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(B) Using $LMVT$ for $f$ in the interval $[1, 2]$:There exists $c \in (1, 2)$ such that $\frac{f(2) - f(1)}{2 - 1} = f'(c)$.Since $f'(c) \le 2$,we hav

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$4$

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(B) Using $LMVT$ for $f$ in the interval $[1, 2]$:There exists $c \in (1, 2)$ such that $\frac{f(2) - f(1)}{2 - 1} = f'(c)$.Since $f'(c) \le 2$,we have $f(2) - f(1) \le 2(1)$,which implies $f(2) - 2 \le 2$,so $f(2) \le 4$....$(1)$Again,using $LMVT$ for $f$ in the interval $[2, 4]$:There exists $d \in (2, 4)$ such that $\frac{f(4) - f(2)}{4 - 2} = f'(d)$.Since $f'(d) \le 2$,we have $f(4) - f(2) \le 2(2)$,which implies $8 - f(2) \le 4$,so $f(2) \ge 4$....$(2)$From equations $(1)$ and $(2)$,we conclude that $f(2) = 4$.

Suppose that $f$ is differentiable for all $x$ and that $f'(x) \le 2$ for all $x$. If $f(1) = 2$ and $f(4) = 8$,then $f(2)$ has the value equal to

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