Class 12: Maxima and Minima

$\displaystyle \text{Find the points of local maxima or local minima, if any, of the following functions, } \\ \text{using the first derivative test. Also, find the local maximum or local minimum values, } \\ \text{as the case may be:}$

$\displaystyle \textbf{Question 1: }~f(x)=(x-5)^4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=(x-5)^{4}$
$\displaystyle \Rightarrow f'(x)=4(x-5)^{3}$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow 4(x-5)^{3}=0$
$\displaystyle \Rightarrow x=5$
$\displaystyle \text{Now, for } x<5,\ f'(x)<0 \text{ and for } x>5,\ f'(x)>0$
$\displaystyle \text{Hence, } f'(x)\text{ changes from negative to positive as } x \text{ increases through } 5$
$\displaystyle \Rightarrow x=5 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=5 \text{ is } (5-5)^{4}=0$

$\displaystyle \textbf{Question 2:}~f(x)=x^3-3x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=(x-5)^{4}$
$\displaystyle \Rightarrow f'(x)=4(x-5)^{3}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 4(x-5)^{3}=0$
$\displaystyle \Rightarrow x=5$
$\displaystyle \text{Now, for } x<5,\ f'(x)<0 \text{ and for } x>5,\ f'(x)>0$
$\displaystyle \text{Hence, } f'(x)\text{ changes from negative to positive as } x \text{ increases through } 5$
$\displaystyle \Rightarrow x=5 \text{ is the point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=5 \text{ is } (5-5)^{4}=0$

$\displaystyle \textbf{Question 3:}~f(x)=x^3(x-1)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x^{3}(x-1)^{2}$
$\displaystyle \Rightarrow f'(x)=3x^{2}(x-1)^{2}+2x^{3}(x-1)$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 3x^{2}(x-1)^{2}+2x^{3}(x-1)=0$
$\displaystyle \Rightarrow x^{2}(x-1)\left[3(x-1)+2x\right]=0$
$\displaystyle \Rightarrow x^{2}(x-1)(5x-3)=0$
$\displaystyle \Rightarrow x=0,\ 1,\ \frac{3}{5}$
$\displaystyle \text{Now, } f'(x)\text{ does not change sign at } x=0$
$\displaystyle \Rightarrow x=0 \text{ is a point of inflexion}$
$\displaystyle \text{For } x<1,\ f'(x)<0 \text{ and for } x>1,\ f'(x)>0$
$\displaystyle \Rightarrow x=1 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=1 \text{ is } 1^{3}(1-1)^{2}=0$
$\displaystyle \text{For } x<\frac{3}{5},\ f'(x)>0 \text{ and for } x>\frac{3}{5},\ f'(x)<0$
$\displaystyle \Rightarrow x=\frac{3}{5} \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=\frac{3}{5} \text{ is } \left(\frac{3}{5}\right)^{3}\left(\frac{3}{5}-1\right)^{2} \\ =\frac{27}{125}\times\frac{4}{25} =\frac{108}{3125}$

$\displaystyle \textbf{Question 4:}~f(x)=(x-1)(x+2)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=(x-1)(x+2)^{2}$
$\displaystyle \Rightarrow f'(x)=(x+2)^{2}+2(x+2)(x-1)$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow (x+2)^{2}+2(x+2)(x-1)=0$
$\displaystyle \Rightarrow (x+2)\left[(x+2)+2(x-1)\right]=0$
$\displaystyle \Rightarrow (x+2)(3x)=0$
$\displaystyle \Rightarrow x=-2,\ 0$
$\displaystyle \text{Now, for } x<-2,\ f'(x)>0 \text{ and for } -2<x<0,\ f'(x)<0$
$\displaystyle \Rightarrow x=-2 \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=-2 \text{ is } (-2-1)(-2+2)^{2}=0$
$\displaystyle \text{For } x<0,\ f'(x)<0 \text{ and for } x>0,\ f'(x)>0$
$\displaystyle \Rightarrow x=0 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=0 \text{ is } (0-1)(0+2)^{2}=-4$

