Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2].

Perhaps Rolle's Theorem, Mean Value Theorem, or Intermediate Value Theorem hold clues? ...Other than simply using my TI-84, I have no idea how to accomplish this.

See response:

http://www.jiskha.com/display.cgi?id=1298948844

To show that the equation has at most one root in the interval [-2,2], we can make use of the Intermediate Value Theorem.

The Intermediate Value Theorem states that if a function f(x) is continuous over a closed interval [a, b], and if f(a) and f(b) have opposite signs, then there exists at least one value c in the interval (a, b) such that f(c) = 0.

To apply the Intermediate Value Theorem, we need to check the signs of the function values at the endpoints of the interval [-2, 2].

Let's first substitute x = -2 into the equation:
f(-2) = (-2)^3 - 15(-2) + c = -8 + 30 + c = 22 + c

Next, let's substitute x = 2 into the equation:
f(2) = (2)^3 - 15(2) + c = 8 - 30 + c = -22 + c

Now, we need to consider the signs of f(-2) and f(2) to determine if they are opposite, which would satisfy the requirements of the Intermediate Value Theorem.

Let's analyze these cases:
1. If f(-2) > 0 and f(2) < 0:
- This means that f(-2) has a positive value and f(2) has a negative value.
- Therefore, there must exist a root of the equation in the interval [-2, 2].

2. If f(-2) < 0 and f(2) > 0:
- This means that f(-2) has a negative value and f(2) has a positive value.
- Again, there must exist a root of the equation in the interval [-2, 2].

3. If f(-2) and f(2) have the same sign (either both positive or both negative):
- This means that there is no change in sign between f(-2) and f(2).
- In this case, the Intermediate Value Theorem does not guarantee the existence of a root in the interval [-2, 2].

Therefore, to conclude that the equation has at most one root in the interval [-2, 2], we need to rule out the possibility of case 3, where f(-2) and f(2) have the same sign.

To show this, we need the specific value of c. Unfortunately, without more information about the equation or the value of c, we cannot determine its exact nature or solve for c.