Question

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.

Answer 1

$f(x)=\begin{cases}x^3&\text{for }x\le1\\mx+b&\text{for }x>1\end{cases}$

$f$ must be continuous in order to be differentiable, so we need to have

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$

By its definition, $f(1)=1^3=1$, and

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}x^3=1$

$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}(mx+b)=m+b$

so that $\boxed{m+b=1}$.

We want the derivative to exist at $x=1$, which requires that we pick an appropriate value for $f'(1)$ so that $f'(x)$ is also continuous. At the moment, we know

$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)$ such that

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1^+}f'(x)$

We have

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}3x^2=3$

$\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}m=m$

so that $\boxed{m=3}$ (which means we need to pick $f'(1)=3$) and so $m+b=1\implies\boxed{b=-2}$