Class 12: Maxima and Minima

$\displaystyle \textbf{Question 1: }~\text{Find the points of local maxima or local minima and corresponding} \\ \text{local maximum and local minimum values of each of the following functions. } \\ \text{Also, find the points of inflection, if any:}$

$\displaystyle \textbf{(i)}~f(x)=x^4-62x^2+120x+9$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x^{4}-62x^{2}+120x+9$
$\displaystyle \Rightarrow f'(x)=4x^{3}-124x+120$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 4x^{3}-124x+120=0$
$\displaystyle \Rightarrow x^{3}-31x+30=0$
$\displaystyle \Rightarrow (x-1)(x^{2}+x-30)=0$
$\displaystyle \Rightarrow (x-1)(x+6)(x-5)=0$
$\displaystyle \Rightarrow x=-6,\ 1,\ 5$
$\displaystyle \text{Now, } f''(x)=12x^{2}-124$
$\displaystyle \text{At } x=1:\ f''(1)=12(1)^{2}-124=-112<0$
$\displaystyle \Rightarrow x=1 \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=1 \text{ is } 1^{4}-62(1)^{2}+120(1)+9=68$
$\displaystyle \text{At } x=5:\ f''(5)=12(5)^{2}-124=176>0$
$\displaystyle \Rightarrow x=5 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=5 \text{ is } 5^{4}-62(5)^{2}+120(5)+9=-316$
$\displaystyle \text{At } x=-6:\ f''(-6)=12(6)^{2}-124=308>0$
$\displaystyle \Rightarrow x=-6 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=-6 \text{ is } (-6)^{4}-62(6)^{2}+120(-6)+9=-1647$

$\displaystyle \textbf{(ii)}~f(x)=x^3-6x^2+9x+15$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x^{3}-6x^{2}+9x+15$
$\displaystyle \Rightarrow f'(x)=3x^{2}-12x+9$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 3x^{2}-12x+9=0$
$\displaystyle \Rightarrow x^{2}-4x+3=0$
$\displaystyle \Rightarrow (x-1)(x-3)=0$
$\displaystyle \Rightarrow x=1,\ 3$
$\displaystyle \text{Thus, } x=1 \text{ and } x=3 \text{ are the possible points of local extrema}$
$\displaystyle \text{Now, } f''(x)=6x-12$
$\displaystyle \text{At } x=1:\ f''(1)=6(1)-12=-6<0$
$\displaystyle \Rightarrow x=1 \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=1 \text{ is } 1^{3}-6(1)^{2}+9(1)+15=19$
$\displaystyle \text{At } x=3:\ f''(3)=6(3)-12=6>0$
$\displaystyle \Rightarrow x=3 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=3 \text{ is } 3^{3}-6(3)^{2}+9(3)+15=15$

$\displaystyle \textbf{(iii)}~f(x)=(x-1)(x+2)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=(x-1)(x+2)^{2}$
$\displaystyle =(x-1)(x^{2}+4x+4)$
$\displaystyle =x^{3}+4x^{2}+4x-x^{2}-4x-4$
$\displaystyle =x^{3}+3x^{2}-4$
$\displaystyle \Rightarrow f'(x)=3x^{2}+6x$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 3x^{2}+6x=0$
$\displaystyle \Rightarrow 3x(x+2)=0$
$\displaystyle \Rightarrow x=0,\ -2$
$\displaystyle \text{Thus, } x=0 \text{ and } x=-2 \text{ are the possible points of local extrema}$
$\displaystyle \text{Now, } f''(x)=6x+6$
$\displaystyle \text{At } x=0:\ f''(0)=6(0)+6=6>0$
$\displaystyle \Rightarrow x=0 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=0 \text{ is } (0-1)(0+2)^{2}=-4$
$\displaystyle \text{At } x=-2:\ f''(-2)=6(-2)+6=-6<0$
$\displaystyle \Rightarrow x=-2 \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=-2 \text{ is } (-2-1)(-2+2)^{2}=0$

