Class 12: Maxima and Minima

$\displaystyle \textbf{Question 1: }~\text{Find the absolute maximum and the absolute minimum values of} \\ \text{the following functions in the given intervals:}$

$\displaystyle \textbf{(i)}~f(x)=4x-\dfrac{x^2}{2}\text{ in }[-2,45]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=4x-\frac{x^{2}}{2}$
$\displaystyle \Rightarrow f'(x)=4-x$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow 4-x=0$
$\displaystyle \Rightarrow x=4$
$\displaystyle \text{Thus, } x=4 \text{ is the critical point of } f.$
$\displaystyle \text{Now, evaluate } f(x) \text{ at the end points and the critical point.}$
$\displaystyle f(-2)=4(-2)-\frac{(-2)^2}{2}=-8-2=-10$
$\displaystyle f(4)=4(4)-\frac{4^2}{2}=16-8=8$
$\displaystyle f(4.5)=4(4.5)-\frac{(4.5)^2}{2}=18-10.125=7.875$
$\displaystyle \text{Hence, the absolute maximum value of } f(x) \text{ is } 8 \text{ at } x=4,$
$\displaystyle \text{and the absolute minimum value of } f(x) \text{ is } -10 \text{ at } x=-2.$

$\displaystyle \textbf{(ii)}~f(x)=(x-1)^2+3\text{ in }[-3,1]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=(x-1)^2+3$
$\displaystyle \Rightarrow f'(x)=2(x-1)$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow 2(x-1)=0$
$\displaystyle \Rightarrow x=1$
$\displaystyle \text{Thus, } x=1 \text{ is the critical point of } f.$
$\displaystyle \text{Now, evaluate } f(x) \text{ at the end points and the critical point.}$
$\displaystyle f(-3)=(-3-1)^2+3=16+3=19$
$\displaystyle f(1)=(1-1)^2+3=3$
$\displaystyle \text{Hence, the absolute maximum value of } f(x) \text{ is } 19 \text{ at } x=-3,$
$\displaystyle \text{and the absolute minimum value of } f(x) \text{ is } 3 \text{ at } x=1.$

$\displaystyle \textbf{(iii)}~f(x)=3x^4-8x^3+12x^2-48x+25\text{ in }[0,3]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=3x^{4}-8x^{3}+12x^{2}-48x+25$
$\displaystyle \Rightarrow f'(x)=12x^{3}-24x^{2}+24x-48$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow 12x^{3}-24x^{2}+24x-48=0$
$\displaystyle \Rightarrow x^{3}-2x^{2}+2x-4=0$
$\displaystyle \Rightarrow x^{2}(x-2)+2(x-2)=0$
$\displaystyle \Rightarrow (x-2)(x^{2}+2)=0$
$\displaystyle \Rightarrow x=2$
$\displaystyle \text{No real root exists for } x^{2}+2=0$
$\displaystyle \text{Thus, } x=2 \text{ is the only critical point of } f.$
$\displaystyle \text{Now, evaluate } f(x) \text{ at the end points and the critical point.}$
$\displaystyle f(0)=3(0)^{4}-8(0)^{3}+12(0)^{2}-48(0)+25=25$
$\displaystyle f(2)=3(2)^{4}-8(2)^{3}+12(2)^{2}-48(2)+25=-39$
$\displaystyle f(3)=3(3)^{4}-8(3)^{3}+12(3)^{2}-48(3)+25=16$
$\displaystyle \text{Hence, the absolute maximum value of } f(x) \text{ is } 25 \text{ at } x=0,$
$\displaystyle \text{and the absolute minimum value of } f(x) \text{ is } -39 \text{ at } x=2.$

