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Determine whether the Mean Value theorem can be applied to f on the closed interval [a, b]. (Select all that apply.) f(x) = 3x3, [1, 2] Yes, the Mean Value Theorem can be applied. No, because f is not continuous on the closed interval [a, b]. No, because f is not differentiable in the open interval (a, b). None of the above. If the Mean Value Theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = f(b) − f(a) b − a . (Enter your answers as a comma-separated list. If the Mean Value Theorem cannot be applied, enter NA.) c = 27

Answer :

Answer:

Yes, the Mean Value Theorem can be applied.

The value of c is c=1.5275

Step-by-step explanation:

The mean value theorem states that for a given two endpoints of an interval and a continuos function f, there is at least one point within this interval at which the tangent to the function f (first derivative) is parallel to the secant of the endpoints of the interval.

For the function f(x) and the interval:

[tex]f(x) = 3x^3,\,\,\,[1,2][/tex]

the condition of continuity is met, so the mean value theorem can be applied.

To calculate the value c such that:

[tex]f '(c) = \dfrac{f(b) - f(a)}{ b - a}[/tex]

we have to calculate the first derivative of f(x)

[tex]\dfrac{df}{dx}=3*3x^{3-1}=9x^2[/tex]

Now we calculate the secant slope:

[tex]f '(c) = \dfrac{f(b) - f(a)}{ b - a}= \dfrac{f(2) - f(1)}{ 2 - 1}=(3*2^3-3*1^3)=(24-3)=21[/tex]

Now we can write:

[tex]f '(c) = 9c^2=21\\\\c^2=21/9=2.33\\\\c=\sqrt{2.33}=1.5275[/tex]

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