Question

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

Answer 1

Given that $f\left(1\right)=-2$ and $f^{\prime }\left(x\right)\ge 2$ $\forall$ $x\in [1,6]$
Lagrange's mean value theorem states that if $f\left(x\right)$ be continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists some $c$ between $a$ and $b$ such that $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Given that $f$ is differentiable for all $x$. Therefore Lagrange's mean value theorem can be applied.
Therefore, $f^{\prime }\left(c\right)=\frac{f\left(6\right)-f\left(1\right)}{6-1}\ge 2$
$\Rightarrow f\left(6\right)-f\left(1\right)\ge 10$
$\Rightarrow f\left(6\right)-(-2)\ge 10$
$\Rightarrow f\left(6\right)\ge 10-2$
$\Rightarrow f\left(6\right)\ge 8$