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

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:

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

To calculate the value c such that:

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

Now we calculate the secant slope:

Now we can write:

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

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step-by-step explanation: