For the function to be differentiable, its derivative has to exist everywhere, which means the derivative itself must be continuous. Differentiating gives
\(f'(x)=\begin{cases}24x^2&\text{for }x<1\\?&\text{for }x=1\\B&\text{for }x>1\end{cases}\)
The question mark is a placeholder, and if the derivative is to be continuous, then the question mark will have the same value as the limit as \(x\to1\) from either side.
\(\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}24x^2=24\)
\(\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}B=B\)
So the derivative will be continuous as long as \(B=24\)
For the function to be differentiable everywhere, we need to require that \(f(x)\) is itself continuous, which means the following limits should be the same:
\(\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}8x^3=8\)
\(\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}Bx+C=24+C\)
\(24+C=8\implies C=-16\)
So, the function should be
\(f(x)=\begin{cases}8x^3&\text{for }x\le1\\24x-16&\text{for }x>1\end{cases}\)
with derivative
\(f'(x)=\begin{cases}24x^2&\text{for }x<1\\24&\text{for }x\ge1\end{cases}\)