$\displaystyle \textbf{(iv)}~f(x)=\dfrac{2}{x}-\dfrac{2}{x^2},\;x>0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=\frac{2}{x}-\frac{2}{x^{2}}$
$\displaystyle \Rightarrow f(x)=2x^{-1}-2x^{-2}$
$\displaystyle \Rightarrow f'(x)=-2x^{-2}+4x^{-3} =\frac{4}{x^{3}}-\frac{2}{x^{2}}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow \frac{4}{x^{3}}-\frac{2}{x^{2}}=0$
$\displaystyle \Rightarrow 4-2x=0$
$\displaystyle \Rightarrow x=2$
$\displaystyle \text{Thus, } x=2 \text{ is the possible point of local extremum}$
$\displaystyle \text{Now, } f''(x)=-\frac{12}{x^{4}}+\frac{4}{x^{3}}$
$\displaystyle \text{At } x=2:\ f''(2)=-\frac{12}{16}+\frac{4}{8} =-\frac{12}{16}+\frac{8}{16} =-\frac{1}{4}<0$
$\displaystyle \Rightarrow x=2 \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=2 \text{ is } \frac{2}{2}-\frac{2}{2^{2}} =1-\frac{1}{2} =\frac{1}{2}$

$\displaystyle \textbf{(v)}~f(x)=xe^x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=xe^{x}$
$\displaystyle \Rightarrow f'(x)=e^{x}+xe^{x}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow e^{x}+xe^{x}=0$
$\displaystyle \Rightarrow e^{x}(1+x)=0$
$\displaystyle \Rightarrow x=-1$
$\displaystyle \text{Thus, } x=-1 \text{ is the possible point of local extremum}$
$\displaystyle \text{Now, } f''(x)=e^{x}+e^{x}+xe^{x}$
$\displaystyle \Rightarrow f''(x)=e^{x}(x+2)$
$\displaystyle \text{At } x=-1:\ f''(-1)=e^{-1}(1)>0$
$\displaystyle \Rightarrow x=-1 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=-1 \text{ is } (-1)e^{-1}=-\frac{1}{e}$

$\displaystyle \textbf{(vi)}~f(x)=\dfrac{x}{2}+\dfrac{2}{x},\;x>0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=\frac{x}{2}+\frac{2}{x},\ x\ne 0$
$\displaystyle \Rightarrow f'(x)=\frac{1}{2}-\frac{2}{x^{2}}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow \frac{1}{2}-\frac{2}{x^{2}}=0$
$\displaystyle \Rightarrow \frac{1}{2}=\frac{2}{x^{2}}$
$\displaystyle \Rightarrow x^{2}=4$
$\displaystyle \Rightarrow x=\pm 2$
$\displaystyle \text{Since } x>0,\ \text{we consider } x=2$
$\displaystyle \text{Now, } f''(x)=\frac{4}{x^{3}}$
$\displaystyle \text{At } x=2:\ f''(2)=\frac{4}{(2)^{3}}=\frac{1}{2}>0$
$\displaystyle \Rightarrow x=2 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=2 \text{ is } \frac{2}{2}+\frac{2}{2}=2$