$\displaystyle \textbf{(iv)}~f(x)=(x-2)\sqrt{x-1}\text{ in }[1,9]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=(x-2)\sqrt{x-1}$
$\displaystyle \Rightarrow f'(x)=\sqrt{x-1}+\frac{x-2}{2\sqrt{x-1}}$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow \sqrt{x-1}+\frac{x-2}{2\sqrt{x-1}}=0$
$\displaystyle \Rightarrow 2(x-1)+(x-2)=0$
$\displaystyle \Rightarrow 2x-2+x-2=0$
$\displaystyle \Rightarrow 3x-4=0$
$\displaystyle \Rightarrow x=\frac{4}{3}$
$\displaystyle \text{Thus, } x=\frac{4}{3} \text{ is the critical point of } f.$
$\displaystyle \text{Now, evaluate } f(x) \text{ at the end points and the critical point.}$
$\displaystyle f(1)=(1-2)\sqrt{1-1}=0$
$\displaystyle f\!\left(\frac{4}{3}\right)=\left(\frac{4}{3}-2\right)\sqrt{\frac{4}{3}-1} = -\frac{2}{3}\cdot\frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}$
$\displaystyle f(9)=(9-2)\sqrt{9-1}=14\sqrt{2}$
$\displaystyle \text{Hence, the absolute maximum value of } f(x) \text{ is } 14\sqrt{2} \text{ at } x=9,$
$\displaystyle \text{and the absolute minimum value of } f(x) \text{ is } -\frac{2}{3\sqrt{3}} \text{ at } x=\frac{4}{3}.$

$\displaystyle \textbf{Question 2: }~\text{Find the maximum value of }2x^3-24x+107\text{ in the interval }[1,3]. \\ \text{ Find the maximum value of the same function in }[-3,-1].$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=2x^{3}-24x+107$
$\displaystyle \Rightarrow f'(x)=6x^{2}-24$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow 6x^{2}-24=0$
$\displaystyle \Rightarrow x^{2}=4$
$\displaystyle \Rightarrow x=\pm 2$
$\displaystyle \text{Thus, the critical points of } f \text{ are } x=-2 \text{ and } x=2.$
$\displaystyle \text{Now, consider the interval } [1,3].$
$\displaystyle \text{Evaluate } f(x) \text{ at the end points and the critical point in } [1,3].$
$\displaystyle f(1)=2(1)^{3}-24(1)+107=85$
$\displaystyle f(2)=2(2)^{3}-24(2)+107=75$
$\displaystyle f(3)=2(3)^{3}-24(3)+107=89$
$\displaystyle \text{Hence, the absolute maximum value of } f(x) \text{ in } [1,3] \text{ is } 89 \text{ at } x=3.$
$\displaystyle \text{Now, consider the interval } [-3,-1].$
$\displaystyle \text{Evaluate } f(x) \text{ at the end points and the critical point in } [-3,-1].$
$\displaystyle f(-3)=2(-3)^{3}-24(-3)+107=125$
$\displaystyle f(-2)=2(-2)^{3}-24(-2)+107=139$
$\displaystyle f(-1)=2(-1)^{3}-24(-1)+107=129$
$\displaystyle \text{Hence, the absolute maximum value of } f(x) \text{ in } [-3,-1] \text{ is } 139 \text{ at } x=-2.$

$\displaystyle \textbf{Question 3: }~\text{Find the absolute maximum and minimum values of the function } \\ f\text{ given by }f(x)=\cos^2 x+\sin x,\;x\in[0,\pi].$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=\cos^{2}x+\sin x$
$\displaystyle \Rightarrow f'(x)=2\cos x(-\sin x)+\cos x$
$\displaystyle \Rightarrow f'(x)=-2\sin x\cos x+\cos x$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow -2\sin x\cos x+\cos x=0$
$\displaystyle \Rightarrow \cos x(2\sin x-1)=0$
$\displaystyle \Rightarrow \sin x=\frac{1}{2} \text{ or } \cos x=0$
$\displaystyle \Rightarrow x=\frac{\pi}{6} \text{ or } \frac{\pi}{2} \quad [x\in(0,\pi)]$
$\displaystyle \text{Thus, the critical points of } f \text{ in } [0,\pi] \text{ are } 0,\ \frac{\pi}{6},\ \frac{\pi}{2} \text{ and } \pi.$
$\displaystyle \text{Now, evaluate } f(x) \text{ at the end points and the critical points.}$
$\displaystyle f(0)=\cos^{2}(0)+\sin(0)=1$
$\displaystyle f\!\left(\frac{\pi}{6}\right)=\cos^{2}\!\left(\frac{\pi}{6}\right)+\sin\!\left(\frac{\pi}{6}\right)=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}$
$\displaystyle f\!\left(\frac{\pi}{2}\right)=\cos^{2}\!\left(\frac{\pi}{2}\right)+\sin\!\left(\frac{\pi}{2}\right)=1$
$\displaystyle f(\pi)=\cos^{2}(\pi)+\sin(\pi)=1$
$\displaystyle \text{Hence, the absolute maximum value of } f(x) \text{ in } [0,\pi] \text{ is } \frac{5}{4} \text{ at } x=\frac{\pi}{6},$
$\displaystyle \text{and the absolute minimum value of } f(x) \text{ in } [0,\pi] \text{ is } 1 \text{ at } x=0,\ \frac{\pi}{2},\ \pi.$

