Question

Suppose, that $f$ is differentiable for all $x$ and that $f^{\prime }(x)\le 2$ 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

Answer: (B)

$LMV$ Theorem for $f$ in $[1,2]$
$\forall c\in \left(1,2\right)\frac{f\left(2\right)-f\left(1\right)}{2-1}$
$=f^{\prime }\left(c\right)\le 2$
$f\left(2\right)-f\left(1\right)\le 2$
$\Rightarrow f\left(2\right)\le 4...(1)$
Again, using $LMV$ Theorem in $\left[2,4\right]$
$\forall d\in \left(2,4\right)\frac{f\left(4\right)-f\left(2\right)}{4-2}$
$=f^{\prime }\left(d\right)\le 2$
$\therefore f\left(4\right)-f\left(2\right)\le 4$
$\Rightarrow 8-f\left(2\right)\le 4$
$\Rightarrow 4\le f\left(2\right)$
$\Rightarrow f\left(2\right)\ge 4$
From $\left(1\right)$ and $\left(2\right)$, we get $f\left(2\right)=4$$LMV$ Theorem for $f$ in $[1,2]$
$\forall c\in \left(1,2\right)\frac{f\left(2\right)-f\left(1\right)}{2-1}$
$=f^{\prime }\left(c\right)\le 2$
$f\left(2\right)-f\left(1\right)\le 2$
$\Rightarrow f\left(2\right)\le 4...(1)$
Again, using $LMV$ Theorem in $\left[2,4\right]$
$\forall d\in \left(2,4\right)\frac{f\left(4\right)-f\left(2\right)}{4-2}$
$=f^{\prime }\left(d\right)\le 2$
$\therefore f\left(4\right)-f\left(2\right)\le 4$
$\Rightarrow 8-f\left(2\right)\le 4$
$\Rightarrow 4\le f\left(2\right)$
$\Rightarrow f\left(2\right)\ge 4...(2)$
From $\left(1\right)$ and $\left(2\right)$, we get $f\left(2\right)=4$