Question

If \( f(x) = \begin{cases} -2 & \text{if } x \le -1 \\ 2x & \text{if } -1 < x \le 1 \\ 2 & \text{if } x > 1 \end{cases} \), then test the continuity of the function at \( x = -1 \) and at \( x = 1 \).

Hint:

When dealing with piecewise functions, be very careful to select the correct piece of the function definition corresponding to the limit you are evaluating (left-hand, right-hand) or the point you are evaluating.

Answer 1

Step 1: Understanding the Concept:
For a function to be continuous at a point \( c \), the left-hand limit (LHL), the right-hand limit (RHL), and the value of the function at that point, \( f(c) \), must all be equal.
\[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \] Step 2: Key Formula or Approach:
We will test the continuity at \( x = -1 \) and \( x = 1 \) by calculating the LHL, RHL, and the function's value at each point.
Step 3: Detailed Explanation or Calculation:
Testing continuity at \( x = -1 \):
1. Left-Hand Limit (LHL):
For \( x<-1 \), the function is \( f(x) = -2 \).
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-2) = -2 \] 2. Right-Hand Limit (RHL):
For \( x>-1 \), the function is \( f(x) = 2x \).
\[ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (2x) = 2(-1) = -2 \] 3. Value of the function at \( x = -1 \):
For \( x = -1 \), the function is \( f(x) = -2 \).
\[ f(-1) = -2 \] Since LHL = RHL = \( f(-1) = -2 \), the function is continuous at \( x = -1 \).
Testing continuity at \( x = 1 \):
1. Left-Hand Limit (LHL):
For \( x<1 \), the function is \( f(x) = 2x \).
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x) = 2(1) = 2 \] 2. Right-Hand Limit (RHL):
For \( x>1 \), the function is \( f(x) = 2 \).
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2) = 2 \] 3. Value of the function at \( x = 1 \):
For \( x = 1 \), the function is \( f(x) = 2x \).
\[ f(1) = 2(1) = 2 \] Since LHL = RHL = \( f(1) = 2 \), the function is continuous at \( x = 1 \).
Step 4: Final Answer:
The function \( f(x) \) is continuous at both \( x = -1 \) and \( x = 1 \).