Page 3 - LN

⇒ f(x ) < f x
2
1
So x < x ⇒ f(x ) < f x for all x ,x ∈ R
2
1
1
1
2
2
So f(x) =2x + 1 is the function strictly increasing on R.

2
Example:- Show that the function f(x) = x - x +1 is neither increasing nor decreasing in the
interval (0,1)

2
Ans:- f(x) = x – x + 1

⇒ f (x) = 2x – 1
1
f (x) = 0
1

⇒ 2x - 1 = 0

1
⇒ x =
2

1
1
2
Choose interval (0, ) f (x )< 0 , So the function(x) = x - x + 1 is decreasing
2
1
2
1
In the interval ( , 1) f (x) > 0 , so the function f(x) = x - x + 1 is st.increasing function
2
2
So the function f(x) = x - x +1 is neither increasing nor decreasing in the interval (0,1)

Working rule to find the interval in which function is increasing or decreasing :-

Let f(x) be a function in given interval

1
Step-1: find f (x)
1
Step-2: put f (x) = 0 and find the value of x in R

Step-3: divide the number line (R) into disjoint sub intervals with the help of values of x
obtained in step-2

1
1
Step-4: in each subinterval f (x) > 0 or f (x)< 0
1
1)If f (x) < 0, then f(x) is st . increasing in that interval.