude that the absolute maximum value of $f$ on $\left[ 0,\pi  \right]$ is $\sqrt{2}$ occurring at $x=\frac{\pi }{4}$. Also, the absolute minimum value of $f$ on $\left[ 0,\pi  \right]$ is $-1$ occurring at $x=\pi $.

Ans: The given function is $f\left( x \right)=4x-\frac{1}{2}{{x}^{2}}$ 

$\therefore {f}'\left( x \right)=4x-\frac{1}{2}{{x}^{2}}$

Now,

${f}'\left( x \right)=0\Rightarrow x=4$ 

Then, we evaluate the value of $f$ at critical point $x=4$ and at the endpoints of the interval $\left[ -2,\frac{9}{2} \right]$.

  $  f\left( 4 \right)=16-\frac{1}{2}\left( 16 \right)=8 $  

 $  f\left( -2 \right)=-8-\frac{1}{2}\left( 4 \right)=-10 $  

 $  f\left( \frac{9}{2} \right)=4\left( \frac{9}{2} \right)-\frac{1}{2}{{\left( \frac{9}{2} \right)}^{2}}=18-\frac{81}{8}=7.875 $  

Hence, we can conclude that the absolute maximum value of $f$ on $\left[ -2,\frac{9}{2} \right]$ is $8$ occurring at $x=4$. Also, the absolute minimum value of $f$ on $\left[ -2,\frac{9}{2} \right]$ is $-10$ occurring at $x=-2$.

Ans: The given function is $f\left( x \right)={{\left( x-1 \right)}^{2}}+3$ 

$\therefore {f}'\left( x \right)=2\left( x-1 \right)$

Now,

${f}'\left( x \right)=0\Rightarrow 2\left( x-1 \right)=0,x=1$ 

Then, we evaluate the value of x at critical point $x=1$ and at the endpoints of the interval $\left[ -3,1 \right]$.

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

 $  f\left( -3 \right)={{\left( -3-1 \right)}^{2}}+3=19 $  

Hence, we can conclude that the absolute maximum value of f on $\left[ -3,1 \right]$ is $19$ taking place at $x=-3$. Also, the absolute minimum value of f on $\left[ -3,1 \right]$ is $3$ taking place at $x=1$.

Ans: The profit function is given as $p\left( x \right)=41-24x-18{{x}^{2}}$.

$  \therefore {p}'\left( x \right)=-24-36x $  

$  {p}''\left( x \right)=-36 $  

$  Now, $  

$  {p}'\left( x \right)=0 $  

$  \Rightarrow x=\frac{-24}{36}=-\frac{2}{3} $  

$  Also, $  

$  {p}''\left( \frac{-2}{3} \right)=-36<0 $  

By second derivatives test, $x=-\frac{2}{3}$ is the point of local maximum of $p$

Therefore, Maximum profit = $p\left( -\frac{2}{3} \right)$ 

$  =41-24\left( -\frac{2}{3} \right)-18{{\left( -\frac{2}{3} \right)}^{2}} $  

$  =41+16-8 $  

$  =49 $  

Hence, the maximum profit that the company can make is $49$ units.

Ans:  Let’s take f(x) = $3x^{4}-8x^{3}+ 12x^{2}-48x+25$ on [0,3]

${f}'\left ( x \right ) = 12x^{3}-24x^{2}+24x-48$

Now ${f}'\left ( x \right ) = 0$

$12x^{3}-24x^{2}+24x-48 = 0$

$x^{3}-2x^{2}+2x-4=0$

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

$x=2 or x=\pm \sqrt{2}$

As $ x=\pm \sqrt{2}$ is imaginary, therefore this value is not considered.

At x=2, f(2) = 3(16)-8(8)+12(4)-48(2)+25 = -39

At x=0, f(0) = 3(0)-8(0)+12(0)-48(0)+25 = 25

At x=3, f(3) = 3(81)-8(27)+12(9)-48(3)+25 = 16

Therefore, the minimum value of the given function is -39 and maximum value is 25.

