Let $$S$$, be the set of all functions $$f : [0, 1] \rightarrow R$$, which are continuous on [0, 1], and differentiable on (0, 1). Then for every $$f$$ in $$S$$, there exists $$c \in (0, 1)$$, depending on $$f$$, such that.

Given :

• f is continuous on [0,1]
• f differentiable on (0,1)

So, we can apply Lagrange’s Mean Value Theorem (LMVT).

As we can say that f is also continuous on [c,1] and also differentiable on (c,1).

So we can apply LMVT on [c,1].

Apply LMVT on interval [c,1]:
since f satisfies conditions,

there exists some points let it be b such that :

   $$f'\left(c\right)\ =\ \frac{f\left(1\right)-f\left(c\right)}{1-c}$$

Hence the correct answer is

$$f'\left(c\right)\ =\ \frac{f\left(1\right)-f\left(c\right)}{1-c}$$