Show that the function f(x)= |x| + |x-1| is continous at function for every value of x
To show that the function f(x) = |x| + |x-1| is continuous for every value of x, we can use the fact that the sum, product, and composition of continuous functions are continuous.
First, let's analyze the function in intervals.
For x < 0, both |x| and |x-1| are equal to -x. So, f(x) = |x| + |x-1| = -x + (-x) = -2x. This is a linear function and is continuous for x < 0.
For 0 ≤ x < 1, |x| is equal to x, and |x-1| is equal to 1-x. So, f(x) = |x| + |x-1| = x + (1-x) = 1. This is a constant function and is continuous for 0 ≤ x < 1.
For x ≥ 1, both |x| and |x-1| are equal to x. So, f(x) = |x| + |x-1| = x + x = 2x. This is a linear function and is continuous for x ≥ 1.
Since f(x) is continuous for x < 0, 0 ≤ x < 1, and x ≥ 1, and these intervals cover every value of x, we can conclude that the function f(x) = |x| + |x-1| is continuous for every value of x.
To show that the function f(x) = |x| + |x-1| is continuous for every value of x, we need to prove that it is continuous at the two break points: x = 0 and x = 1.
1. Continuity at x = 0:
We need to show that the limit as x approaches 0 of f(x) is equal to the value of f(0).
As x approaches 0 from the left side, i.e., x < 0,
| x | = -x and | x - 1 | = -(x - 1).
Therefore, f(x) = -x + (x - 1) = -1.
As x approaches 0 from the right side, i.e., x > 0,
| x | = x and | x - 1 | = x - 1.
Therefore, f(x) = x + (x - 1) = 2x - 1.
So, f(x) approaches -1 from the left side and 2x - 1 from the right side as x approaches 0.
Now, let's evaluate f(0) = |0| + |0 - 1| = 1.
Since the limit as x approaches 0 of f(x) (-1) is equal to f(0) (1), f(x) is continuous at x = 0.
2. Continuity at x = 1:
We need to show that the limit as x approaches 1 of f(x) is equal to the value of f(1).
As x approaches 1 from the left side, i.e., x < 1,
| x | = -x and | x - 1 | = -(x - 1).
Therefore, f(x) = -x + (x - 1) = -1.
As x approaches 1 from the right side, i.e., x > 1,
| x | = x and | x - 1 | = x - 1.
Therefore, f(x) = x + (x - 1) = 2x - 1.
So, f(x) approaches -1 from the left side and 2x - 1 from the right side as x approaches 1.
Now, let's evaluate f(1) = |1| + |1 - 1| = 1.
Since the limit as x approaches 1 of f(x) (-1) is equal to f(1) (1), f(x) is continuous at x = 1.
Since f(x) is continuous at the two break points x = 0 and x = 1, it is continuous for every value of x.