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If $f: [x_1,x_2] \to \mathbb{R}$ is differentiable, show for some $c \in (x_1,x_2)$ that $$ \frac{1}{x_1-x_2} \left| \begin{matrix} x_1 & x_2 \\ f(x_1) & f(x_2) \end{matrix} \right|=f(c)-cf'(c) $$

My attempt: Actually taking the determinant, multiplying by a negative, and carrying across the denominator on the left gives $$ x_1f(x_2)-x_2f(x_1)=(-f(c)+cf'(c)) \cdot (x_2-x_1) $$ and this screams Mean Value Theorem. So I took the function $g(x)=(x_2+x_1-x)f(x)$ which is clearly differentiable on $[x_1,x_2]$, then by the Mean Value Theorem, we know there is a $c \in (x_1,x_2)$ such that $$ g(x_2)-g(x_1)=g'(c)(x_2-x_1) $$ But for our function $g(x)$, we know $g(x_2)=x_1f(x_2)$ and $g(x_1)=x_2f(x_1)$. Moreover, $g'(x)= - f(x) + (x_2+x_1-x)f'(x)$. Then this gives $$ x_1f(x_2)-x_2f(x_1)=(-f(c)+(x_2+x_1-c)f'(c))(x_2-x_1) $$ which is so close to what we wanted to show that I do not see how this could not be the correct approach. Have I missed something or is the result false?

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    $\begingroup$ It is D. Pompeiu's mean value theorem: Sur une proposition analogue au théorème des accroissements finis, Mathematica (Cluj, Rumania), 22 (1946), 143-146 . You should add the condition $0 \notin [x_1,x_2]$ . $\endgroup$ – Tony Piccolo Jan 1 '16 at 9:41
  • $\begingroup$ See also abs/math/0305231v1 for a proof (Theorem 5, page 3). (the auxiliary function considered is $x\mapsto xf(1/x)$, where the hypothesis $0\notin [x_1,x_2]$ comes in handy). $\endgroup$ – Clement C. Jan 1 '16 at 9:58
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Using Cauchy's Mean Value Theorem with functions $\frac{f(x)}x$ and $\frac1x$, we get $$ \frac{x_1f(x_2)-x_2f(x_1)}{x_1-x_2}=\frac{\frac{f(x_2)}{x_2}-\frac{f(x_1)}{x_1}}{\frac1{x_2}-\frac1{x_1}} =\frac{\frac{cf'(c)-f(c)}{c^2}}{-\frac1{c^2}}=f(c)-cf'(c) $$ for some $c\in[x_1,x_2]$. For this, I believe we may need that $0\not\in[x_1,x_2]$.

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  • $\begingroup$ A very clear solution! +1. However, what would make one think to use those functions other than just sheer trial and error? $\endgroup$ – Kyle L Jan 3 '16 at 22:01
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    $\begingroup$ @KyleL: The fact that $\left(\frac{f(x)}x\right)'=\frac{\color{#C00000}{xf'(x)-f(x)}}{x^2}$ lead me to consider $\frac{f(x)}x$. $\endgroup$ – robjohn Jan 4 '16 at 14:05

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