$\displaystyle \textbf{(vii)}~f(x)=(x+1)(x+2)^{1/3},\;x\ge -2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=(x+1)(x+2)^{\tfrac{1}{3}}$
$\displaystyle \Rightarrow f'(x)=(x+2)^{\tfrac{1}{3}}+\frac{1}{3}(x+1)(x+2)^{-\tfrac{2}{3}}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow (x+2)^{\tfrac{1}{3}}+\frac{1}{3}(x+1)(x+2)^{-\tfrac{2}{3}}=0$
$\displaystyle \Rightarrow (x+2)^{-\tfrac{2}{3}} \left[(x+2)+\frac{1}{3}(x+1)\right]=0$
$\displaystyle \Rightarrow (x+2)+\frac{1}{3}(x+1)=0$
$\displaystyle \Rightarrow 3(x+2)+(x+1)=0$
$\displaystyle \Rightarrow 4x+7=0$
$\displaystyle \Rightarrow x=-\frac{7}{4}$
$\displaystyle \text{Thus, } x=-\frac{7}{4} \text{ is the possible point of local extremum}$
$\displaystyle \text{Now, } f''(x)=\frac{2}{3}(x+2)^{-\tfrac{2}{3}} -\frac{2}{9}(x+1)(x+2)^{-\tfrac{5}{3}}$
$\displaystyle \text{At } x=-\frac{7}{4}:$
$\displaystyle f''\!\left(-\frac{7}{4}\right) =\frac{2}{3}\left(\frac{1}{4}\right)^{-\tfrac{2}{3}} +\frac{1}{18}\left(\frac{1}{4}\right)^{-\tfrac{5}{3}}>0$
$\displaystyle \Rightarrow x=-\frac{7}{4} \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=-\frac{7}{4} \text{ is }$
$\displaystyle f\!\left(-\frac{7}{4}\right) =\left(-\frac{7}{4}+1\right)\left(-\frac{7}{4}+2\right)^{\tfrac{1}{3}} =-\frac{3}{4}\left(\frac{1}{4}\right)^{\tfrac{1}{3}} =-\frac{3}{4^{\tfrac{4}{3}}}$

$\displaystyle \textbf{(viii)}~f(x)=x\sqrt{32-x^2},\;-5\le x\le5$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x\sqrt{32-x^{2}}$
$\displaystyle \Rightarrow f'(x)=\sqrt{32-x^{2}} -\frac{x^{2}}{\sqrt{32-x^{2}}}$
$\displaystyle \Rightarrow f'(x)=\frac{32-2x^{2}}{\sqrt{32-x^{2}}}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow \frac{32-2x^{2}}{\sqrt{32-x^{2}}}=0$
$\displaystyle \Rightarrow 32-2x^{2}=0$
$\displaystyle \Rightarrow x^{2}=16$
$\displaystyle \Rightarrow x=\pm 4$
$\displaystyle \text{Thus, } x=4 \text{ and } x=-4 \text{ are the possible points of local extrema}$
$\displaystyle \text{Now, } f''(x)=\frac{-4x}{\sqrt{32-x^{2}}} -\frac{x(32-2x^{2})}{(32-x^{2})^{3/2}}$
$\displaystyle \text{At } x=4:\ f''(4)<0$
$\displaystyle \Rightarrow x=4 \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=4 \text{ is } 4\sqrt{32-16}=16$
$\displaystyle \text{At } x=-4:\ f''(-4)>0$
$\displaystyle \Rightarrow x=-4 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=-4 \text{ is } -4\sqrt{32-16}=-16$

$\displaystyle \textbf{(ix)}~f(x)=x^3-2ax^2+a^2x,\;a>0,\;x\in R$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x^{3}-2ax^{2}+a^{2}x$
$\displaystyle \Rightarrow f'(x)=3x^{2}-4ax+a^{2}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 3x^{2}-4ax+a^{2}=0$
$\displaystyle \Rightarrow 3x^{2}-3ax-ax+a^{2}=0$
$\displaystyle \Rightarrow 3x(x-a)-a(x-a)=0$
$\displaystyle \Rightarrow (3x-a)(x-a)=0$
$\displaystyle \Rightarrow x=\frac{a}{3},\ a$
$\displaystyle \text{Thus, } x=\frac{a}{3} \text{ and } x=a \text{ are the possible points of local extrema}$
$\displaystyle \text{Now, } f''(x)=6x-4a$
$\displaystyle \text{At } x=a:\ f''(a)=6a-4a=2a>0$
$\displaystyle \Rightarrow x=a \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=a \text{ is } a^{3}-2a(a)^{2}+a^{2}(a)=0$
$\displaystyle \text{At } x=\frac{a}{3}:\ f''\!\left(\frac{a}{3}\right)=6\left(\frac{a}{3}\right)-4a=-2a<0$
$\displaystyle \Rightarrow x=\frac{a}{3} \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=\frac{a}{3} \text{ is }$
$\displaystyle f\!\left(\frac{a}{3}\right) =\left(\frac{a}{3}\right)^{3} -2a\left(\frac{a}{3}\right)^{2} +a^{2}\left(\frac{a}{3}\right) =\frac{4a^{3}}{27}$

