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 $b&gt;a$, 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 $c_1 = \sqrt{\frac{a^2+ab+b^2}{3}} &gt; 0$ and $c_2 = -\sqrt{\frac{a^2+ab+b^2}{3}} &lt; 0$. We need to prove that: at least one of the following two inequalities holds:
$$(1) \quad a&lt;c_1&lt;b$$
$$(2)\quad a&lt;c_2&lt;b$$

Consider 3 cases:


$0\le a &lt;b$: In this case $(1)$ is true because $c_1^2-a^2 = \frac{1}{3}(b-a)(b+2a) &gt;0$ and $b^2-c_1^2 = \frac{1}{3}(b-a)(2b+a) &gt;0$.
$a &lt; b \le 0$: Similarly, in this case $(2)$ is true because $a^2-c_2^2 &gt;0$ and $c_2^2-b^2 &gt;0$. 
$a\le 0 &lt;b$: in this case $a&lt;c_1$ and $c_2&lt;b$ are already true, we only need to show that at least one of the following two are true: $c_1&lt;b$ or $c_2&gt;a$, or equivalently: $c_1^2&lt;b^2$ or $c_2^2 &lt; a^2$. 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.
When $0&lt;a&lt;b$, we have $$a^2&lt;ab&lt;b^2,$$ then

$$3a^2&lt;a^2+ab+b^2&lt;3b^2,$$ or
$$a^2&lt;c^2&lt;b^2.$$

When $a&lt;0&lt;-a&lt;b$,

$$-ab&lt;a^2+b^2$$ and $$0&lt;a^2+ab+b^2&lt;3b^2,$$
or $$a&lt;0&lt;c&lt;b.$$

Other cases are handled similarly.