Proving Unique Solution for f(x) = 0 on (a,b) with f'(x) ≠ 0

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The discussion centers on proving the existence and uniqueness of a solution to the equation f(x) = 0 within the interval (a, b), given that f is continuous on [a, b], differentiable on (a, b), and that f'(x) ≠ 0 throughout (a, b) with f(a) and f(b) having opposite signs. The proof requires the application of the Intermediate Value Theorem to establish the existence of at least one solution. Additionally, to demonstrate uniqueness, one must assume the contrary and analyze the implications, confirming that no more than one solution can exist in the specified interval.

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Assume that f is continuous on [a,b] and differentiable on (a,b). Assume also that f′(x) ≠ 0 on (a,b) and f(a) and f(b) have different signs. Show that the equation f (x) = 0 has a unique solution in (a, b).


I'm not really sure how to even start this proof. Do I need to use the Intermediate Value Theorem? Any help would be great!
 
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msell2 said:
Assume that f is continuous on [a,b] and differentiable on (a,b). Assume also that f′(x) ≠ 0 on (a,b) and f(a) and f(b) have different signs. Show that the equation f (x) = 0 has a unique solution in (a, b).


I'm not really sure how to even start this proof. Do I need to use the Intermediate Value Theorem? Any help would be great!

You actually need to prove two things: (1) there is a solution x in (a,b); and (2) there is only one such solution.

Do you know how to get (1)?

For (2): assume the contrary and see what happens.
 

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