\displaystyle \text{Question 1: 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{Then, } x_1 < x_2

\displaystyle \text{Implies that } \log_e x_1 < \log_e x_2

\displaystyle \text{Implies that } f(x_1) < f(x_2)

\displaystyle \text{Therefore, } x_1 < x_2 \text{ implies that } f(x_1) < f(x_2) \ \forall  \ x_1, x_2 \in ( 0, \infty)

\displaystyle \text{Therefore, } f(x) \text{ is increasing on }  ( 0, \infty).

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\displaystyle \text{Question 2: 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{Here, } x_1 < x_2

\displaystyle  \text{Implies that } \log_a x_1 < \log_a x_2

\displaystyle \text{Implies that } f(x_1) < f(x_2) \therefore x_1 < x_2

\displaystyle  \text{Implies that } f(x_1) < f(x_2) \ \forall  \ x_1, x_2 \in ( 0, \infty)

\displaystyle \text{Therefore, } f(x) \text{ is increasing on }  ( 0, \infty).

\displaystyle  \text{Case 2: Let } 0 < a < 1

\displaystyle  \text{Here, } x_1 < x_2

\displaystyle  \text{Implies that } \log_a x_1 > \log_a x_2

\displaystyle \text{Implies that } f(x_1) > f(x_2) \therefore x_1 < x_2

\displaystyle  \text{Implies that } f(x_1) > f(x_2) \ \forall  \ x_1, x_2 \in ( 0, \infty)

\displaystyle \text{Therefore, } f(x) \text{ is decreasing on }  ( 0, \infty).

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\displaystyle \text{Question 3: Prove that } f(x) = a x + b, \text{ where } a, b \text{  our constants and } a > 0 \\ \text{ is an increasing function on }  R.   

\displaystyle  \text{Answer:}  

\displaystyle  \text{Here, } f(x) = a x + b

\displaystyle \text{Let } x_1, x_2 \in R \text{ such that }  x_1 < x_2

\displaystyle \text{Then, } x_1 < x_2

\displaystyle \text{Implies that } ax_1 < ax_2  \because a > o

\displaystyle \text{Implies that } a x_1 + b < a x_2 + b

\displaystyle \text{Implies that } f(x_1) < f(x_2) \therefore x_1 < x_2

\displaystyle  \text{Implies that } f(x_1) < f(x_2) \ \forall  \ x_1, x_2 \in R

\displaystyle \text{Therefore, } f(x) \text{ is increasing on }  R.

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\displaystyle \text{Question 4: Prove that } f(x) = a x + b, \text{ where } a, b \text{  our constants and } a < 0 \\ \text{ is a decreasing function on }  R.   

\displaystyle  \text{Answer:}  

\displaystyle  \text{Here, } f(x) = a x + b

\displaystyle \text{Let } x_1, x_2 \in R \text{ such that }  x_1 < x_2

\displaystyle \text{Then, } x_1 < x_2

\displaystyle \text{Implies that } ax_1 > ax_2  \because a < o

\displaystyle \text{Implies that } a x_1 + b > a x_2 + b

\displaystyle \text{Implies that } f(x_1) > f(x_2) \therefore x_1 < x_2

\displaystyle  \text{Implies that } f(x_1) > f(x_2) \ \forall  \ x_1, x_2 \in R

\displaystyle \text{Therefore, } f(x) \text{ is decreasing on }  R.

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\displaystyle \text{Question 5:  Show that } f(x) = \frac{1}{x} \text{  is a decreasing function on } (0, \infty)  

\displaystyle  \text{Answer:}  

\displaystyle  \text{Here, } f(x) = \frac{1}{x}

\displaystyle \text{Let } x_1, x_2 \in (0, \infty) \text{ such that }  x_1 < x_2

\displaystyle \text{Then, } x_1 < x_2

\displaystyle \text{Implies that } \frac{1}{x_1} > \frac{1}{x_2}

\displaystyle \text{Implies that } f(x_1) > f(x_2) \therefore x_1 < x_2

\displaystyle  \text{Implies that } f(x_1) > f(x_2) \ \forall  \ x_1, x_2 \in (0, \infty)

\displaystyle \text{Therefore, } f(x) \text{ is decreasing on }  \in (0, \infty).

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\displaystyle \text{Question 6: Show that } f(x) = \frac{1}{1+x^2} \text{ decreasing in the interval } [0, \infty) \\ \text{ and increasing in the interval } (-\infty, 0].  

\displaystyle  \text{Answer:}  

\displaystyle  \text{Here, } 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 \text{Then, } x_1 < x_2

\displaystyle \text{Implies }  x_1^2 < x_2^2

\displaystyle \text{Implies } 1+ x_1^2 < 1+ x_2^2

\displaystyle \text{Implies } \frac{1}{1+x^2} > \frac{1}{1+x^2}

\displaystyle  \text{Implies that } f(x_1) > f(x_2) \ \forall  \ x_1, x_2 \in (0, \infty)

\displaystyle \text{Therefore, } f(x) \text{ is decreasing on }  \in (0, \infty).

