Show that $f$ is differentiable at $p$ iff there is a function $\phi$ that is continuous at $p$ and fulfills $f(x)=f(p)+(x-p)\phi(x)$

Question

Let $I\subset \mathbb{R}$ be an interval, $p\in I$, and $f : I\to \mathbb{R}$ a function.

Prove that $f$ is differentiable at $p$ if and only if there is a function $\phi:I\to \mathbb{R}$ that is continuous at $p$ and $$f(x)=f(p)+(x-p)\phi(x)$$ is fulfilled. Moreover if such a function $\phi$ exists then $$f'(p)=\phi(p).$$ HINT: This is largely a restatement of the usual definition of derivative.

I know that for a function to be differentiable then the derivatives from the left and right must be equal. Just not sure where to start with this one.

Answer 1

IF

Assume such a function $\varphi$ exists.

Then:

$\begin{array}{rcl} f'(p) &=& \displaystyle \lim_{h \to 0} \frac {f(p+h) - f(p)} {h} \\ &=& \displaystyle \lim_{h \to 0} \frac {f(p+h) - f(p)} {(p+h)-p} \\ &=& \displaystyle \lim_{h \to 0} \varphi(p+h) \\ &=& \displaystyle \varphi(p) \\ \end{array}$

ONLY IF

Let $f$ be differentiable at $p$.

Then, let $\varphi : x \mapsto \begin{cases} \dfrac {f(x) - f(p)} {x-p} & x \ne p \\ f'(p) & x = p \end{cases}$.

Now, to prove that $\varphi$ is continuous at $p$:

$\begin{array}{rcl} \displaystyle \lim_{x \to p} \varphi(x) &=& \displaystyle \lim_{x \to p} \frac {f(x) - f(p)} {x-p} \\ &=& \displaystyle \lim_{h \to 0} \frac {f(p+h) - f(p)} {(p+h)-p} \\ &=& \displaystyle \lim_{h \to 0} \frac {f(p+h) - f(p)} {h} \\ &=& f'(p) \\ &=& \varphi(p) \end{array}$