$\displaystyle \textbf{(x)}~f(x)=x+\dfrac{a^2}{x},\;a>0,\;x\ne0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x+\frac{a^{2}}{x},\ x\ne 0$
$\displaystyle \Rightarrow f'(x)=1-\frac{a^{2}}{x^{2}}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 1-\frac{a^{2}}{x^{2}}=0$
$\displaystyle \Rightarrow x^{2}=a^{2}$
$\displaystyle \Rightarrow x=\pm a$
$\displaystyle \text{Thus, } x=a \text{ and } x=-a \text{ are the possible points of local extrema}$
$\displaystyle \text{Now, } f''(x)=\frac{2a^{2}}{x^{3}}$
$\displaystyle \text{At } x=a:\ f''(a)=\frac{2a^{2}}{a^{3}}=\frac{2}{a}>0$
$\displaystyle \Rightarrow x=a \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=a \text{ is } a+\frac{a^{2}}{a}=2a$
$\displaystyle \text{At } x=-a:\ f''(-a)=\frac{2a^{2}}{(-a)^{3}} =-\frac{2}{a}<0$
$\displaystyle \Rightarrow x=-a \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=-a \text{ is } -a+\frac{a^{2}}{-a}=-2a$

$\displaystyle \textbf{(xi)}~f(x)=x\sqrt{2-x^2},\;-\sqrt{2}\le x\le\sqrt{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x\sqrt{2-x^{2}}$
$\displaystyle \Rightarrow f'(x)=\sqrt{2-x^{2}}-\frac{x^{2}}{\sqrt{2-x^{2}}}$
$\displaystyle \Rightarrow f'(x)=\frac{2-2x^{2}}{\sqrt{2-x^{2}}}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow \frac{2-2x^{2}}{\sqrt{2-x^{2}}}=0$
$\displaystyle \Rightarrow 2-2x^{2}=0$
$\displaystyle \Rightarrow x^{2}=1$
$\displaystyle \Rightarrow x=\pm 1$
$\displaystyle \text{Thus, } x=1 \text{ and } x=-1 \text{ are the possible points of local extrema}$
$\displaystyle \text{Now, } f''(x)=\frac{-4x}{\sqrt{2-x^{2}}} -\frac{x(2-2x^{2})}{(2-x^{2})^{3/2}}$
$\displaystyle \text{At } x=1:\ f''(1)<0$
$\displaystyle \Rightarrow x=1 \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=1 \text{ is } 1\sqrt{2-1}=1$
$\displaystyle \text{At } x=-1:\ f''(-1)>0$
$\displaystyle \Rightarrow x=-1 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=-1 \text{ is } -1\sqrt{2-1}=-1$

