ziphumulegn

2022-07-07

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

${f}^{\prime}({x}_{1})=(b+a)\frac{{f}^{\prime}({x}_{2})}{2{x}_{2}}=({b}^{2}+ba+{a}^{2})\frac{{f}^{\prime}({x}_{3})}{3{x}_{3}^{2}}$

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!

${f}^{\prime}({x}_{1})=(b+a)\frac{{f}^{\prime}({x}_{2})}{2{x}_{2}}=({b}^{2}+ba+{a}^{2})\frac{{f}^{\prime}({x}_{3})}{3{x}_{3}^{2}}$

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!

iskakanjulc

Beginner2022-07-08Added 18 answers

By the mean value theorem,

$\begin{array}{}\text{(1)}& \mathrm{\exists}{x}_{1}\in (a,b)\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{1em}{0ex}}\frac{f(b)-f(a)}{b-a}={f}^{\prime}({x}_{1})\phantom{\rule{thinmathspace}{0ex}}.\end{array}$

Consider $g:[{a}^{2},{b}^{2}]\to \mathbb{R}$ defined by $g(x)=f(\sqrt{x})$. Applying the mean value theorem to $g$, we infer that there is some $t\in ({a}^{2},{b}^{2})$ such that

$\frac{g({b}^{2})-g({a}^{2})}{{b}^{2}-{a}^{2}}={g}^{\prime}(t)\phantom{\rule{thinmathspace}{0ex}},$

or equivalently, writing ${x}_{2}:=\sqrt{t}$,

$\begin{array}{}\text{(2)}& \frac{f(b)-f(a)}{(b-a)(b+a)}=\frac{{f}^{\prime}(\sqrt{t})}{2\sqrt{t}}=\frac{{f}^{\prime}({x}_{2})}{2{x}_{2}}\phantom{\rule{thinmathspace}{0ex}}.\end{array}$

Similarly, apply the mean value theorem to $h:[{a}^{3},{b}^{3}]\to \mathbb{R}$ defined by $h(x)=f({x}^{1/3})$. We deduce that there is some $s\in ({a}^{3},{b}^{3})$ such that

$\frac{h({b}^{3})-h({a}^{3})}{{b}^{3}-{a}^{3}}={h}^{\prime}(s)\phantom{\rule{thinmathspace}{0ex}},$

or equivalently, writing ${x}_{3}:={s}^{1/3}$,

$\begin{array}{}\text{(3)}& \frac{f(b)-f(a)}{(b-a)({b}^{2}+ba+{a}^{2})}=\frac{{f}^{\prime}({s}^{1/3})}{3{s}^{2/3}}=\frac{{f}^{\prime}({x}_{3})}{3{x}_{3}^{2}}\phantom{\rule{thinmathspace}{0ex}}.\end{array}$

$\begin{array}{}\text{(1)}& \mathrm{\exists}{x}_{1}\in (a,b)\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{1em}{0ex}}\frac{f(b)-f(a)}{b-a}={f}^{\prime}({x}_{1})\phantom{\rule{thinmathspace}{0ex}}.\end{array}$

Consider $g:[{a}^{2},{b}^{2}]\to \mathbb{R}$ defined by $g(x)=f(\sqrt{x})$. Applying the mean value theorem to $g$, we infer that there is some $t\in ({a}^{2},{b}^{2})$ such that

$\frac{g({b}^{2})-g({a}^{2})}{{b}^{2}-{a}^{2}}={g}^{\prime}(t)\phantom{\rule{thinmathspace}{0ex}},$

or equivalently, writing ${x}_{2}:=\sqrt{t}$,

$\begin{array}{}\text{(2)}& \frac{f(b)-f(a)}{(b-a)(b+a)}=\frac{{f}^{\prime}(\sqrt{t})}{2\sqrt{t}}=\frac{{f}^{\prime}({x}_{2})}{2{x}_{2}}\phantom{\rule{thinmathspace}{0ex}}.\end{array}$

Similarly, apply the mean value theorem to $h:[{a}^{3},{b}^{3}]\to \mathbb{R}$ defined by $h(x)=f({x}^{1/3})$. We deduce that there is some $s\in ({a}^{3},{b}^{3})$ such that

$\frac{h({b}^{3})-h({a}^{3})}{{b}^{3}-{a}^{3}}={h}^{\prime}(s)\phantom{\rule{thinmathspace}{0ex}},$

or equivalently, writing ${x}_{3}:={s}^{1/3}$,

$\begin{array}{}\text{(3)}& \frac{f(b)-f(a)}{(b-a)({b}^{2}+ba+{a}^{2})}=\frac{{f}^{\prime}({s}^{1/3})}{3{s}^{2/3}}=\frac{{f}^{\prime}({x}_{3})}{3{x}_{3}^{2}}\phantom{\rule{thinmathspace}{0ex}}.\end{array}$

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