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: Write the given function clearly.
We are given the rational function:
\[ f(x)=\frac{x}{x^2-4} \] We have to find the points where this function is discontinuous.

Step 2: Recall the continuity rule for rational functions.
A rational function is continuous everywhere in its domain. It becomes discontinuous only at those points where its denominator becomes zero.

Step 3: Set the denominator equal to zero.
So we solve:
\[ x^2-4=0 \]

Step 4: Factorize the expression.
\[ x^2-4=(x-2)(x+2) \] Therefore,
\[ (x-2)(x+2)=0 \]

Step 5: Find the critical values.
From the factorized form, we get:
\[ x=2 \quad \text{or} \quad x=-2 \]

Step 6: Interpret these values.
At \(x=2\) and \(x=-2\), the denominator becomes zero, so the function is not defined there. Hence, the function is discontinuous at these two points.

Step 7: State the final answer.
Thus, the set of all points where the function is discontinuous is:
\[ \boxed{\{-2,2\}} \] which matches option \((5)\).