$\displaystyle \textbf{(xii)}~f(x)=x+\sqrt{1-x},\;x\le1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x+\sqrt{1-x},\ x\le 1$
$\displaystyle \Rightarrow f'(x)=1-\frac{1}{2\sqrt{1-x}}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 1-\frac{1}{2\sqrt{1-x}}=0$
$\displaystyle \Rightarrow \frac{1}{2\sqrt{1-x}}=1$
$\displaystyle \Rightarrow \sqrt{1-x}=\frac{1}{2}$
$\displaystyle \Rightarrow 1-x=\frac{1}{4}$
$\displaystyle \Rightarrow x=\frac{3}{4}$
$\displaystyle \text{Thus, } x=\frac{3}{4} \text{ is the possible point of local extremum}$
$\displaystyle \text{Now, } f''(x)=-\frac{1}{4(1-x)^{3/2}}$
$\displaystyle \text{At } x=\frac{3}{4}:\ f''\!\left(\frac{3}{4}\right) =-\frac{1}{4\left(\frac{1}{4}\right)^{3/2}} =-2<0$
$\displaystyle \Rightarrow x=\frac{3}{4} \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=\frac{3}{4} \text{ is }$
$\displaystyle f\!\left(\frac{3}{4}\right) =\frac{3}{4}+\sqrt{1-\frac{3}{4}} =\frac{3}{4}+\frac{1}{2} =\frac{5}{4}$

$\displaystyle \textbf{Question 2: }~\text{Find the local extremum values of the following functions:}$

$\displaystyle \textbf{(i)}~f(x)=(x-1)(x-2)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=(x-1)(x-2)^{2}$
$\displaystyle =(x-1)(x^{2}-4x+4)$
$\displaystyle =x^{3}-4x^{2}+4x-x^{2}+4x-4$
$\displaystyle =x^{3}-5x^{2}+8x-4$
$\displaystyle \Rightarrow f'(x)=3x^{2}-10x+8$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 3x^{2}-10x+8=0$
$\displaystyle \Rightarrow 3x^{2}-6x-4x+8=0$
$\displaystyle \Rightarrow (x-2)(3x-4)=0$
$\displaystyle \Rightarrow x=2,\ \frac{4}{3}$
$\displaystyle \text{Thus, } x=2 \text{ and } x=\frac{4}{3} \text{ are the possible points of local extrema}$
$\displaystyle \text{Now, } f''(x)=6x-10$
$\displaystyle \text{At } x=2:\ f''(2)=12-10=2>0$
$\displaystyle \Rightarrow x=2 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=2 \text{ is } (2-1)(2-2)^{2}=0$
$\displaystyle \text{At } x=\frac{4}{3}:\ f''\!\left(\frac{4}{3}\right)=8-10=-2<0$
$\displaystyle \Rightarrow x=\frac{4}{3} \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=\frac{4}{3} \text{ is } \left(\frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)^{2} =\frac{1}{3}\cdot\frac{4}{9} =\frac{4}{27}$

$\displaystyle \textbf{(ii)}~f(x)=x\sqrt{1-x},\;x\le1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=x\sqrt{1-x}$
$\displaystyle \Rightarrow f'(x)=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}$
$\displaystyle \text{For local maxima or minima, we must have } f'(x)=0$
$\displaystyle \Rightarrow \sqrt{1-x}-\frac{x}{2\sqrt{1-x}}=0$
$\displaystyle \Rightarrow \sqrt{1-x}=\frac{x}{2\sqrt{1-x}}$
$\displaystyle \Rightarrow 2(1-x)=x$
$\displaystyle \Rightarrow 2-2x=x$
$\displaystyle \Rightarrow 3x=2$
$\displaystyle \Rightarrow x=\frac{2}{3}$
$\displaystyle \text{Thus, } x=\frac{2}{3} \text{ is a possible point of local maxima or local minima.}$
$\displaystyle \text{Now,}$
$\displaystyle f''(x)= -\frac{1}{\sqrt{1-x}}-\frac{1}{2}\left(\frac{2-x}{(1-x)\sqrt{1-x}}\right)$
$\displaystyle \text{At } x=\frac{2}{3}:$
$\displaystyle f''\!\left(\frac{2}{3}\right)= -\sqrt{3}-\frac{4\sqrt{3}}{3}<0$
$\displaystyle \text{So, } x=\frac{2}{3} \text{ is the point of local maximum.}$
$\displaystyle \text{The local maximum value is given by}$
$\displaystyle f\!\left(\frac{2}{3}\right)=\frac{2}{3}\sqrt{1-\frac{2}{3}}=\frac{2}{3\sqrt{3}}$

