Let `alpha in R`. Prove that a function `f: RtoR` is differentiable at `alpha` if and only if there is a function `g:RtoR` which is continuous at `alpha` and satisfies `f (x) -f(alpha) = g (x) (x-alpha), AA x in R.`

Answer

To prove that a function \( f: \mathbb{R} \to \mathbb{R} \) is differentiable at \( \alpha \) if and only if there exists a function \( g: \mathbb{R} \to \mathbb{R} \) that is continuous at \( \alpha \) and satisfies the equation:

\[ f(x) - f(\alpha) = g(x)(x - \alpha) \quad \forall x \in \mathbb{R}, \]

we will break the proof into two parts: the "if" part and the "only if" part.

### Part 1: If \( f \) is differentiable at \( \alpha \), then such a \( g \) exists.

1. **Assume \( f \) is differentiable at \( \alpha \)**:
By definition, the derivative of \( f \) at \( \alpha \) is given by:

\[ f'(\alpha) = \lim_{x \to \alpha} \frac{f(x) - f(\alpha)}{x - \alpha}. \]

2. **Define \( g(x) \)**:
We can define the function \( g(x) \) as follows:

\[ g(x) = \frac{f(x) - f(\alpha)}{x - \alpha} \quad \text{for } x \neq \alpha. \]

Note that \( g(x) \) is not defined at \( x = \alpha \) since it leads to a \( \frac{0}{0} \) form.

3. **Show that \( g(x) \) approaches \( f'(\alpha) \) as \( x \to \alpha \)**:
Since \( f \) is differentiable at \( \alpha \), we have:

\[ \lim_{x \to \alpha} g(x) = f'(\alpha). \]

4. **Continuity of \( g \) at \( \alpha \)**:
We can extend \( g(x) \) to be continuous at \( \alpha \) by defining:

\[ g(\alpha) = f'(\alpha). \]

Thus, \( g(x) \) is continuous at \( \alpha \).

5. **Conclusion for Part 1**:
Therefore, if \( f \) is differentiable at \( \alpha \), there exists a function \( g \) that is continuous at \( \alpha \) such that:

\[ f(x) - f(\alpha) = g(x)(x - \alpha). \]

### Part 2: If such a \( g \) exists, then \( f \) is differentiable at \( \alpha \).

1. **Assume \( g \) is continuous at \( \alpha \)**:
We have:

\[ f(x) - f(\alpha) = g(x)(x - \alpha). \]

2. **Rearranging the equation**:
Dividing both sides by \( x - \alpha \) (for \( x \neq \alpha \)) gives:

\[ \frac{f(x) - f(\alpha)}{x - \alpha} = g(x). \]

3. **Taking the limit as \( x \to \alpha \)**:
Since \( g \) is continuous at \( \alpha \), we can take the limit:

\[ \lim_{x \to \alpha} \frac{f(x) - f(\alpha)}{x - \alpha} = \lim_{x \to \alpha} g(x) = g(\alpha). \]

4. **Defining the derivative**:
Therefore, we can define:

\[ f'(\alpha) = g(\alpha). \]

5. **Conclusion for Part 2**:
This shows that \( f \) is differentiable at \( \alpha \).

### Final Conclusion:
We have shown both directions of the proof. Thus, we conclude that a function \( f \) is differentiable at \( \alpha \) if and only if there exists a function \( g \) that is continuous at \( \alpha \) and satisfies the equation:

\[ f(x) - f(\alpha) = g(x)(x - \alpha) \quad \forall x \in \mathbb{R}. \]