Applications of Derivatives
Second Derivative of Piecewise Function
nta_pyq_2024_apr
Grade 12
Question:
If $f(x)=\begin{cases}x^3\sin\left(\dfrac{1}{x}\right), & x\neq0\\ 0, & x=0\end{cases}$, then:
$f''\left(\dfrac{2}{\pi}\right)=\dfrac{24-\pi^2}{2\pi}$
$f''\left(\dfrac{2}{\pi}\right)=\dfrac{12-\pi^2}{2\pi}$
$f''(0)=1$
$f''(0)=0$
Step-by-Step Solution
Key Concept: $f'(x)=3x^2\sin(1/x)-x\cos(1/x)$. $f''(x)=6x\sin(1/x)-3\cos(1/x)-\cos(1/x)-\frac{\sin(1/x)}{x}$... evaluate at $x=2/\pi$.
$f''(2/\pi)=\frac{12}{\pi}-\frac{\pi}{2}=\frac{24-\pi^2}{2\pi}$.
Correct Answer: 1