$\displaystyle \textbf{(iii)}~f(x)=-(x-1)^3(x+1)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=-(x-1)^3(x+1)^2$
$\displaystyle \Rightarrow f'(x)=-\left[3(x-1)^2(x+1)^2+2(x+1)(x-1)^3\right]$
$\displaystyle \text{For local maxima or minima, we must have } f'(x)=0$
$\displaystyle \Rightarrow -3(x-1)^2(x+1)^2-2(x+1)(x-1)^3=0$
$\displaystyle \Rightarrow (x-1)^2(x+1)\left[-3(x+1)-2(x-1)\right]=0$
$\displaystyle \Rightarrow (x-1)^2(x+1)\left[-5x-1\right]=0$
$\displaystyle \Rightarrow x=1,\,-1,\,-\frac{1}{5}$
$\displaystyle \text{Thus, } x=1,\,-1 \text{ and } x=-\frac{1}{5} \text{ are the possible points of local maxima or local minima.}$
$\displaystyle \text{Now,}$
$\displaystyle f''(x)=-\left[6(x-1)(x+1)^2+12(x+1)(x-1)^2+2(x-1)^3\right]$
$\displaystyle \text{At } x=1:$
$\displaystyle f''(1)=0$
$\displaystyle \text{So, } x=1 \text{ is a point of inflexion.}$
$\displaystyle \text{At } x=-1:$
$\displaystyle f''(-1)=16>0$
$\displaystyle \text{So, } x=-1 \text{ is the point of local minimum.}$
$\displaystyle \text{The local minimum value is given by}$
$\displaystyle f(-1)=-( -1-1)^3(-1+1)^2=0$
$\displaystyle \text{At } x=-\frac{1}{5}:$
$\displaystyle f''\!\left(-\frac{1}{5}\right)=-\frac{336}{125}<0$
$\displaystyle \text{So, } x=-\frac{1}{5} \text{ is the point of local maximum.}$
$\displaystyle \text{The local maximum value is given by}$
$\displaystyle f\!\left(-\frac{1}{5}\right)=-\left(-\frac{6}{5}\right)^3\left(\frac{4}{5}\right)^2=\frac{3456}{3125}$

$\displaystyle \textbf{Question 3: }~\text{The function }y=a\log x+bx^2+x\text{ has extreme values at } \\ x=1\text{ and }x=2.\text{ Find }a\text{ and }b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=a\log x+bx^2+x$
$\displaystyle \Rightarrow f'(x)=\frac{a}{x}+2bx+1$
$\displaystyle \text{Since } f'(x) \text{ has extreme values at } x=1 \text{ and } x=2,$
$\displaystyle \Rightarrow f'(1)=0$
$\displaystyle \Rightarrow \frac{a}{1}+2b(1)+1=0$
$\displaystyle \Rightarrow a+2b+1=0$
$\displaystyle \Rightarrow a=-1-2b \qquad \text{(1)}$
$\displaystyle \Rightarrow f'(2)=0$
$\displaystyle \Rightarrow \frac{a}{2}+2b(2)+1=0$
$\displaystyle \Rightarrow a+8b=-2$
$\displaystyle \Rightarrow a=-2-8b \qquad \text{(2)}$
$\displaystyle \text{From equations (1) and (2),}$
$\displaystyle -1-2b=-2-8b$
$\displaystyle \Rightarrow 6b=-1$
$\displaystyle \Rightarrow b=-\frac{1}{6}$
$\displaystyle \text{Substituting } b=-\frac{1}{6} \text{ in equation (1),}$
$\displaystyle a=-1+\frac{1}{3}=-\frac{2}{3}$

