Find the values of m and b that make the following function differentiable. the piecewise function f of x equals x cubed when x is less than or equal to one or mx plus b when x is greater than one.

[tex]f(x)=\begin{cases}x^3&\text{for }x\le1\\mx+b&\text{for }x>1\end{cases}[/tex][tex]f[/tex] must be continuous in order to be differentiable, so we need to have[tex]\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)[/tex]By its definition, [tex]f(1)=1^3=1[/tex], and[tex]\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}x^3=1[/tex][tex]\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}(mx+b)=m+b[/tex]so that [tex]\boxed{m+b=1}[/tex].We want the derivative to exist at [tex]x=1[/tex], which requires that we pick an appropriate value for [tex]f'(1)[/tex] so that [tex]f'(x)[/tex] is also continuous. At the moment, we know[tex]f'(x)=\begin{cases}3x^2&\text{for }x<1\\m&\text{for }x>1\end{cases}{/tex]so we need to pick [tex]f'(1)[/tex] such that[tex]\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1^+}f'(x)[/tex]We have[tex]\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}3x^2=3[/tex][tex]\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}m=m[/tex]so that [tex]\boxed{m=3}[/tex] (which means we need to pick [tex]f'(1)=3[/tex]) and so [tex]m+b=1\implies\boxed{b=-2}[/tex]