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]</p><p>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}`