Verify Mean Value Theorem for $\,f\left(x\right)=x^3\,$ I am trying to verify the Mean-Value Theorem for $\,f\left(x\right)=x^3.\,$ So far I have:
$$f'\left(c\right)=\frac{b^3-a^3}{b-a}=b^2+ab+a^2=3c^2$$
Question: since we know that $b>a$, how can we show that at least one solutions of $c$ lies between $a$ and $b$? 

I could not find a simpler solution than the following one.
Mean Value Theorem: There exists $c\in (a,b)$ such that $3c^2=a^2+ab+b^2$. 
Denote $c_1 = \sqrt{\frac{a^2+ab+b^2}{3}} > 0$ and $c_2 = -\sqrt{\frac{a^2+ab+b^2}{3}} < 0$. We need to prove that: at least one of the following two inequalities holds:
$$(1) \quad a<c_1<b$$
$$(2)\quad a<c_2<b$$
Consider 3 cases:


*

*$0\le a <b$: In this case $(1)$ is true because $c_1^2-a^2 = \frac{1}{3}(b-a)(b+2a) >0$ and $b^2-c_1^2 = \frac{1}{3}(b-a)(2b+a) >0$.

*$a < b \le 0$: Similarly, in this case $(2)$ is true because $a^2-c_2^2 >0$ and $c_2^2-b^2 >0$. 

*$a\le 0 <b$: in this case $a<c_1$ and $c_2<b$ are already true, we only need to show that at least one of the following two are true: $c_1<b$ or $c_2>a$, or equivalently: $c_1^2<b^2$ or $c_2^2 < a^2$. This is easy by contradiction: if $c_1^2 \ge b^2$ AND $c_2^2 \ge a^2$ then $c_1^2+c_2^2 \ge a^2+b^2 \Leftrightarrow 2(a^2+ab+b^2)\ge 3(a^2+b^2) \Leftrightarrow 0 \ge (a-b)^2$, obviously false.
We are done :)
P/s: a remark for those who are trying to apply the intermediate value theorem:
If we denote $g(x) = 3x^2 - (a^2+ab+b^2)$ then we have $g(a)g(b) = -(a-b)^2(2a+b)(2b+a)$, not always negative.
