Please find the error in this proof and briefly explain why it is incorrect.

If this would help, this is the textbook for the course: https://cdn.discordapp.com/attachments/885190741343764511/885300638504845312/Calculus_One_and_Several_Variables.pdf

Find the error in the following "Proof" and explain in a short paragraph why the reasoning is incorrect. Statement: If f is a function that is differentiable at the point w then the following are equivalent 1. f' (w) = m 2. there is a line y = max + b containing the point (w, f(w)) and there is o > 0 such that either: . f(x) mx + b for all r E (w - 6, w + 8). Proof: To prove that (2) implies (1) begin by observing that since it is known that the line y = ma + b contains the point (w, f(w)) it is possible to find b by evaluating y at w. This gives f(w) = mw + b and hence b = f(w) - mw and so y = mx + f(w) -mw. (5) The other hypothesis is that f is differentiable at w and so lim f(I) - f(w) = f'(w) T - W (6) and lim f(x) - f(w) - = f'(w) - W (7) Two cases must now be considered. Assume first that there is o > 0 such that f(x) wthen 1/(x - w) > 0 and so f(x) - f(w) m(I -w) I- W - W and therefore by Equation (6) f'(w) = lim f(x) - f(w) mc -w lim : m T - - W I MO C - - W On the other hand, if x lim = m I - W C- W It follows that f' (w) = m in this case. Now assume that there is o > 0 such that f(x) > mx + bfor all x E (w - 6, w + 6). It must again be shown that f'(w) = m. Note that by Equation (5) f(x) - f(w) > mx + f(w) - mu - f(w) = m(z - w). Hence, if x > w then 1/(x - w) > 0 and so f(I) - f(w) mc-w) 2 T- W - 20 and therefore by Equation (6) f'(w) = lim f(x) - f(w) mc-w) > lim = m I- W C - W On the other hand, if x