Question

The set of all points where the function $f(x)=\frac{x}{x^2-4},\ x\in\mathbb{R}$ is discontinuous, is:

• A. \(\{0,2\}\)
• B. \(\{0,4\}\)
• C. \((0,-2,2)\)
• D. \((2,4)\)
• E. \(\{-2,2\}\)

Hint:

For rational functions, discontinuity occurs exactly where the denominator becomes zero. Always factor the denominator completely.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
A rational function, which is a ratio of two polynomials \( f(x) = \frac{P(x)}{Q(x)} \), is continuous everywhere except at the points where the denominator \( Q(x) \) is equal to zero. At these points, the function is undefined, leading to a discontinuity.
Step 2: Key Formula or Approach:
To find the points of discontinuity, we need to find the roots of the denominator. We set the denominator equal to zero and solve for \( x \).
\[ x^2 - 4 = 0 \] Step 3: Detailed Explanation:
We are given the function \( f(x) = \frac{x}{x^2-4} \).
The denominator is \( Q(x) = x^2 - 4 \).
Set the denominator to zero to find the points where the function is undefined:
\[ x^2 - 4 = 0 \] This can be solved by adding 4 to both sides:
\[ x^2 = 4 \] Taking the square root of both sides gives two solutions:
\[ x = \sqrt{4} \quad \text{or} \quad x = -\sqrt{4} \] \[ x = 2 \quad \text{or} \quad x = -2 \] The function is discontinuous at \( x=2 \) and \( x=-2 \) because division by zero occurs at these points. These are infinite discontinuities.
Step 4: Final Answer:
The set of all points where the function is discontinuous is \{-2, 2\}. This corresponds to option (E).