[tex]\bf f(x)=
\begin{cases}
x^3&x\le 1\\
mx+b&x\ \textgreater \ 1
\end{cases}[/tex]
now, a function may well be continuous, but it may not be differentiable.
graph wise, differentiability means, a smooth curve, with no abrupt spikes or drops or cuts.
so for this piece-wise to be differentiable, the subfunctions must meet at junctions smoothly, with no abrupt changes, so mx + b, must blend in with x³ smoothly.
now, that can only occur if mx + b has the same slope as x³ at that point where they meet. The point where they meet is at x = 1, well, let's check what is the slope of x³ at x = 1.
[tex]\bf \cfrac{d}{dx}[x^3]\implies \left. 3x^2\cfrac{}{}\right|_{x=1}\implies 3(1)^2\implies 3[/tex]
therefore "m" for mx+b, must be 3 then.
now, what's the value of x³ at x = 1? well is just (1)³, which is 1.
so for 3x + b, when x = 1, it must also equals 1 as well, that way it meets x³, thus
3x + b
3(1) + b = 1
b = -2
check the graph below.
