At what values of x does the graph of f(x) = x^4 - 4x^3 have points of inflection?
Answer and explanation
Correct answer: C. x = 0 and x = 2
- Step 1: Find first derivative: f'(x) = 4x^3 - 12x^2.
- Step 2: Find second derivative: f''(x) = 12x^2 - 24x = 12x(x - 2).
- Step 3: Set f''(x) = 0 => x = 0 or x = 2.
- Step 4: Check signs: for x < 0, f'' > 0; for 0 < x < 2, f'' < 0; for x > 2, f'' > 0.
- Step 5: Concavity changes at both x = 0 and x = 2, confirming two inflection points.
Answer
x = 0 and x = 2.
Why not the others: 'x = 0 and x = 3' confuses inflection points with critical points of f(x); 'x = 2 only' ignores the sign change at x = 0; 'x = -2 and x = 2' makes a sign error.
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