Using the principle of the meanvalue theorem on the indicated interval, how do you find all numbers c that satisfy the conclusion of the theorem for #f(x) = 5(1 + 2x)^(1/2)# in the interval [1,4]?
Solve the equation that is the conclusion to the Mean Value Theorem. Choose only the solutions in the interval
The algebra is tedious.
So we need to solve:
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To find all numbers ( c ) that satisfy the conclusion of the Mean Value Theorem for ( f(x) = 5(1 + 2x)^{1/2} ) on the interval ([1,4]), follow these steps:

Compute the derivative of ( f(x) ) with respect to ( x ). [ f'(x) = \frac{d}{dx}[5(1 + 2x)^{1/2}] ]

Evaluate ( f'(x) ) at each endpoint of the interval ([1,4]). [ f'(1) \text{ and } f'(4) ]

Apply the Mean Value Theorem: [ f'(c) = \frac{f(b)  f(a)}{b  a} ] where ( c ) is a number in the interval ([1,4]) such that ( f'(c) ) equals the average rate of change of ( f(x) ) over the interval ([1,4]).

Solve for ( c ) by setting ( f'(c) ) equal to the average rate of change of ( f(x) ) over the interval. [ f'(c) = \frac{f(4)  f(1)}{4  1} ]

Solve for ( c ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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