Question

Consider a differential function $f\left(x\right)$ on the set of real numbers such that $f(-1)=0$ and $|f^{\prime }\left(x\right)|\le 2$. Given these conditions, which one of the following inequalties is necessarily ture for all $xϵ[-2,2]$?

• A. $f\left(x\right)\le 2|x+1|$
• B. $f\left(x\right)\le \frac{1}{2}|x+1|$
• C. $f\left(x\right)\le 2|x|$
• D. $f\left(x\right)\le \frac{1}{2}|x|$

Answer 1

The correct option is A $f\left(x\right)\le 2|x+1|$

Conditions given
$f(-1)=0$ and |f'(x)|\leq 2\)
Consider option (1)
$f\left(x\right)=2|x+1|$



$f(-1)=0,$ satisfied
$f^{\prime }\left(x\right)|\le 2,$ satisfied
Consider opiton (2)
$f\left(x\right)=\frac{1}{2}|x+1|$


$f(-1)=0$, satisfied
If $|f^{\prime }\left(x\right)|\le 2$, not satisifed
Consider option (3)



$f(-1)=0$, satisfied
It $|f^{\prime }\left(x\right)\le 2,$ satisfied
consider option (4)
$f\left(x\right)=\frac{1}{2}|x|$



$f(-1)=0$, not satisfied
$|f^{\prime }\left(x\right)\le 2,$ not satisfied.