Evaluate the nature of the point for the curve , given that .
Evaluate the nature of the point $(0,0)$ for the curve $y = x^4$, given that $f''(0) = 0$.
- aThe function at the point $(0,0)$ is undefined because the second derivative test fails.
- bIt is a point of local maximum because the first derivative is changing its sign from positive to negative as $x$ increases through 0.
- cIt is a local minimum because the second derivative is positive for all $x \neq 0$.
- dIt is a point of neither local maximum nor local minimum because the second derivative is zero.
Answer
Answer
AIWritten by AI - it can contain mistakes.
Correct option: c
(c) It is a local minimum because the second derivative is positive for all $x \neq 0$.
Given $y = x^4$, we have $\frac{dy}{dx} = 4x^3$ and $\frac{d^2y}{dx^2} = 12x^2$.
At $x = 0$, $\frac{d^2y}{dx^2} = 0$, but for all $x \neq 0$, $\frac{d^2y}{dx^2} = 12x^2 > 0$, which confirms that the curve is concave upward around $x = 0$.
Furthermore, $\frac{dy}{dx}$ changes from negative for $x < 0$ to positive for $x > 0$, confirming that $(0,0)$ is a local minimum.
From ISC 2027 Specimen Mathematics Paper 1, question 1(vii).