Ans: Let 

$  f\left( x \right)=\sin 2x $  

$  \therefore {f}'\left( x \right)=2\cos 2x $  

$  Now, $  

$  {f}'\left( x \right)=0\Rightarrow \cos 2x=0 $  

$  \Rightarrow 2x=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2} $  

$  \Rightarrow x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4} $  

Then, we evaluate the values of $f$ at critical points 

$x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ and at the endpoints of

the interval $\left[ 0,2\pi  \right]$.

$  f\left( \frac{\pi }{4} \right)=\sin \frac{\pi }{2}=1, $  

$  f\left( \frac{3\pi }{4} \right)=\sin \frac{3\pi }{2}=-1, $  

$  f\left( \frac{5\pi }{4} \right)=\sin \frac{5\pi }{2}=1, $  

$  f\left( \frac{7\pi }{4} \right)=\sin \frac{7\pi }{2}=-1, $  

$  f\left( 0 \right)=\sin 0=0, $  

$  f\left( 2\pi  \right)=\sin 2\pi =0 $  

Hence, we can conclude that the absolute maximum value of 

$f\left[ 0,2\pi  \right]$ is occurring at $x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}.$

Ans: Let $f\left( x \right)=\sin x+\cos x$ 

$  \therefore {f}'\left( x \right)=\cos x-\sin x $  

$  {f}'\left( x \right)=0\Rightarrow \sin x=\cos x $  

$  \Rightarrow \tan x=1\Rightarrow x=\frac{\pi }{4},\frac{5\pi }{4},... $  

$  {f}'\left( x \right)=-\sin x-\cos x=-\left( \sin x+\cos x \right) $  

Now, when $\left( \sin x+\cos x \right)$ is positive, i.e., 

when sin x and cos x are both positive, f” (x) will be negative.

We also know that in the first quadrant, both sin x and cos x are positive. Then, ${f}''\left( x \right)$ will be 

negative when $x\in \left( 0,\frac{\pi }{2} \right)$.

As a result, we consider $x=\frac{\pi }{4}$ 

${f}''\left( \frac{\pi }{4} \right)=-\left( \sin \frac{\pi }{4}+\cos \frac{\pi }{4} \right)=-\left( \frac{2}{\sqrt{2}} \right)=-\sqrt{2}<0$ 

By second derivative test, $f$ will be the maximum at $x=\frac{\pi }{4}$ and the 

maximum value of $f$ is  $f\left( \frac{\pi }{4} \right)=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}.$

Ans: Let $f\left( x \right)=2{{x}^{3}}-24x-107$    

$  \therefore {f}'\left( x \right)=6{{x}^{2}}-24=6\left( {{x}^{2}}-4 \right) $  

$  Now, $  

$  {f}'\left( x \right)=0\Rightarrow 6\left( {{x}^{2}}-4 \right)=0 $  

$  \Rightarrow {{x}^{2}}=4\Rightarrow x=\pm 2 $  

We first consider the interval $\left[ 1,3 \right]$.

Then, we evaluate the value of $f$ at the critical point $x\in \left[ 1,3 \right]$ and at the endpoints of the interval $\left[ 1,3 \right]$.

$  f\left( 2 \right)=2\left( 8 \right)-24\left( 2 \right)+107=75 $  

$  f\left( 1 \right)=2\left( 1 \right)-24\left( 1 \right)+107=85 $  

$  f\left( 3 \right)=2\left( 27 \right)-24\left( 3 \right)+107=89 $  

Hence, the absolute maximum value of $f\left( x \right)$ in the interval $\left[ 1,3 \right]$ is

$89$ occurring at $x=3$.

Next, we consider the interval $\left[ -3,-1 \right]$.

Evaluate the value of $f$ at the critical point $x\in \left[ 1,3 \right]$ and at 

the endpoints of the interval $\left[ -3,-1 \right]$.

$  f\left( -3 \right)=2\left( -27 \right)-24\left( -3 \right)+107=125 $  

$  f\left( -1 \right)=2\left( -1 \right)-24\left( -1 \right)+107=129 $  

$  f\left( -2 \right)=2\left( -8 \right)-24\left( -2 \right)+107=139 $  

Hence, the absolute maximum value of $f\left( x \right)$ in the interval $\left[ -3,-1 \right]$ is

$139$ occurring at $x=-2$.