The function f(x) = \( \frac{4 - x^2}{4x - x^3} \), is-
discontinuous at only one point in its domain.
discontinuous at two points in its domain.
discontinuous at three points in its domain.
continuous everywhere in its domain.
Step-by-Step Solution
Key Concept: A function is discontinuous at points where it is undefined (denominator = 0) or where limits don't exist. We must find all points excluded from the domain by factoring the denominator completely.
<p><strong>Step 1: Identify the denominator and find where it equals zero.</strong></p><p>Given: f(x) = (4 - x²)/(4x - x³)</p><p>Denominator: 4x - x³ = x(4 - x²)</p><p><strong>Step 2: Factor completely.</strong></p><p>4x - x³ = x(4 - x²) = x(2 - x)(2 + x)</p><p><strong>Step 3: Find zeros of the denominator.</strong></p><p>Setting x(2 - x)(2 + x) = 0</p><p>We get: x = 0, x = 2, x = -2</p><p><strong>Step 4: Determine the domain.</strong></p><p>The function is undefined (and hence discontinuous) at x = 0, x = 2, and x = -2.</p><p>Domain of f(x): ℝ \ {-2, 0, 2}</p><p><strong>Step 5: Check if discontinuities are removable.</strong></p><p>Numerator: 4 - x² = (2 - x)(2 + x)</p><p>So f(x) = [(2-x)(2+x)]/[x(2-x)(2+x)]</p><p>At x = ±2, the factors (2-x) and (2+x) cancel, leaving removable discontinuities (holes).</p><p>At x = 0, no cancellation occurs—this is a non-removable discontinuity.</p><p>However, all three points (x = -2, 0, 2) are excluded from the domain, making the function discontinuous at all three points.</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C