discuss the continuity of function f(x)= |x|+|x-1| in interval [-1,2]
discuss the continuity of function f(x)= |x|+|x-1| in interval [-1,2]
It can be directly concluded by observation that the function will be continuous across R but further justification is given below. Note that both |x| and |x-1| are continuous across R though not differentiable at every value of x.
f(x) = |x| + |x-1|
It can be observe that the critical points for the modulus bound terms are 0 and 1. I.e. at x = 0, |x| changes its sign and at x = 1, |x-1| changes its sign.
For x ÃŽ[-1, 0]
f(x) = -x -(x-1) =-2x+1
This is a linear function and is continuous.
For x = 0
LHL = RHL = f(0) = 1
For x ÃŽ[0, 1]
f(x) = x -(x-1) =1
This is a constant linear function and is continuous.
For x = 1
LHL = RHL = f(1) = 1
For x ÃŽ[1, 2]
f(x) = x+(x-1) =2x-1
This is a constant linear function and is continuous.
Hence, f(x) is continuous in the interval [-1,2].
It can be directly concluded by observation that the function will be continuous across R but further justification is given below. Note that both |x| and |x-1| are continuous across R though not differentiable at every value of x.
f(x) = |x| + |x-1|
It can be observe that the critical points for the modulus bound terms are 0 and 1. I.e. at x = 0, |x| changes its sign and at x = 1, |x-1| changes its sign.
For x ÃŽ[-1, 0]
f(x) = -x -(x-1) =-2x+1
This is a linear function and is continuous.
For x = 0
LHL = RHL = f(0) = 1
For x ÃŽ[0, 1]
f(x) = x -(x-1) =1
This is a constant linear function and is continuous.
For x = 1
LHL = RHL = f(1) = 1
For x ÃŽ[1, 2]
f(x) = x+(x-1) =2x-1
This is a constant linear function and is continuous.
Hence, f(x) is continuous in the interval [-1,2].
