Question

14.
Show that f(x) = 8x+2 is strictly increasing on R (without using the Derivative test)
*
15.
Show that f(x) =
is strictly decreasing on R (without using the Derivative test)

Answer 1

Step-by-step explanation:

x1>x2

8x1>8x2

8x1+2>8x2+2

f(x1)>f(x2)

x1>x2

implies

f(x1)>f(x2)

f(x) is strictly increasing function

Answer 2

Answer:

$f(x)$ is a increasing function

Step-by-step explanation:

Given $f(x)=8x+2$

$f(x)$  is said to be a increasing function when $f(x_{2})-f(x_{1})\ge 0$.

Let $x_{1} \ \&\ x_{2}$ be the two real numbers and $x_{2}=x_{1} +a$

Then  $f(x_{1})=8x_{1}+2$

         $f(x_{2})=f(x_{1}+a)=8(x_{1}+a)+2$

                                      $=8x_{1}+8a+2$

                                      $=f(x_{1})+8a$

 $\implies f(x_{2})-f(x_{1})=8a$  $\implies f(x_{2})-f(x_{1})\ge 0$

Therefore, $f(x)$ is a increasing function.

Second part, the function is not given in the question.