Question

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x) = x , [0, 25]

Answer 1

Answer:

c is all the points in the open interval (0,25)

Step-by-step explanation:

Here given is a function

$f(x) =x$, which is continuous  in the interval [0,25] and differentiable in (0,25)

Mean value theorem says there exists at least one c in the interval (0,25) such that

$f'(c) = \frac{f(25)-f(0)}{25-0}$

We have

$f(25)=25 and f(0) = 0\\f'(c) = 1$

For the given function

$f'(x) =1$

Hence we have c equals all the points in the open interval (0,25)

Answer 2

The number c that satisfies the conclusion on the given interval is;

c = 1

We are given the function; f(x) = x

• According to the mean value theorem, the function is continuous on [0,25] and differentiable on (0, 25).

• According to mean value theorem, for the coordinates (a, b), there will exist a value of c in a way that;

f'(c) = [f(b) - f(a)]/(b - a)

Plugging in the relevant values gives us;

f'(c) =  [f(25) - f(0)]/(25 - 0)

Since f(x) = x, then f(25) = 25 and f(0) = 0

Thus; f'(c) = (25 - 0)/25

f'(c) = 1

Now, f(x) = x

Thus, f'(x) = 1

f'(c) = 1

Thus, c = 1

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