# Let f be continuous on [a,b] and differential on (a,b). Prove that if a >= 0 there are x1, x2, x3 ∈

Let f be continuous on [a,b] and differential on (a,b). Prove that if a >= 0 there are x1, x2, x3 ∈ (a,b) such that
${f}^{\prime }\left({x}_{1}\right)=\left(b+a\right)\frac{{f}^{\prime }\left({x}_{2}\right)}{2{x}_{2}}=\left({b}^{2}+ba+{a}^{2}\right)\frac{{f}^{\prime }\left({x}_{3}\right)}{3{x}_{3}^{2}}$
I think this problem will use the generalized mean value theorem to solve. However, I don't know how to apply it. Can you suggest a solution? Thank you very much!
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iskakanjulc
By the mean value theorem,
$\begin{array}{}\text{(1)}& \mathrm{\exists }{x}_{1}\in \left(a,b\right)\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{1em}{0ex}}\frac{f\left(b\right)-f\left(a\right)}{b-a}={f}^{\prime }\left({x}_{1}\right)\phantom{\rule{thinmathspace}{0ex}}.\end{array}$
Consider $g:\left[{a}^{2},{b}^{2}\right]\to \mathbb{R}$ defined by $g\left(x\right)=f\left(\sqrt{x}\right)$. Applying the mean value theorem to $g$, we infer that there is some $t\in \left({a}^{2},{b}^{2}\right)$ such that
$\frac{g\left({b}^{2}\right)-g\left({a}^{2}\right)}{{b}^{2}-{a}^{2}}={g}^{\prime }\left(t\right)\phantom{\rule{thinmathspace}{0ex}},$
or equivalently, writing ${x}_{2}:=\sqrt{t}$,
$\begin{array}{}\text{(2)}& \frac{f\left(b\right)-f\left(a\right)}{\left(b-a\right)\left(b+a\right)}=\frac{{f}^{\prime }\left(\sqrt{t}\right)}{2\sqrt{t}}=\frac{{f}^{\prime }\left({x}_{2}\right)}{2{x}_{2}}\phantom{\rule{thinmathspace}{0ex}}.\end{array}$
Similarly, apply the mean value theorem to $h:\left[{a}^{3},{b}^{3}\right]\to \mathbb{R}$ defined by $h\left(x\right)=f\left({x}^{1/3}\right)$. We deduce that there is some $s\in \left({a}^{3},{b}^{3}\right)$ such that
$\frac{h\left({b}^{3}\right)-h\left({a}^{3}\right)}{{b}^{3}-{a}^{3}}={h}^{\prime }\left(s\right)\phantom{\rule{thinmathspace}{0ex}},$
or equivalently, writing ${x}_{3}:={s}^{1/3}$,
$\begin{array}{}\text{(3)}& \frac{f\left(b\right)-f\left(a\right)}{\left(b-a\right)\left({b}^{2}+ba+{a}^{2}\right)}=\frac{{f}^{\prime }\left({s}^{1/3}\right)}{3{s}^{2/3}}=\frac{{f}^{\prime }\left({x}_{3}\right)}{3{x}_{3}^{2}}\phantom{\rule{thinmathspace}{0ex}}.\end{array}$