$\displaystyle \textbf{Question 4: }~\text{Show that }\dfrac{\log x}{x}\text{ has a maximum value at }x=e.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here,}$
$\displaystyle f(x)=\frac{\log x}{x}$
$\displaystyle \Rightarrow f'(x)=\frac{1-\log x}{x^{2}}$
$\displaystyle \text{For local maxima or minima, we must have } f'(x)=0$
$\displaystyle \Rightarrow \frac{1-\log x}{x^{2}}=0$
$\displaystyle \Rightarrow 1-\log x=0$
$\displaystyle \Rightarrow \log x=1$
$\displaystyle \Rightarrow \log x=\log e$
$\displaystyle \Rightarrow x=e$
$\displaystyle \text{Now,}$
$\displaystyle f''(x)=\frac{-3+2\log x}{x^{3}}$
$\displaystyle f''(e)=\frac{-3+2\log e}{e^{3}}=-\frac{1}{e^{3}}<0$
$\displaystyle \text{So, } x=e \text{ is the point of local maximum.}$

$\displaystyle \textbf{Question 5: }~\text{Find the maximum and minimum values of the function } \\ f(x)=\dfrac{4}{x+2}+x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=\frac{4}{x+2}+x$
$\displaystyle \Rightarrow f'(x)=-\frac{4}{(x+2)^2}+1$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow -\frac{4}{(x+2)^2}+1=0$
$\displaystyle \Rightarrow \frac{4}{(x+2)^2}=1$
$\displaystyle \Rightarrow (x+2)^2=4$
$\displaystyle \Rightarrow x+2=\pm 2$
$\displaystyle \Rightarrow x=0 \text{ or } x=-4$
$\displaystyle \text{Thus, } x=0 \text{ and } x=-4 \text{ are the possible points of local maxima or local minima.}$
$\displaystyle \text{Now,}$
$\displaystyle f''(x)=\frac{8}{(x+2)^3}$
$\displaystyle \text{At } x=0:$
$\displaystyle f''(0)=\frac{8}{(2)^3}=1>0$
$\displaystyle \text{So, } x=0 \text{ is a point of local minimum.}$
$\displaystyle \text{The local minimum value is given by}$
$\displaystyle f(0)=\frac{4}{0+2}+0=2$
$\displaystyle \text{At } x=-4:$
$\displaystyle f''(-4)=\frac{8}{(-2)^3}=-1<0$
$\displaystyle \text{So, } x=-4 \text{ is a point of local maximum.}$
$\displaystyle \text{The local maximum value is given by}$
$\displaystyle f(-4)=\frac{4}{-4+2}-4=-6$

$\displaystyle \textbf{Question 6: }~\text{Find the maximum and minimum values of } \\ f(x)=\tan x-2x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=\tan x-2x$
$\displaystyle \Rightarrow f'(x)=\sec^2 x-2$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow \sec^2 x-2=0$
$\displaystyle \Rightarrow \sec^2 x=2$
$\displaystyle \Rightarrow \sec x=\pm\sqrt{2}$
$\displaystyle \Rightarrow x=\frac{\pi}{4} \text{ or } \frac{3\pi}{4}$
$\displaystyle \text{Thus, } x=\frac{\pi}{4} \text{ and } x=\frac{3\pi}{4} \text{ are the possible points of local maxima or local minima.}$
$\displaystyle \text{Now,}$
$\displaystyle f''(x)=2\sec^2 x\tan x$
$\displaystyle \text{At } x=\frac{\pi}{4}:$
$\displaystyle f''\!\left(\frac{\pi}{4}\right)=2\sec^2\!\left(\frac{\pi}{4}\right)\tan\!\left(\frac{\pi}{4}\right)=4>0$
$\displaystyle \text{So, } x=\frac{\pi}{4} \text{ is a point of local minimum.}$
$\displaystyle \text{The local minimum value is given by}$
$\displaystyle f\!\left(\frac{\pi}{4}\right)=\tan\!\left(\frac{\pi}{4}\right)-2\cdot\frac{\pi}{4}=1-\frac{\pi}{2}$
$\displaystyle \text{At } x=\frac{3\pi}{4}:$
$\displaystyle f''\!\left(\frac{3\pi}{4}\right)=2\sec^2\!\left(\frac{3\pi}{4}\right)\tan\!\left(\frac{3\pi}{4}\right)=-4<0$
$\displaystyle \text{So, } x=\frac{3\pi}{4} \text{ is a point of local maximum.}$
$\displaystyle \text{The local maximum value is given by}$
$\displaystyle f\!\left(\frac{3\pi}{4}\right)=\tan\!\left(\frac{3\pi}{4}\right)-2\cdot\frac{3\pi}{4}=-1-\frac{3\pi}{2}$