$\displaystyle \textbf{Question 5:}~f(x)=\dfrac{1}{x^2+2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=\frac{1}{x^{2}+2}$
$\displaystyle \Rightarrow f'(x)=\frac{-2x}{(x^{2}+2)^{2}}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow \frac{-2x}{(x^{2}+2)^{2}}=0$
$\displaystyle \Rightarrow -2x=0$
$\displaystyle \Rightarrow x=0$
$\displaystyle \text{Now, for } x<0,\ f'(x)>0$
$\displaystyle \text{and for } x>0,\ f'(x)<0$
$\displaystyle \text{Hence, } f'(x)\text{ changes from positive to negative as } x \text{ increases through } 0$
$\displaystyle \Rightarrow x=0 \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=0 \text{ is } \frac{1}{2}$

$\displaystyle \textbf{Question 6:}~f(x)=x^3-6x^2+9x+15$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x^{3}-6x^{2}+9x+15$
$\displaystyle \Rightarrow f'(x)=3x^{2}-12x+9$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 3x^{2}-12x+9=0$
$\displaystyle \Rightarrow x^{2}-4x+3=0$
$\displaystyle \Rightarrow (x-1)(x-3)=0$
$\displaystyle \Rightarrow x=1,\ 3$
$\displaystyle \text{Now, for } x<1,\ f'(x)>0,\ \text{for } 1<x<3,\ f'(x)<0, \text{ and for } x>3,\ f'(x)>0$
$\displaystyle \Rightarrow f'(x)\text{ changes from positive to negative as } x \text{ increases through } 1$
$\displaystyle \Rightarrow x=1 \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=1 \text{ is } 1^{3}-6(1)^{2}+9(1)+15=19$
$\displaystyle \Rightarrow f'(x)\text{ changes from negative to positive as } x \text{ increases through } 3$
$\displaystyle \Rightarrow x=3 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=3 \text{ is } 3^{3}-6(3)^{2}+9(3)+15=15$

$\displaystyle \textbf{Question 7:}~f(x)=\sin 2x,\;0<x<\pi$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=\sin 2x$
$\displaystyle \Rightarrow f'(x)=2\cos 2x$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 2\cos 2x=0$
$\displaystyle \Rightarrow \cos 2x=0$
$\displaystyle \Rightarrow 2x=\frac{\pi}{2},\ \frac{3\pi}{2}$
$\displaystyle \Rightarrow x=\frac{\pi}{4},\ \frac{3\pi}{4}$
$\displaystyle \text{Now, } f'(x)>0 \text{ for } x<\frac{\pi}{4} \text{ and } f'(x)<0 \text{ for } x>\frac{\pi}{4}$
$\displaystyle \Rightarrow x=\frac{\pi}{4} \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=\frac{\pi}{4} \text{ is } \sin\!\left(\frac{\pi}{2}\right)=1$
$\displaystyle \text{Further, } f'(x)<0 \text{ for } x<\frac{3\pi}{4} \text{ and } f'(x)>0 \text{ for } x>\frac{3\pi}{4}$
$\displaystyle \Rightarrow x=\frac{3\pi}{4} \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=\frac{3\pi}{4} \text{ is } \sin\!\left(\frac{3\pi}{2}\right)=-1$

