\displaystyle \textbf{Question 1: }\text{Prove that the function }f(x)=\log_e x\text{ is increasing on }(0,\infty).
\displaystyle \text{Answer:}
\displaystyle \text{Let }x_1,x_2\in(0,\infty)\text{ such that }x_1<x_2.
\displaystyle \text{Since }e^x\text{ is strictly increasing on }\mathbb{R},\text{ its inverse function }\log_e x\text{ is also strictly increasing.}
\displaystyle \therefore x_1<x_2\Rightarrow \log_e x_1<\log_e x_2.
\displaystyle \therefore x_1<x_2\Rightarrow f(x_1)<f(x_2),\ \forall\,x_1,x_2\in(0,\infty).
\displaystyle \therefore f(x)\text{ is strictly increasing on }(0,\infty).
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Prove that the function }f(x)=\log_a x\text{ is increasing on }(0,\infty)\text{ if }a>1\text{ and decreasing on }(0,\infty)\text{ if }0<a<1.
\displaystyle \text{Answer:}
\displaystyle f(x)=\log_a x
\displaystyle \text{Let }x_1,x_2\in(0,\infty)\text{ such that }x_1<x_2.
\displaystyle \text{Case 1: Let }a>1.
\displaystyle \text{Since }a^x\text{ is strictly increasing on }\mathbb{R},\text{ its inverse function }\log_a x\text{ is also strictly increasing.}
\displaystyle \therefore x_1<x_2\Rightarrow \log_a x_1<\log_a x_2.
\displaystyle \therefore x_1<x_2\Rightarrow f(x_1)<f(x_2),\ \forall\,x_1,x_2\in(0,\infty).
\displaystyle \therefore f(x)\text{ is strictly increasing on }(0,\infty).
\displaystyle \text{Case 2: Let }0<a<1.
\displaystyle \text{Since }a^x\text{ is strictly decreasing on }\mathbb{R},\text{ its inverse function }\log_a x\text{ is also strictly decreasing.}
\displaystyle \therefore x_1<x_2\Rightarrow \log_a x_1>\log_a x_2.
\displaystyle \therefore x_1<x_2\Rightarrow f(x_1)>f(x_2),\ \forall\,x_1,x_2\in(0,\infty).
\displaystyle \therefore f(x)\text{ is strictly decreasing on }(0,\infty).
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove that }f(x)=ax+b,\text{ where }a,b\text{ are constants and }a>0,\text{ is an increasing function on }\mathbb{R}.
\displaystyle \text{Answer:}
\displaystyle f(x)=ax+b
\displaystyle \text{Let }x_1,x_2\in\mathbb{R}\text{ such that }x_1<x_2.
\displaystyle \Rightarrow ax_1<ax_2\qquad[\because\,a>0]
\displaystyle \Rightarrow ax_1+b<ax_2+b
\displaystyle \Rightarrow f(x_1)<f(x_2)
\displaystyle \therefore x_1<x_2\Rightarrow f(x_1)<f(x_2),\ \forall\,x_1,x_2\in\mathbb{R}.
\displaystyle \therefore f(x)\text{ is strictly increasing on }\mathbb{R}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Prove that }f(x)=ax+b,\text{ where }a,b\text{ are constants and }a<0,\text{ is a decreasing function on }\mathbb{R}.
\displaystyle \text{Answer:}
\displaystyle f(x)=ax+b
\displaystyle \text{Let }x_1,x_2\in\mathbb{R}\text{ such that }x_1<x_2.
\displaystyle \Rightarrow ax_1>ax_2\qquad[\because\,a<0]
\displaystyle \Rightarrow ax_1+b>ax_2+b
\displaystyle \Rightarrow f(x_1)>f(x_2)
\displaystyle \therefore x_1<x_2\Rightarrow f(x_1)>f(x_2),\ \forall\,x_1,x_2\in\mathbb{R}.
\displaystyle \therefore f(x)\text{ is strictly decreasing on }\mathbb{R}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Show that }f(x)=\frac{1}{x}\text{ is a decreasing function on }(0,\infty).
\displaystyle \text{Answer:}
\displaystyle f(x)=\frac{1}{x}
\displaystyle \text{Let }x_1,x_2\in(0,\infty)\text{ such that }x_1<x_2.
\displaystyle \Rightarrow \frac{1}{x_1}>\frac{1}{x_2}
\displaystyle \Rightarrow f(x_1)>f(x_2)
\displaystyle \therefore x_1<x_2\Rightarrow f(x_1)>f(x_2),\ \forall\,x_1,x_2\in(0,\infty).
\displaystyle \therefore f(x)\text{ is strictly decreasing on }(0,\infty).
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Show that }f(x)=\frac{1}{1+x^2}\text{ decreases in the interval }[0,\infty)\text{ and}
\displaystyle \text{increases in the interval }(-\infty,0].
\displaystyle \text{Answer:}
\displaystyle f(x)=\frac{1}{1+x^2}
\displaystyle \text{Case 1: Let }x_1,x_2\in(0,\infty)\text{ such that }x_1<x_2.
