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 $b$? 
 A: 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.
A: If $0\leq a<b$ then $3a^2<a^2+ab+b^2<3b^2$. The continuity of $x\mapsto x^2$ then implies that there is a $c\in\ ]a,b[\ $ with $3c^2=a^2+ab+b^2$. The case $a<b\leq0$ is similar.
In the case $a<0<b$ we argue as follows (using the continuity of $f'$): One has
$$\mu:={b^3-a^3\over b-a}={0-a\over b-a}f'\left({a\over\sqrt{3}}\right)+{b-0\over b-a}f'\left({b\over\sqrt{3}}\right)\ .$$
This shows that $\mu$ is a weighted mean of $f'\left({a\over\sqrt{3}}\right)$ and $f'\left({b\over\sqrt{3}}\right)$, and therefore lies in the interval with these endpoints. Applying the IVT to $f'$ then shows that there is a $c\in\ \bigl]{a\over\sqrt{3}},{b\over\sqrt{3}}\bigr[\ $ with $f'(c)=\mu$.
Update
The distinction of cases can be avoided in the following (somewhat  tricky) way: Assume $a<b$ and put
$$c_-:=\left({1\over2}+{1\over2\sqrt{3}}\right) a+\left({1\over2}-{1\over2\sqrt{3}}\right)b,\qquad c_+:=\left({1\over2}-{1\over2\sqrt{3}}\right) a+\left({1\over2}+{1\over2\sqrt{3}}\right)b\ .$$
Then both $c_-$ and $c_+$ are weighted means of $a$ and $b$, and one in fact has
$$a<c_-<c_+<b\ .$$
It so happens that
$$\mu=a^2+ab+b^2={1\over2}\bigl(f'(c_-)+f'(c_+)\bigr)$$
(check this!), which implies that $\mu$ lies in the interval with endpoints $f'(c_-)$ and $f'(c_+)$. Applying the IVT to $f'$ then shows that there is a $c\in\ ]c_-,c_+[\ $ with $f'(c)=\mu$.