$\displaystyle \textbf{Question 8:}~f(x)=\sin x-\cos x,\;0<x<2\pi$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=\sin x-\cos x$
$\displaystyle \Rightarrow f'(x)=\cos x+\sin x$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow \cos x+\sin x=0$
$\displaystyle \Rightarrow \cos x=-\sin x$
$\displaystyle \Rightarrow \tan x=-1$
$\displaystyle \Rightarrow x=\frac{3\pi}{4},\ \frac{7\pi}{4}$
$\displaystyle \text{Now, } f'(x)>0 \text{ for } x<\frac{3\pi}{4} \text{ and } f'(x)<0 \text{ for } x>\frac{3\pi}{4}$
$\displaystyle \Rightarrow x=\frac{3\pi}{4} \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=\frac{3\pi}{4} \text{ is } \sin\!\left(\frac{3\pi}{4}\right)-\cos\!\left(\frac{3\pi}{4}\right)=\sqrt{2}$
$\displaystyle \text{Further, } f'(x)<0 \text{ for } x<\frac{7\pi}{4} \text{ and } f'(x)>0 \text{ for } x>\frac{7\pi}{4}$
$\displaystyle \Rightarrow x=\frac{7\pi}{4} \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=\frac{7\pi}{4} \text{ is } \sin\!\left(\frac{7\pi}{4}\right)-\cos\!\left(\frac{7\pi}{4}\right)=-\sqrt{2}$

$\displaystyle \textbf{Question 9:}~f(x)=\cos x,\;0<x<\pi$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=\cos x$
$\displaystyle \Rightarrow f'(x)=-\sin x$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow -\sin x=0$
$\displaystyle \Rightarrow \sin x=0$
$\displaystyle \Rightarrow x=0 \text{ or } \pi$
$\displaystyle \text{Since } 0<x<\pi,\ \text{no critical point lies in } (0,\pi)$
$\displaystyle \text{Hence, } f(x)\text{ has no local maximum or minimum in } (0,\pi)$

$\displaystyle \textbf{Question 10:}~f(x)=\sin 2x-x,\;-\dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=\sin 2x-x$
$\displaystyle \Rightarrow f'(x)=2\cos 2x-1$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 2\cos 2x-1=0$
$\displaystyle \Rightarrow \cos 2x=\frac{1}{2}$
$\displaystyle \Rightarrow 2x=\pm\frac{\pi}{3}$
$\displaystyle \Rightarrow x=\pm\frac{\pi}{6}$
$\displaystyle \text{Now, } f'(x)>0 \text{ for } x<\frac{\pi}{6} \text{ and } f'(x)<0 \text{ for } x>\frac{\pi}{6}$
$\displaystyle \Rightarrow x=\frac{\pi}{6} \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=\frac{\pi}{6} \text{ is } \sin\!\left(\frac{\pi}{3}\right)-\frac{\pi}{6} =\frac{\sqrt{3}}{2}-\frac{\pi}{6}$
$\displaystyle \text{Further, } f'(x)<0 \text{ for } x>-\frac{\pi}{6} \text{ and } f'(x)>0 \text{ for } x<-\frac{\pi}{6}$
$\displaystyle \Rightarrow x=-\frac{\pi}{6} \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=-\frac{\pi}{6} \text{ is } \sin\!\left(-\frac{\pi}{3}\right)+\frac{\pi}{6} =\frac{\pi}{6}-\frac{\sqrt{3}}{2}$

$\displaystyle \textbf{Question 11:}~f(x)=2\sin x-x,\;-\dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=2\sin x-x$
$\displaystyle \Rightarrow f'(x)=2\cos x-1$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 2\cos x-1=0$
$\displaystyle \Rightarrow \cos x=\frac{1}{2}$
$\displaystyle \Rightarrow x=\pm\frac{\pi}{3}$
$\displaystyle \text{Now, } f'(x)>0 \text{ for } x<\frac{\pi}{3} \text{ and } f'(x)<0 \text{ for } x>\frac{\pi}{3}$
$\displaystyle \Rightarrow x=\frac{\pi}{3} \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=\frac{\pi}{3} \text{ is } 2\sin\!\left(\frac{\pi}{3}\right)-\frac{\pi}{3} =\sqrt{3}-\frac{\pi}{3}$
$\displaystyle \text{Further, } f'(x)<0 \text{ for } x>-\frac{\pi}{3} \text{ and } f'(x)>0 \text{ for } x<-\frac{\pi}{3}$
$\displaystyle \Rightarrow x=-\frac{\pi}{3} \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=-\frac{\pi}{3} \text{ is } 2\sin\!\left(-\frac{\pi}{3}\right)+\frac{\pi}{3} =\frac{\pi}{3}-\sqrt{3}$

