Determine whether the Mean Value Theorem can be applied to f on the closed interval [a, b]. 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 . If the Mean Value Theorem cannot be applied, explain why not. f(x) = x^3 + 2x, [-1,1]

There is a value of in (-1, 1), .

Step-by-step explanation:

Let for , we need to prove that is continuous and differentiable to apply the Mean Value Theorem. Given that is a polynomical function, its domain comprises all real numbers and therefore, function is continuous.

If is differentiable, then exists for all value of . By definition of derivative, we obtain the following expression:

(Eq. 2)

The derivative of a cubic function is quadratic function, which is also a polynomic function. Hence, the function is differentiable at the given interval.

According to the Mean Value Theorem, the following relationship is fulfilled:

(Eq. 3)

If we know that , and , then we expand the definition as follows:

There is a value of in the interval (-1, 1), .

c

step-by-step explanation:

i think because what else can be a sample of every member

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .