Q. Let f(x) and g(x) be differentiable functions such that f'(x)g(x)\neq f(x)g'(x) for any real x. Show that between any two real solutions of f(x)=0, there is atleast one real solution of g(x)=0 .
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1 Answers
21Assume g(x)\ne 0 in between two real roots say m,n of f(x)= 0. Consider h(x)= \frac{f(x)}{g(x)} .h(m)= h(n) = 0. So by Rolles
theorem h'(x)= \frac{f'(x)g(x) - g'(x)f(x)}{g^2(x)} must be zero some where in between m,n. But that contradicts f'(x)g(x)\neq f(x)g'(x). To be more straightforward the derivative of h(x) can not exist in [m,n], which can only be made possible if we have g(x)= 0 at least once in [m,n]