$\displaystyle \textbf{Question 12:}~f(x)=x\sqrt{1-x},\;x>0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x\sqrt{1-x}$
$\displaystyle \Rightarrow f'(x)=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}$
$\displaystyle \Rightarrow f'(x)=\frac{2(1-x)-x}{2\sqrt{1-x}}$
$\displaystyle \Rightarrow f'(x)=\frac{2-3x}{2\sqrt{1-x}}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow \frac{2-3x}{2\sqrt{1-x}}=0$
$\displaystyle \Rightarrow 2-3x=0$
$\displaystyle \Rightarrow x=\frac{2}{3}$
$\displaystyle \text{Now, for } x<\frac{2}{3},\ f'(x)>0 \text{ and for } x>\frac{2}{3},\ f'(x)<0$
$\displaystyle \Rightarrow x=\frac{2}{3} \text{ is a point of local maximum}$
$\displaystyle \text{The local maximum value of } f(x) \text{ at } x=\frac{2}{3} \text{ is } \frac{2}{3}\sqrt{1-\frac{2}{3}} =\frac{2}{3\sqrt{3}} =\frac{2\sqrt{3}}{9}$

$\displaystyle \textbf{Question 13:}~f(x)=x^3(2x-1)^3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=x^{3}(2x-1)^{3}$
$\displaystyle \Rightarrow f'(x)=3x^{2}(2x-1)^{3} + x^{3}\cdot 3(2x-1)^{2}\cdot 2$
$\displaystyle \Rightarrow f'(x)=3x^{2}(2x-1)^{2} \big[(2x-1)+2x\big]$
$\displaystyle \Rightarrow f'(x)=3x^{2}(2x-1)^{2}(4x-1)$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow 3x^{2}(2x-1)^{2}(4x-1)=0$
$\displaystyle \Rightarrow x=0,\ \frac{1}{2},\ \frac{1}{4}$
$\displaystyle \text{Now, } f'(x)\text{ does not change sign at } x=0 \text{ and } x=\frac{1}{2}$
$\displaystyle \Rightarrow x=0 \text{ and } x=\frac{1}{2} \text{ are points of inflexion}$
$\displaystyle \text{For } x<\frac{1}{4},\ f'(x)<0 \text{ and for } x>\frac{1}{4},\ f'(x)>0$
$\displaystyle \Rightarrow x=\frac{1}{4} \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=\frac{1}{4} \text{ is } \left(\frac{1}{4}\right)^{3} \left(2\cdot\frac{1}{4}-1\right)^{3} =-\frac{1}{512}$

$\displaystyle \textbf{Question 14:}~f(x)=\dfrac{x}{2}+\dfrac{2}{x},\;x>0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } f(x)=\frac{x}{2}+\frac{2}{x},\ x\ne 0$
$\displaystyle \Rightarrow f'(x)=\frac{1}{2}-\frac{2}{x^{2}}$
$\displaystyle \text{For a local maximum or a local minimum, we require } f'(x)=0$
$\displaystyle \Rightarrow \frac{1}{2}-\frac{2}{x^{2}}=0$
$\displaystyle \Rightarrow \frac{1}{2}=\frac{2}{x^{2}}$
$\displaystyle \Rightarrow x^{2}=4$
$\displaystyle \Rightarrow x=\pm 2$
$\displaystyle \text{Now, } f'(x)<0 \text{ for } 0<x<2 \text{ and } f'(x)>0 \text{ for } x>2$
$\displaystyle \Rightarrow x=2 \text{ is a point of local minimum}$
$\displaystyle \text{The local minimum value of } f(x) \text{ at } x=2 \text{ is } \frac{2}{2}+\frac{2}{2}=2$