Find all values of c that satisfy the Mean Value Theorem for integrals for

f(x)= x^{2} on the interval [-3, 3]

f(-3) = f(3) = 0

The only horizontal tangent is at x=0.

To apply the Mean Value Theorem for integrals, we need to first verify that the function f(x) = x^2 satisfies the conditions of the theorem on the given interval [-3, 3].

The conditions of the Mean Value Theorem for integrals are:

1. The function f(x) must be continuous on the closed interval [a, b].
2. The function f(x) must be integrable on the closed interval [a, b].

In this case, f(x) = x^2 satisfies both conditions, as it is a polynomial function and is continuous and integrable on the interval [-3, 3]. Therefore, we can proceed to find the values of c that satisfy the Mean Value Theorem.

According to the Mean Value Theorem for integrals, there exists a value c in the interval [a, b] such that:

∫[a, b] f(x) dx = f(c) * (b - a)

In this case, a = -3, b = 3, and f(x) = x^2. Substituting these values into the equation, we get:

∫[-3, 3] x^2 dx = f(c) * (3 - (-3))

Integrating x^2 with respect to x, we get:

[1/3 * x^3] from -3 to 3 = f(c) * 6

Simplifying further, we have:

[1/3 * 3^3] - [1/3 * (-3)^3] = f(c) * 6

[1/3 * 27] - [1/3 * (-27)] = f(c) * 6

9 - (-9) = f(c) * 6

18 = f(c) * 6

Dividing both sides by 6, we get:

f(c) = 18/6

f(c) = 3

To find the values of c that satisfy the equation f(c) = 3, we can observe that f(x) = x^2 is an increasing function on the interval [-3, 3]. Therefore, we can say that there exists a value c in the interval (-3, 3) such that f(c) = 3.

Hence, the values of c that satisfy the Mean Value Theorem for the function f(x) = x^2 on the interval [-3, 3] are all the values in the open interval (-3, 3).