Question

Show that \( f(x) = |x| \) is continuous for all values of x.

Hint:

For piecewise functions, the only points where continuity might fail are the "break points" where the definition of the function changes. If the function is continuous at these points, and the individual pieces are continuous functions, then the entire function is continuous.

Answer 1

Step 1: Understanding the Concept:
To show that the absolute value function is continuous everywhere, we can analyze it as a piecewise function and check for continuity at the point where its definition changes, which is at \( x=0 \). For all other points, the function is a simple polynomial (\( y=x \) or \( y=-x \)), which is known to be continuous.
Step 2: Key Formula or Approach:
The piecewise definition of \( |x| \) is:
\[ f(x) = |x| = \begin{cases} x & \text{if } x \ge 0
-x & \text{if } x<0 \end{cases} \] For continuity at a point \( c \), we must show \( \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \).
Step 3: Detailed Explanation or Calculation:
Case 1: \( c>0 \)
For any \( c>0 \), in a small neighborhood around c, \( f(x) = x \). \[ \lim_{x \to c} f(x) = \lim_{x \to c} x = c \] Also, \( f(c) = c \). Since the limit equals the function value, \( f(x) \) is continuous for all \( c>0 \).
Case 2: \( c<0 \)
For any \( c<0 \), in a small neighborhood around c, \( f(x) = -x \). \[ \lim_{x \to c} f(x) = \lim_{x \to c} (-x) = -c \] Also, \( f(c) = -c \). Since the limit equals the function value, \( f(x) \) is continuous for all \( c<0 \).
Case 3: \( c = 0 \)
We need to check the left-hand limit, right-hand limit, and the function value at \( x=0 \).
- Left-Hand Limit (LHL): For \( x \to 0^- \), we use \( f(x) = -x \). \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0 \] - Right-Hand Limit (RHL): For \( x \to 0^+ \), we use \( f(x) = x \). \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x) = 0 \] - Function Value: \( f(0) = 0 \).
Since LHL = RHL = \( f(0) = 0 \), the function is continuous at \( x=0 \).
Step 4: Final Answer:
Since the function is continuous for \( x>0 \), \( x<0 \), and at \( x = 0 \), it is continuous for all real values of x. Hence proved.