Question

Suppose that $f\left(0\right)=-3$ and $f^{\prime }\left(x\right)\le 5$ for all values of $x$. Then the largest value which $f\left(2\right)$ can assume is $\dots \dots \dots$.

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

Answer: (C)

$7$

By Lagrange's mean value theorem in $[0,2]$
$\frac{f\left(2\right)-f\left(0\right)}{2-0}=f^{\prime }\left(c\right)$ where $cϵ(0,2)$
$\Rightarrow \frac{f\left(2\right)+3}{2}\le 5$
$\Rightarrow f\left(2\right)\le 7$