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Is this correct and can anything further be ascribed:

  1. $f(x), g(x)$ are continuous on $[a,b]$ - so, no holes on line segment ab on x-axis
  2. $f(x), g(x)$ are differentiable on $(a,b)$ - so, all points have a computable slope on ab barring 'a' and 'b'
  3. $g(a) \neq g(b)$, so $g$ is always rising or falling with no point where slope is $0$
  4. there is a point $c$ with slope proportional to $f(b) - f(a) = \delta{y}$ the 'rise' in 'rise/run = slope' or $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$$
hrkrshnn
  • 6,357
  • I did not understand 3 and 4. – hrkrshnn Jun 15 '14 at 11:37
  • http://math.stackexchange.com/questions/114694/proving-cauchys-generalized-mean-value-theorem?rq=1 – cactus314 Jun 15 '14 at 12:06
  • @boywholived if $g(a) \neq $g(b) then you can't have two points with the same height on the y-axis. If $g(a) = 10 and $g(b) = 10, then they will have the same height (10). Therefore the graph is always rising or falling - you are forced to go ever upwards or ever downwards.. Correct? –  Jun 16 '14 at 01:35
  • @boywholived f(b) - f(a) is how much the graph rises.. the inclination and that's proportional to the slope (f'(c)). Greater the slope, greater the vertical offset.. which means that somewhere between a and b (on the x-axis) there is a point 'c' with slope f'(c) which is proportional to f(b) - f(a)..or to put it another way: wht if you had points g(a) = g(b) - that would imply there would have to be a point where slope is 0 (Rolle's Theorem) - why, because g(a) - g(b) = 0. So if f(b) - f(a) = x, then there has to be a point f'(c) proportional to x. –  Jun 16 '14 at 01:48

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