Proof using mean-value theorem

In summary, the mean-value theorem states that if f is continuous at x and x+h and differentiable between these two numbers, then there exists a number c in the interval (x, x+h) such that f'(c) is equal to the average rate of change of f on the interval [x, x+h]. This can also be expressed as f(x+h) - f(x) = f'(x+ah)h, where 0 < a < 1. By applying the MVT, we can use this expression to show the relationship between the derivative of f and the difference in its values at x and x+h.
  • #1
javi438
15
0
use the mean-value theorem to show that if f is continuous at x and at x+h and is differentiable between these 2 numbers, then f(x+h) - f(x) = f'(x+ah)h for some number a between 0 and 1.

mvt: if f is diff'ble on (a,b) and continuous on [a,b] then there is at least one number c in (a,b) for which f'(c)=[f(b)-f(a)]/(b-a)

any help will be appreciateddd..i don't know where to start :(
 
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  • #2
Try thinking of the expression

f(x+h) = f(x) = hf'(x+ah) where 0 < a < 1

as another form of expressing the mean value theorem . You know that f is continuous on [x, x+h] and differentiable on (x, x+h). So now apply the MVT. I don't want to give much more info yet because I'd be giving up the whole proof. Give it a try and see how far you can get with it.
 
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What is the mean-value theorem used for?

The mean-value theorem is used to prove the existence of a point where the slope of a function is equal to the average rate of change of the function over an interval.

What is the formula for the mean-value theorem?

The formula for the mean-value theorem is f'(c) = (f(b) - f(a))/(b - a), where c is a point in the interval (a, b) where the slope of the function is equal to the average rate of change.

How is the mean-value theorem applied in calculus?

In calculus, the mean-value theorem is used to prove the existence of a point where the derivative of a function is equal to its average rate of change over an interval. This is useful in finding critical points, where the derivative is equal to zero, and in proving the existence of extrema on a function.

Can the mean-value theorem be used to find the exact values of functions?

No, the mean-value theorem only proves the existence of a point where the slope of a function is equal to its average rate of change. It does not provide the exact values of the function at that point.

Are there any limitations to the mean-value theorem?

Yes, the mean-value theorem can only be applied to continuous functions on a closed interval. It also assumes that the function is differentiable on the open interval (a, b) and that the derivative is continuous on the closed interval [a, b].

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