Question

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

Answer 1

LMV Theorem for f in [1, 2]

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

$=f^{\prime }\left(c\right)\le 2$

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

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

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

$=f^{\prime }\left(d\right)\le 2$
$\therefore f\left(4\right)-f\left(2\right)\le 4$
$\Rightarrow 8-f\left(2\right)\le 4$

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

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

From $\left(1\right)and\left(2\right),wegetf\left(2\right)=4LMVTheoremforfin[1,2]$

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

$=f^{\prime }\left(c\right)\le 2$
$f\left(2\right)-f\left(1\right)\le 2$

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

Again, using LMV Theorem in $[2,4]$

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

$$=f^{\prime }\left(d\right)\le 2$$

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

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

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

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

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