\displaystyle \textbf{Question 1: }\text{Find the maximum and minimum values, if any, without}
\displaystyle \text{using derivatives of the following function:}

\displaystyle \textbf{(i) }f(x)=4x^2-4x+4,\quad x\in\mathbb{R}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=4x^2-4x+4
\displaystyle f(x)=(4x^2-4x+1)+3
\displaystyle f(x)=(2x-1)^2+3
\displaystyle \text{Since }(2x-1)^2\geq0,\quad \forall\,x\in\mathbb{R},
\displaystyle (2x-1)^2+3\geq3
\displaystyle \therefore f(x)\geq3,\quad \forall\,x\in\mathbb{R}
\displaystyle \text{Equality holds when }2x-1=0
\displaystyle \Rightarrow x=\frac{1}{2}
\displaystyle \therefore \text{The minimum value of }f(x)\text{ is }3\text{, attained at }x=\frac{1}{2}.
\displaystyle \text{Also, }f(x)\text{ can be made arbitrarily large as }|x|\text{ increases.}
\displaystyle \therefore \text{The maximum value of }f(x)\text{ does not exist.}
\displaystyle \\

 

\displaystyle \textbf{(ii) }f(x)=-(x-1)^2+2,\quad x\in\mathbb{R}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=-(x-1)^2+2
\displaystyle \text{Since }(x-1)^2\ge0,\quad \forall\,x\in\mathbb{R},
\displaystyle -(x-1)^2\le0
\displaystyle \therefore f(x)=-(x-1)^2+2\le2,\quad \forall\,x\in\mathbb{R}
\displaystyle \text{Equality holds when }x-1=0
\displaystyle \Rightarrow x=1
\displaystyle \therefore \text{The maximum value of }f(x)\text{ is }2\text{, attained at }x=1.
\displaystyle \text{Also, }f(x)\text{ can be made arbitrarily small as }|x|\text{ increases.}
\displaystyle \therefore \text{The minimum value of }f(x)\text{ does not exist.}

 

\displaystyle \textbf{(iii) }f(x)=|x+2|,\quad x\in\mathbb{R}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=|x+2|
\displaystyle \text{Since }|x+2|\ge0,\quad \forall\,x\in\mathbb{R},
\displaystyle \therefore f(x)\ge0,\quad \forall\,x\in\mathbb{R}
\displaystyle \text{Equality holds when }x+2=0
\displaystyle \Rightarrow x=-2
\displaystyle \therefore \text{The minimum value of }f(x)\text{ is }0\text{, attained at }x=-2.
\displaystyle \text{Also, }f(x)\text{ can be made arbitrarily large as }|x|\text{ increases.}
\displaystyle \therefore \text{The maximum value of }f(x)\text{ does not exist.}

 

\displaystyle \textbf{(iv) }f(x)=\sin2x+5,\quad x\in\mathbb{R}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\sin2x+5
\displaystyle \text{Since }-1\le\sin2x\le1,\quad \forall\,x\in\mathbb{R},
\displaystyle -1+5\le\sin2x+5\le1+5
\displaystyle \therefore 4\le f(x)\le6
\displaystyle \text{Hence, the minimum value of }f(x)\text{ is }4\text{ and the maximum value is }6.
\displaystyle \text{Now, }f(x)=6\Rightarrow\sin2x=1
\displaystyle \Rightarrow 2x=\frac{\pi}{2}+2n\pi,\;n\in\mathbb{Z}
\displaystyle \Rightarrow x=\frac{\pi}{4}+n\pi,\;n\in\mathbb{Z}
\displaystyle \text{Thus, }f(x)\text{ attains its maximum value }6\text{ at }x=\frac{\pi}{4}+n\pi,\;n\in\mathbb{Z}.
\displaystyle \text{Also, }f(x)=4\Rightarrow\sin2x=-1
\displaystyle \Rightarrow 2x=\frac{3\pi}{2}+2n\pi,\;n\in\mathbb{Z}
\displaystyle \Rightarrow x=\frac{3\pi}{4}+n\pi,\;n\in\mathbb{Z}
\displaystyle \text{Thus, }f(x)\text{ attains its minimum value }4\text{ at }x=\frac{3\pi}{4}+n\pi,\;n\in\mathbb{Z}.

