Question

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

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

Answer 1

Answer: (B)

$f\left(6\right)\ge 8$

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, $f{}^{\prime }\left(c\right)=\frac{f\left(6\right)-f\left(1\right)}{6-1}\ge 2$ (Since, $[a,b]=[1,6]$)

$⟹f\left(6\right)-f\left(1\right)\ge 2\left(5\right)$

$⟹f\left(6\right)-(-2)\ge 10$

$⟹f\left(6\right)\ge 10-2$

$⟹f\left(6\right)\ge 8$