\displaystyle \Rightarrow x_1^2<x_2^2
\displaystyle \Rightarrow 1+x_1^2<1+x_2^2
\displaystyle \Rightarrow \frac{1}{1+x_1^2}>\frac{1}{1+x_2^2}
\displaystyle \Rightarrow f(x_1)>f(x_2),\ \forall\,x_1,x_2\in(0,\infty).
\displaystyle \therefore f(x)\text{ is strictly decreasing on }[0,\infty).
\displaystyle \text{Case 2: Let }x_1,x_2\in(-\infty,0]\text{ such that }x_1<x_2.
\displaystyle \Rightarrow x_1^2>x_2^2
\displaystyle \Rightarrow 1+x_1^2>1+x_2^2
\displaystyle \Rightarrow \frac{1}{1+x_1^2}<\frac{1}{1+x_2^2}
\displaystyle \Rightarrow f(x_1)<f(x_2),\ \forall\,x_1,x_2\in(-\infty,0].
\displaystyle \therefore f(x)\text{ is strictly increasing on }(-\infty,0].
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Show that }f(x)=\frac{1}{1+x^2}\text{ is neither increasing nor decreasing on }\mathbb{R}.
\displaystyle \text{Answer:}
\displaystyle f(x)=\frac{1}{1+x^2}
\displaystyle \mathbb{R}\text{ can be divided into the intervals }(0,\infty)\text{ and }(-\infty,0].
\displaystyle \text{Case 1: Let }x_1,x_2\in(0,\infty)\text{ such that }x_1<x_2.
\displaystyle \Rightarrow x_1^2<x_2^2
\displaystyle \Rightarrow 1+x_1^2<1+x_2^2
\displaystyle \Rightarrow \frac{1}{1+x_1^2}>\frac{1}{1+x_2^2}
\displaystyle \Rightarrow f(x_1)>f(x_2),\ \forall\,x_1,x_2\in(0,\infty).
\displaystyle \therefore f(x)\text{ is strictly decreasing on }(0,\infty).
\displaystyle \text{Case 2: Let }x_1,x_2\in(-\infty,0]\text{ such that }x_1<x_2.
\displaystyle \Rightarrow x_1^2>x_2^2
\displaystyle \Rightarrow 1+x_1^2>1+x_2^2
\displaystyle \Rightarrow \frac{1}{1+x_1^2}<\frac{1}{1+x_2^2}
\displaystyle \Rightarrow f(x_1)<f(x_2),\ \forall\,x_1,x_2\in(-\infty,0].
\displaystyle \therefore f(x)\text{ is strictly increasing on }(-\infty,0].
\displaystyle \text{Thus, }f(x)\text{ has different monotonic behaviour on the two intervals.}
\displaystyle \therefore f(x)\text{ is neither increasing nor decreasing on }\mathbb{R}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Without using the derivative, show that the function }f(x)=|x|\text{ is }(a)\text{ strictly}
\displaystyle \text{increasing in }(0,\infty)\text{ and }(b)\text{ strictly decreasing in }(-\infty,0).
\displaystyle \text{Answer:}
\displaystyle f(x)=|x|
\displaystyle \text{(a) Let }x_1,x_2\in(0,\infty)\text{ such that }x_1<x_2.
\displaystyle \Rightarrow |x_1|<|x_2|
\displaystyle \Rightarrow f(x_1)<f(x_2)
\displaystyle \therefore x_1<x_2\Rightarrow f(x_1)<f(x_2),\ \forall\,x_1,x_2\in(0,\infty).
\displaystyle \therefore f(x)\text{ is strictly increasing on }(0,\infty).
\displaystyle \text{(b) Let }x_1,x_2\in(-\infty,0)\text{ such that }x_1<x_2.
\displaystyle \Rightarrow |x_1|>|x_2|
\displaystyle \Rightarrow f(x_1)>f(x_2)
\displaystyle \therefore x_1<x_2\Rightarrow f(x_1)>f(x_2),\ \forall\,x_1,x_2\in(-\infty,0).
\displaystyle \therefore f(x)\text{ is strictly decreasing on }(-\infty,0).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Without using the derivative show that the function }f(x)=7x-3\text{ is a}
\displaystyle \text{strictly increasing function on }\mathbb{R}.
\displaystyle \text{Answer:}
\displaystyle f(x)=7x-3
\displaystyle \text{Let }x_1,x_2\in\mathbb{R}\text{ such that }x_1<x_2.
\displaystyle \Rightarrow 7x_1<7x_2\qquad[\because\,7>0]
\displaystyle \Rightarrow 7x_1-3<7x_2-3
\displaystyle \Rightarrow f(x_1)<f(x_2)
\displaystyle \therefore x_1<x_2\Rightarrow f(x_1)<f(x_2),\ \forall\,x_1,x_2\in\mathbb{R}.
\displaystyle \therefore f(x)\text{ is strictly increasing on }\mathbb{R}.
\displaystyle \\

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