 

\displaystyle \textbf{(v) }f(x)=|\sin4x+3|,\quad x\in\mathbb{R}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=|\sin4x+3|
\displaystyle \text{Since }-1\le\sin4x\le1,\quad \forall\,x\in\mathbb{R},
\displaystyle 2\le\sin4x+3\le4
\displaystyle \text{Also, }\sin4x+3>0,\quad \forall\,x\in\mathbb{R}
\displaystyle \therefore |\sin4x+3|=\sin4x+3
\displaystyle \therefore 2\le f(x)\le4
\displaystyle \text{Hence, the minimum value of }f(x)\text{ is }2\text{ and the maximum value is }4.
\displaystyle \text{Now, }f(x)=4\Rightarrow\sin4x=1
\displaystyle \Rightarrow 4x=\frac{\pi}{2}+2n\pi,\;n\in\mathbb{Z}
\displaystyle \Rightarrow x=\frac{\pi}{8}+\frac{n\pi}{2},\;n\in\mathbb{Z}
\displaystyle \text{Thus, }f(x)\text{ attains its maximum value }4\text{ at }x=\frac{\pi}{8}+\frac{n\pi}{2},\;n\in\mathbb{Z}.
\displaystyle \text{Also, }f(x)=2\Rightarrow\sin4x=-1
\displaystyle \Rightarrow 4x=\frac{3\pi}{2}+2n\pi,\;n\in\mathbb{Z}
\displaystyle \Rightarrow x=\frac{3\pi}{8}+\frac{n\pi}{2},\;n\in\mathbb{Z}
\displaystyle \text{Thus, }f(x)\text{ attains its minimum value }2\text{ at }x=\frac{3\pi}{8}+\frac{n\pi}{2},\;n\in\mathbb{Z}.

 

\displaystyle \textbf{(vi) }f(x)=2x^3+5,\quad x\in\mathbb{R}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=2x^3+5
\displaystyle \text{Since }x^3\text{ is a strictly increasing function on }\mathbb{R},
\displaystyle \text{the function }2x^3+5\text{ is also strictly increasing on }\mathbb{R}.
\displaystyle \text{Also, }f(x)\to-\infty\text{ as }x\to-\infty
\displaystyle \text{and }f(x)\to\infty\text{ as }x\to\infty
\displaystyle \therefore \text{The function }f(x)\text{ has no minimum value.}
\displaystyle \therefore \text{The function }f(x)\text{ has no maximum value.}

 

\displaystyle \textbf{(vii) }f(x)=-|x+1|+3,\quad x\in\mathbb{R}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=-|x+1|+3
\displaystyle \text{Since }|x+1|\ge0,\quad \forall\,x\in\mathbb{R},
\displaystyle -|x+1|\le0
\displaystyle \therefore f(x)=-|x+1|+3\le3,\quad \forall\,x\in\mathbb{R}
\displaystyle \text{Equality holds when }|x+1|=0
\displaystyle \Rightarrow x=-1
\displaystyle \therefore \text{The maximum value of }f(x)\text{ is }3\text{, attained at }x=-1.
\displaystyle \text{Also, }f(x)\text{ can be made arbitrarily small as }|x|\text{ increases.}
\displaystyle \therefore \text{The minimum value of }f(x)\text{ does not exist.}

 

\displaystyle \textbf{(viii) }f(x)=16x^2-16x+28,\quad x\in\mathbb{R}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=16x^2-16x+28
\displaystyle f(x)=4(4x^2-4x+1)+24
\displaystyle f(x)=4(2x-1)^2+24
\displaystyle \text{Since }(2x-1)^2\ge0,\quad \forall\,x\in\mathbb{R},
\displaystyle 4(2x-1)^2+24\ge24
\displaystyle \therefore f(x)\ge24,\quad \forall\,x\in\mathbb{R}
\displaystyle \text{Equality holds when }2x-1=0
\displaystyle \Rightarrow x=\frac{1}{2}
\displaystyle \therefore \text{The minimum value of }f(x)\text{ is }24\text{, attained at }x=\frac{1}{2}.
\displaystyle \text{Also, }f(x)\text{ can be made arbitrarily large as }|x|\text{ increases.}
\displaystyle \therefore \text{The maximum value of }f(x)\text{ does not exist.}

 

\displaystyle \textbf{(ix) }f(x)=x^3-1,\quad x\in\mathbb{R}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=x^3-1
\displaystyle \text{Since }x^3\text{ is a strictly increasing function on }\mathbb{R},
\displaystyle \text{the function }x^3-1\text{ is also strictly increasing on }\mathbb{R}.
\displaystyle \text{Also, }f(x)\to-\infty\text{ as }x\to-\infty
\displaystyle \text{and }f(x)\to\infty\text{ as }x\to\infty
\displaystyle \therefore \text{The function }f(x)\text{ has no minimum value.}
\displaystyle \therefore \text{The function }f(x)\text{ has no maximum value.}

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