Question

Let $f$ be differentiable for all $x$. If $f\left(1\right)=-2$ and $f^{\prime }\left(x\right)\ge 2$ for $x\in [1,6]$

• A. $f\left(6\right)<8$
• B. $f\left(6\right)\ge 8$
• C. $f\left(6\right)>5$
• D. $f\left(6\right)\le 5$

Answer 1

Answer: (B), (C)

The correct options are
B $f\left(6\right)>5$
C $f\left(6\right)\ge 8$

By Lagrange's mean value theorem there is $c\in (1,6)$.

such that $\frac{f\left(6\right)-f\left(1\right)}{6-1}=f^{\prime }\left(c\right)$

$\Rightarrow \frac{f\left(6\right)+2}{5}=f^{\prime }\left(c\right)$ i.e. $f\left(6\right)=5f\left(c\right)-2\ge 5.2-2=8$ so $f\left(6\right)>5$ also.