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Suppose that f is differentiable for all x and that f’(x) ≤ 2 for all x. If f(1) = 2 and f(4) = 8, then f(2) has the value equal to
a) 3

b) 4

c) 6

d) 8
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LMV Theorem for f in [1, 2]

$$\forall\,c \in\left(1, 2\right) \frac{f\left(2\right)-f\left(1\right)}{2-1}$$

$$=f'\left(c\right) \le 2$$

$$f\left(2\right)-f\left(1\right) \le 2$$

$$\Rightarrow f\left(2\right) \le 4 \quad...(1)$$
Again, using LMV Theorem in $$\left[2, 4\right]$$

$$\forall\,d \in\left(2, 4\right) \frac{f\left(4\right)-f\left(2\right)}{4-2}$$

$$=f'\left(d\right) \le 2$$
$$\therefore f\left(4\right)-f\left(2\right) \le 4$$
$$\Rightarrow 8-f\left(2\right) \le 4$$

$$\Rightarrow 4 \le f\left(2\right)$$

$$\Rightarrow f\left(2\right) \ge 4$$

From $$\left(1\right) and \left(2\right), we\ get \ f\left(2\right)=4\ LMV\ Theorem \ for \ f \ in \ [1, 2]$$

$$\forall\,c \in\left(1, 2\right) \frac{f\left(2\right)-f\left(1\right)}{2-1}$$

$$=f'\left(c\right) \le 2$$
$$f\left(2\right)-f\left(1\right) \le 2$$

$$\Rightarrow f\left(2\right) \le 4 \quad...(1)$$

Again, using LMV Theorem in $$\left[2, 4\right]$$

$$\forall\,d \in\left(2, 4\right) \frac{f\left(4\right)-f\left(2\right)}{4-2}$$

$$=f'\left(d\right) \le 2$$

$$\therefore f\left(4\right)-f\left(2\right) \le 4$$

$$\Rightarrow 8-f\left(2\right) \le 4$$

$$\Rightarrow 4 \le f\left(2\right)$$

$$\Rightarrow f\left(2\right) \ge 4 \quad ...(2)$$

From $$\left(1\right)$$ and $$\left(2\right)$$, we get $$f\left(2\right)=4$$

by Expert (10.9k points)