How to make $f(x)$ differentiable.

Question

How to solve \begin{equation*} f(x) = \left\{ \begin{array}{ll} 2x+b &\quad x < a \\ x^{2} & \quad x \geq a\\ \end{array} \right. \end{equation*}

This only uses definition of a derivative, to make $f(x)$ differentiable at $a$.

Answer 1

To make $f$ differentiable at $a$ you need $f'(a)$ to exist, so you need both limits $\lim\limits_{x\to a^+}\dfrac{f(x)-f(a)}{x-a}$ and $\lim\limits_{x\to a^-}\dfrac{f(x)-f(a)}{x-a}$ to exist and equal each other. Now, before you make $f$ differentiable you need to make sure it is continuous, since differentiability implies continuity.

To make $f$ continuous you need to make sure $\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a^-}f(x)=f(a)$. Once you've done it, try to see if you can get differentiability.

Note: Usually, to make $f$ differentiable at $a$, we compare $\lim_{x\to a^+}f'(x)$ and $\lim_{x\to a^-}f'(x)$, hoping they give the same value. This is actually stronger, since it says that $f$ is continuously differentiable. Since you said you have to use the definitions of differentiability, I wrote the exact thing, and not this unnecessarily (but common) stronger way to tackle the problem.

Answer 2

First, we must have $\lim_{x\rightarrow a^{-}}f(x) = \lim_{x\rightarrow a^{+}}f(x)=f(a)$ for continuity. Then:

$$2a + b = a^{2}$$

Next, we must have $\lim_{x\rightarrow a^{-}}f'(x) = \lim_{x\rightarrow a^{+}}f'(x)$ for differentiability. This gives:

$$\frac{d}{dx}(2x + b)\bigg\vert_{a} = \frac{d}{dx}(x^{2})\bigg\vert_{a}$$

$$2 = 2a\Rightarrow a = 1$$

Then,

$2(1) + b = 1$, so $b = -1$, and $\boxed{(a, b) = (1, -1).}$