<s>[INST] Differentiable function to find a+b Let function $f(x)= \begin{cases}ax^3+2, &x<\dfrac 12 \\ bx^2+1, & x\ge \dfrac 12\end{cases}$. If the function is differentiable at $x=\dfrac 12$, then $a+b=$???
 [/INST] $f(n) =
\begin{cases}
ax^3+2  & \text{x<1/2} \\[2ex]
bx^2+1, & \text{x≥1/2}
\end{cases}$
$f(n)' =
\begin{cases}
3ax^2  & \text{left derivative 1/2} \\[2ex]
2bx, & \text{right derivative in 1/2}
\end{cases}$
If f is differentiable at a point x, then f must also be continuous at x.
Since v(x) is a continuous function, then the limit as x approaches t is equal to the value of v(x) at x = t
$\begin{cases}
3a(1/2)^2=b(1/2)^2+1  & \text{left derivative at 1/2} \\[2ex]
2b(1/2)=b(1/2)^2+1, & \text{right derivative at 1/2}
\end{cases}$
$\begin{cases}
a(3/4)=b(1/4)+1  & \text{}\\[2ex]
(1-1/4)b=1, & \text{}
\end{cases}$
$\begin{cases}
a=16/9& \text{}\\[2ex]
b=4/3, & \text{}
\end{cases}$
$a+b=28/9$
</s>