$\displaystyle \textbf{Question 4: }~\text{Find absolute maximum and minimum values of a function } \\ f\text{ given by }f(x)=12x^{4/3}-6x^{1/3},\;x\in[-1,1].$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=12x^{\frac{4}{3}}-6x^{\frac{1}{3}}$
$\displaystyle \Rightarrow f'(x)=16x^{\frac{1}{3}}-2x^{-\frac{2}{3}}$
$\displaystyle \Rightarrow f'(x)=\frac{2(8x-1)}{x^{\frac{2}{3}}}$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow \frac{2(8x-1)}{x^{\frac{2}{3}}}=0$
$\displaystyle \Rightarrow 8x-1=0$
$\displaystyle \Rightarrow x=\frac{1}{8}$
$\displaystyle \text{Thus, } x=\frac{1}{8} \text{ is the critical point of } f.$
$\displaystyle \text{Now, evaluate } f(x) \text{ at the end points and the critical point.}$
$\displaystyle f(-1)=12(-1)^{\frac{4}{3}}-6(-1)^{\frac{1}{3}}=12(1)-6(-1)=18$
$\displaystyle f\!\left(\frac{1}{8}\right)=12\left(\frac{1}{8}\right)^{\frac{4}{3}}-6\left(\frac{1}{8}\right)^{\frac{1}{3}} =\frac{3}{4}-\frac{3}{2}=-\frac{3}{4}$
$\displaystyle f(1)=12(1)^{\frac{4}{3}}-6(1)^{\frac{1}{3}}=6$
$\displaystyle \text{Hence, the absolute maximum value of } f(x) \text{ is } 18 \text{ at } x=-1,$
$\displaystyle \text{and the absolute minimum value of } f(x) \text{ is } -\frac{3}{4} \text{ at } x=\frac{1}{8}.$

$\displaystyle \textbf{Question 5: }~\text{Find the absolute maximum and minimum values of a function } \\ f\text{ given by }f(x)=2x^3-15x^2+36x+1\text{ on the interval }[1,5].$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : } f(x)=2x^{3}-15x^{2}+36x+1$
$\displaystyle \Rightarrow f'(x)=6x^{2}-30x+36$
$\displaystyle \text{For a local maximum or a local minimum, we must have } f'(x)=0$
$\displaystyle \Rightarrow 6x^{2}-30x+36=0$
$\displaystyle \Rightarrow x^{2}-5x+6=0$
$\displaystyle \Rightarrow (x-2)(x-3)=0$
$\displaystyle \Rightarrow x=2 \text{ or } x=3$
$\displaystyle \text{Thus, } x=2 \text{ and } x=3 \text{ are the critical points of } f.$
$\displaystyle \text{Now, evaluate } f(x) \text{ at the end points and the critical points.}$
$\displaystyle f(1)=2(1)^{3}-15(1)^{2}+36(1)+1=24$
$\displaystyle f(2)=2(2)^{3}-15(2)^{2}+36(2)+1=29$
$\displaystyle f(3)=2(3)^{3}-15(3)^{2}+36(3)+1=28$
$\displaystyle f(5)=2(5)^{3}-15(5)^{2}+36(5)+1=56$
$\displaystyle \text{Hence, the absolute maximum value of } f(x) \text{ is } 56 \text{ at } x=5,$
$\displaystyle \text{and the absolute minimum value of } f(x) \text{ is } 24 \text{ at } x=1.$