2

In proving Cauchy's mean value theorem, the first step is to use this function:

$$ h(x)=[f(b)−f(a)]g(x)−[g(b)−g(a)]f(x)$$

I've seen this in many textbooks but none of them actually show how they got this function. I understand every steps in the proof of Cauchy's mean value theorem, except how to derive this equation myself.

Do you guys know how do they get this function? Thanks a lot!

user75521
  • 21
  • 1
    Oftentimes math proofs have "miraculous" steps like this one, that are chosen to make future steps work out. Originally such steps were found by messy calculations, that are not included in the final proof. – vadim123 May 02 '13 at 13:40
  • 2
    Don't think these are born in just seconds , may be minutes , hours or even days. Had you put on some more effort into the proving cauchy's mean value theorem you could have discovered the same step you presently claim is 'miraculous' – hrkrshnn May 02 '13 at 14:34

2 Answers2

2

To prove the theorem, since the fractions create problems for derivatives, cross multiply. Then you need to show that

$$[f(b)-f(a)]g'(x)-[g(b)-g(a)]f'(x)=0 \, \mbox{for some} \, x$$

Now, the left hand side is the derivative of some function, which is that function?

N. S.
  • 134,609
1

The mean value theorem states that $(f'(t),g'(t))$ is somewhere parallel to the line through $(f(a),g(a))$ and $(f(b),g(b))$. A normal vector to this line is $(g(b)-g(a),f(a)-f(b))$, and thus the distance from $(f(t),g(t))$ to the line is a constant plus another constant times $(g(b)-g(a))f(t)+(f(a)-f(b))g(t)$. One good choice of $t$ is where the distance is maximal, hence the preceding expression is stationary.

Harald Hanche-Olsen
  • 32,608
  • 2
  • 61
  • 87