A function \( f(x)=\begin{cases}\displaystyle\frac{kx^{3}}{4} & \text{if }x\le1\\ 8 & \text{if }x\gt1\end{cases} \) for some constant \( k \), then \( \displaystyle\frac{d^{2}f(x)}{dx^{2}} \) at \( x=-1 \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( -\displaystyle\frac{3k}{2} \)
The point x = −1 lies in the branch \( x\le1 \), and near x = −1 the function is \( f(x)=\frac{kx^{3}}{4} \). So
\( \displaystyle f'(x)=\frac{3kx^{2}}{4} \)
\( \displaystyle f''(x)=\frac{3kx}{2} \)
\( \displaystyle f''(-1)=-\frac{3k}{2} \)