Question

PLease help..

Answer 1

what do we know about \(c\) ?

Answer 2

c is a point in (a,b)

Answer 3

@ganeshie8

Answer 4

\[\begin{vmatrix}
1&f(b)-f(a)&f'(c)\\
1&g(b)-g(a)&g'(c)\\
1&h(b)-h(a)&h'(c)
\end{vmatrix}\]
A cofactor expansion along the first column gives
\[\begin{vmatrix}
g(b)-g(a)&g'(c)\\
h(b)-h(a)&h'(c)
\end{vmatrix}-\begin{vmatrix}
f(b)-f(a)&f'(c)\\
h(b)-h(a)&h'(c)
\end{vmatrix}+\begin{vmatrix}
f(b)-f(a)&f'(c)\\
g(b)-g(a)&g'(c)
\end{vmatrix}\]
which simplifies to
\[h'(c)[g(b)-g(a)]-g'(c)[h(b)-h(a)]-h'(c)[f(b)-f(a)]\\
\quad+f'(c)[h(b)-h(a)]+g'(c)[f(b)-f(a)]-g'(c)[f(b)-f(a)]\]
Use the mean value theorem to replace some terms here and there, and they all disappear.

Answer 5

In particular, since \(f,g,h\) are continuous and differentiable over the given interval, there exists \(c\in(a,b)\) such that
\[f'(c)=\frac{f(b)-f(a)}{b-a}\]
and likewise for \(g\) and \(h\).

For instance, the first two terms in the expansion above give
\[\begin{align*}\color{red}{h'(c)}\color{blue}{[g(b)-g(a)]}-g'(c)[h(b)-h(a)]&=\color{red}{\frac{h(b)-h(a)}{b-a}}[g(b)-g(a)]\\\\
&\quad\quad-g'(c)[h(b)-h(a)]\\\\
&=\color{blue}{\frac{g(b)-g(a)}{b-a}}[h(b)-h(a)]\\\\
&\quad\quad-g'(c)[h(b)-h(a)]\\\\
&=\color{blue}{g'(c)}[h(b)-h(a)]-g'(c)[h(b)-h(a)]\\\\
&=0
\end{align*}\]

Answer 6

but is it compulsory that the point c will be same for all the functions f,g, and h?

Answer 7

@SithsAndGiggles

Answer 8

Thats kinda tricky... Alternatively you may use Rolle's thm to reduce the amont of algebra slightly..

Let \(D(x) = \begin{vmatrix}
1&f(b)-f(a)&f(x)\\
1&g(b)-g(a)&g(x)\\
1&h(b)-h(a)&h(x)
\end{vmatrix}\)


It is not hard to see that \(D(a) = D(b) \) (Add second column to third column
for \(D(a)\)). By Rolle's thm there exists some \(c \in (a,b)\) such that \(D'(c) = 0\).

Answer 9

Ha ha ha... Great... Thanks.:).. @ganeshie8