Question

Suppose $f$ is differentiable on $R$ and $a\le f^{\prime }\left(x\right)\le b$ where $x\in R$ where $a,b>0$. If $f\left(0\right)=0$, then

• A. $f\left(x\right)\le min(ax,bx)$
• B. $f\left(x\right)\ge min(ax,bx)$
• C. $a\le f\left(x\right)\le b$
• D. $ax\le f\left(x\right)\le bx$

Answer 1

The correct option is D $ax\le f\left(x\right)\le bx$

For $x>0$. Applying Lagrange's theorem on. $(0,x]$ we have $c\in (0,x)$ such that
$\frac{f\left(x\right)}{x}=\frac{f\left(x\right)-f\left(0\right)}{x-0}=f^{\prime }\left(c\right)$
But,
$a\le f^{\prime }\left(c\right)\le b$ so $a\le \frac{f\left(x\right)}{x}\le b\Rightarrow ax\le f\left(x\right)\le bx$, $x>0$.

Similarly for $x<0$ applying Lagrange's theorem for $(x,0]$, we have $ax\le f\left(x\right)\le bx$.