Question

Determine the values of the constants B and C so that the function given below is differentiable.

f(x)= {8x^3, x less than equal to 1
        {Bx+C, x > 1

Answer 1

For the function to be differentiable, its derivative has to exist everywhere, which means the derivative itself must be continuous. Differentiating gives

$f'(x)=\begin{cases}24x^2&\text{for }x<1\\?&\text{for }x=1\\B&\text{for }x>1\end{cases}$

The question mark is a placeholder, and if the derivative is to be continuous, then the question mark will have the same value as the limit as $x\to1$ from either side.

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}24x^2=24$
$\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}B=B$

So the derivative will be continuous as long as $B=24$

For the function to be differentiable everywhere, we need to require that $f(x)$ is itself continuous, which means the following limits should be the same:

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}8x^3=8$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}Bx+C=24+C$

$24+C=8\implies C=-16$

So, the function should be

$f(x)=\begin{cases}8x^3&\text{for }x\le1\\24x-16&\text{for }x>1\end{cases}$

with derivative

$f'(x)=\begin{cases}24x^2&\text{for }x<1\\24&\text{for }x\ge1\end{cases}$