# Mean Value Theorem and Lower Bound

PROBLEM: I wonder why my answer is wrong.

QUESTION: Suppose $x(t)$ satisfies the hypothesis of the Mean value Theorem (MVT), that is, $x(t)$ is continuous on $A \le t \le B$ and differentiable on $A \lt t \lt B$ and $m$ is the lower bound, then: $$m \le x'(t)$$ for all $t$ such that $$A \lt t \lt B$$

Let us consider the MVT on shorter intervals. The strongest conclusion we can draw is:

MVT says it is differentiable over the interval $A \lt t \lt B$ and since differential is $$\frac{x(b)-x(a)}{b-a}$$ therefore $$m \le \frac{x(b)-x(a)}{b-a}$$ for all $a,b$ over the same interval. Why my answer is wrong ? Why the interval changed to $A \le t \le B$

## 1 Answer

The Mean Value Theorem says that there is a $t\in(a,b)$ (open interval) such that $$x'(t)=\frac{x(b)-x(a)}{b-a}.$$ Hence if $A\leq a<b\leq B$ then $t\in(a,b)\subseteq (A,B)$, and $$m\leq x'(t)=\frac{x(b)-x(a)}{b-a}.$$ So the strongest conclusion is just the last one!

• I guess MIT course on edX just over-complicates things. Like you, they could have simply mentioned (a,b) is an interval inside (A,B) and a,b are two distinct points. Thanks Aug 30 '18 at 6:12