$\displaystyle \textbf{Question 7: }~\text{If }f(x)=x^3+ax^2+bx+c\text{ has a maximum at }x=-1\text{ and minimum at } \\ x=3.\text{ Determine }a,\,b\text{ and }c.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle f(x)=x^{3}+ax^{2}+bx+c$
$\displaystyle \Rightarrow f'(x)=3x^{2}+2ax+b$
$\displaystyle \text{As } f(x) \text{ has a maximum at } x=-1 \text{ and a minimum at } x=3,$
$\displaystyle \Rightarrow f'(-1)=0 \text{ and } f'(3)=0$
$\displaystyle \Rightarrow 3(-1)^{2}+2a(-1)+b=0$
$\displaystyle \Rightarrow 3-2a+b=0 \qquad \text{(1)}$
$\displaystyle \Rightarrow 3(3)^{2}+2a(3)+b=0$
$\displaystyle \Rightarrow 27+6a+b=0 \qquad \text{(2)}$
$\displaystyle \text{Subtracting (1) from (2),}$
$\displaystyle (27+6a+b)-(3-2a+b)=0$
$\displaystyle \Rightarrow 24+8a=0$
$\displaystyle \Rightarrow a=-3$
$\displaystyle \text{Substituting } a=-3 \text{ in (1),}$
$\displaystyle 3-2(-3)+b=0$
$\displaystyle \Rightarrow 3+6+b=0$
$\displaystyle \Rightarrow b=-9$
$\displaystyle \text{And, } c \in R$

$\displaystyle \textbf{Question 8: }~\text{Prove that }f(x)=\sin x+\sqrt{3}\cos x\text{ has maximum value at } \\ x=\dfrac{\pi}{6}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle f(x)=\sin x+\sqrt{3}\cos x$
$\displaystyle \Rightarrow f'(x)=\cos x+\sqrt{3}(-\sin x)$
$\displaystyle \Rightarrow f'(x)=\cos x-\sqrt{3}\sin x$
$\displaystyle \text{For } f(x) \text{ to have a maximum or minimum value, we must have } f'(x)=0$
$\displaystyle \Rightarrow \cos x-\sqrt{3}\sin x=0$
$\displaystyle \Rightarrow \cos x=\sqrt{3}\sin x$
$\displaystyle \Rightarrow \cot x=\sqrt{3}$
$\displaystyle \Rightarrow x=\frac{\pi}{6}$
$\displaystyle \text{Also,}$
$\displaystyle f''(x)=-\sin x-\sqrt{3}\cos x$
$\displaystyle f''\!\left(\frac{\pi}{6}\right)=-\sin\!\left(\frac{\pi}{6}\right)-\sqrt{3}\cos\!\left(\frac{\pi}{6}\right)$
$\displaystyle =-\frac{1}{2}-\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)$
$\displaystyle =-\frac{1}{2}-\frac{3}{2}=-2<0$
$\displaystyle \text{So, } x=\frac{\pi}{6} \text{ is the point of maxima.}$