\displaystyle \text{Case 2: Let } x_1, x_2 \in (-\infty, 0] \text{ such that }  x_1 < x_2

\displaystyle \text{Then, } x_1 < x_2

\displaystyle \text{Implies }  x_1^2 > x_2^2

\displaystyle \text{Implies } 1+ x_1^2 > 1+ x_2^2

\displaystyle \text{Implies } \frac{1}{1+x^2} < \frac{1}{1+x^2}

\displaystyle  \text{Implies that } f(x_1) < f(x_2) \ \forall  \ x_1, x_2 \in (0, \infty)

\displaystyle \text{Therefore, } f(x) \text{ is increasing on }  \in (-\infty, 0].

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\displaystyle \text{Question 7: Show that } f(x) = \frac{1}{1+x^2} \text{ is neither increasing nor decreasing on } R.   

\displaystyle  \text{Answer:}  

\displaystyle  \text{Here, } f(x) = \frac{1}{1+x^2}

\displaystyle R \text{ can be divided into two 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 \text{Then, } x_1 < x_2

\displaystyle \text{Implies }  x_1^2 < x_2^2

\displaystyle \text{Implies } 1+ x_1^2 < 1+ x_2^2

\displaystyle \text{Implies } \frac{1}{1+x^2} > \frac{1}{1+x^2}

\displaystyle  \text{Implies that } f(x_1) > f(x_2) \ \forall  \ x_1, x_2 \in (0, \infty)

\displaystyle \text{Therefore, } f(x) \text{ is decreasing on }  \in (0, \infty).

\displaystyle \text{Case 2: Let } x_1, x_2 \in (-\infty, 0] \text{ such that }  x_1 < x_2

\displaystyle \text{Then, } x_1 < x_2

\displaystyle \text{Implies }  x_1^2 > x_2^2

\displaystyle \text{Implies } 1+ x_1^2 > 1+ x_2^2

\displaystyle \text{Implies } \frac{1}{1+x^2} < \frac{1}{1+x^2}

\displaystyle  \text{Implies that } f(x_1) < f(x_2) \ \forall  \ x_1, x_2 \in (0, \infty)

\displaystyle \text{Therefore, } f(x) \text{ is increasing on }  \in (-\infty, 0].

\displaystyle  \text{Here, } f(x) \text{ is decreasing on } (0, \infty) \text{ and increasing on } (-\infty, 0).

\displaystyle  \text{Therefore, } f(x) \text{ is neither increasing or decreasing on }  R.

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\displaystyle \text{Question 8: Without using the derivative, show that the function } f(x) = |x| \text{ is } \\  \\ \text{(a) strictly increasing in } (0, \infty) \ \ \ \ \text{(b) strictly decreasing in } (-\infty, 0)    

\displaystyle  \text{Answer:}  

\displaystyle  \text{Here, } f(x) = |x|

\displaystyle \text{a) : Let } x_1, x_2 \in (0, \infty) \text{ such that }  x_1 < x_2

\displaystyle \text{Then, } x_1 < x_2

\displaystyle \text{Implies }  |x_1| < |x_2|

\displaystyle \text{Implies }  f(x_1) < f(x_2) \therefore x_1 < x_2

\displaystyle  \text{Implies that } f(x_1) < f(x_2) \ \forall  \ x_1, x_2 \in (0, \infty)

\displaystyle \text{Therefore, } f(x) \text{ is increasing on }  \in (0, \infty).

\displaystyle \text{b) : Let } x_1, x_2 \in (-\infty, 0] \text{ such that }  x_1 < x_2

\displaystyle \text{Then, } x_1 < x_2

\displaystyle \text{Implies }  |x_1| > |x_2|

\displaystyle \text{Implies }  f(x_1) > f(x_2) \therefore x_1 < x_2

\displaystyle  \text{Implies that } f(x_1) > f(x_2) \ \forall  \ x_1, x_2 \in (-\infty, 0]

\displaystyle \text{Therefore, } f(x) \text{ is decreasing on }  \in (-\infty, 0].

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\displaystyle \text{Question 9: Without using the derivative, show that the function } \\ f(x) = 7x-3\text{ is a strictly increasing function on } R.   

\displaystyle  \text{Answer:}  

\displaystyle  \text{Here, } f(x) = 7x-3

\displaystyle \text{Let } x_1, x_2 \in R \text{ such that }  x_1 < x_2

\displaystyle \text{Then, } x_1 < x_2

\displaystyle \text{Implies }  7x_1 - 3 < 7x_2 - 3

\displaystyle \text{Implies }  f(x_1) < f(x_2) \therefore x_1 < x_2

\displaystyle  \text{Implies that } f(x_1) < f(x_2) \ \forall  \ x_1, x_2 \in R

\displaystyle \text{Therefore, } f(x) \text{ is strictly increasing on }  R

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