CONTINUITY OF A ABSOLUTE VALUE FUNCTION AT A POINT
For a function to be continuous at a point,
It must be defined at that point
Its limit must exist at the point
The value of the function at that point must be equal to the value of the limit at that point.
If the function f(x) is continuous at a point x = a, then
Problem 1 :
Check if the function is continuous at x = 0
f(x) = 2x - |x|
Solution :
Step 1 :
Defining the given absolute value function as piece wise function.
Step 2 :
Evaluating the left hand limit :
lim x→0-f(x)=lim x→0- xApplying the limit, we get=0lim x→0-f(x)= 0 ------(1)
Evaluating the right hand limit :
lim x→0+f(x)=lim x→0+ 3xApplying the limit, we get=3(0lim x→0+f(x)= 0 ------(2)
Both left hand limit and right hand limit they are equal.
lim x→0-f(x)=lim x→0+f(x)
So, the function is continuous at x = 0.
Problem 2 :
Discuss the continuity of the function of given by
f(x) = |x - 1| + |x - 2| at x = 1 and x = 2.
Solution :
The condition for three pieces will be
x < 1, 1 ⩽ x < 2 and x ⩾ 2
Choosing one of the value in the first condition (x < 1) and applying the absolute value function, we get negative for x - 1 and for x - 2.
Choosing one of the value in the second condition (1 ⩽ x < 2) and applying the absolute value function, we get positive for x - 1 and negative for x - 2.
Choosing one of the value in the third condition (x ⩾ 2) and applying the absolute value function, we get positive for x - 1 and x - 2.
f(x)=-(x-1)-(x-2)if x <1(x-1)-(x-2)if 1⩽x<2(x-1)+(x-2)if x ⩾2f(x)=-x+1-x+2if x <1x-1-x+2if 1⩽x<2x-1+x-2if x ⩾2f(x)=-2x+3 if x <11 if 1⩽x<22x-3 if x ⩾2