Cauchy Mean Value Theorem
Let \(f(x) \) and \(g(x) \) be two functions which are both:
a) differentiable in [a,b]
b) \(g'(x) \neq 0 \) for any value of \(x\) in [a,b]
then there exists at least one value \(c\) in between \(a\) and \(b\) such that
\[ \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)} \]
Sample Problem
Verify Cauchy Mean Value Theorem for \[ f(x) = x^4, g(x) = x^2 \text{ in } [a,b] \]
Solution
\[ f(x) = x^4, g(x) = x^2 \] \[ f'(x) = 4x^3, g'(x) = 2x \] By Cauchy Mean Value Theorem, \[ \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)} \] \[ \frac{b^4 - a^4}{b^2 - a^2} = \frac{4c^3}{2c} \] \[ 2c^2 = \frac{\cancel{(b^2 - a^2)}(b^2 + a^2)}{\cancel{(b^2 - a^2)}} \] \[ c = \pm \frac{1}{\sqrt{2}}\sqrt{b^2 + a^2} \] \[ \text{Since } c = -\frac{1}{\sqrt{2}}\sqrt{b^2 + a^2} \notin [a,b] \] \[ \boxed{c = \frac{1}{\sqrt{2}}\sqrt{b^2 + a^2} } \]
