Problem 5. Let g : R → R be a function and a ∈ R.(a) Prove that if g is differentiable at a, then there exists function φ : R → R that iscontinuous at a and satisfiesg(x) − g(a) = (x − a) · φ(x) for all x ∈ R.(Hint: We proved a very similar statement within the proof of the chain rule.)(b) Conversely, prove that if there exists function φ : R → R that is continuous at a andsatisfies the above equation, then g is differentiable at a.

(a) If g is differentiable at a, then by definition, the limit as x approaches a of [g(x) - g(a)] / (x - a) exists. Let's denote this limit as g'(a).

Now, define a function φ(x) as follows:

φ(x) = g'(a) when x = a, and φ(x) = [g(x) - g(a)] / (x - a) when x ≠ a.

This function φ(x) is clearly continuous at a because the limit as x approaches a of φ(x) is g'(a), which is the value of φ(a).

Moreover, for all x in R, we can rearrange the definition of φ(x) when x ≠ a to get g(x) - g(a) = (x - a) · φ(x). When x = a, both sides of the equation are 0, so the equation holds for all x in R.

(b) Conversely, suppose there exists a function φ : R → R that is continuous at a and satisfies g(x) - g(a) = (x - a) · φ(x) for all x in R.

We want to show that g is differentiable at a, which means we need to show that the limit as x approaches a of [g(x) - g(a)] / (x - a) exists.

As x approaches a, the right-hand side of the equation becomes (a - a) · φ(a) = 0. Since φ is continuous at a, φ(x) approaches φ(a) as x approaches a. Therefore, the limit as x approaches a of [g(x) - g(a)] / (x - a) = φ(x) exists, which means g is differentiable at a.