If \(y = x^4 - 4x^3 + 7\), then \(\displaystyle\frac{d^2y}{dx^2} = 0\) when
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(x = 0\) and \(x = 2\)
Differentiating twice by the power rule:
\( \displaystyle \frac{dy}{dx} = 4x^3 - 12x^2 \)
\( \displaystyle \frac{d^2y}{dx^2} = 12x^2 - 24x \)
\( = 12x(x - 2) \)
This is zero when x = 0 or x = 2.
Why the other options are incorrect
Option (2) \(x = 0\) and \(x = 3\) — Incorrect. These make the first derivative \(4x^2(x - 3)\) zero, not the second.
Option (3) \(x = 2\) only — Incorrect. At x = 0 the factor 12x is also zero, so x = 0 is a second value.
Option (4) \(x = 1\) and \(x = 2\) — Incorrect. At x = 1, \(\frac{d^2y}{dx^2} = 12 - 24 = -12 \ne 0\).