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 x^3 - 15x + c = 0 has at most one root in the interval [-2, 2], we can indeed make use of the Intermediate Value Theorem (IVT) and some basic algebraic reasoning.

Before proceeding, note that the given equation is a polynomial of degree 3, which means it is a cubic equation. Cubic equations can have at most three distinct roots.

1. Begin by substituting the endpoints of the interval [-2, 2] into the equation.

For x = -2:
(-2)^3 - 15(-2) + c = -8 + 30 + c = 22 + c

For x = 2:
(2)^3 - 15(2) + c = 8 - 30 + c = -22 + c

2. The IVT states that if a function is continuous on a closed interval [a, b] and takes on values f(a) and f(b) with opposite signs, then there exists at least one root (or zero-crossing) between a and b.

In our case, we want to demonstrate that there is at most one root in the interval [-2, 2]. To prove this, we need to show that the function does not change signs within this interval.

3. By substituting the endpoints of the interval into the equation, we obtained expressions of the form 22 + c and -22 + c. Now, consider the following cases:

Case 1: c > 22
If c is greater than 22, then both 22 + c and -22 + c will be positive. (Remember, we are assuming that c is a constant value.) Therefore, the function does not change signs between x = -2 and x = 2. Hence, there is no root in this interval.

Case 2: c < -22
If c is less than -22, then both 22 + c and -22 + c will be negative. Consequently, the function does not change signs between x = -2 and x = 2. Therefore, there is no root in this interval.

Case 3: -22 ≤ c ≤ 22
If c falls within this range, then one of the expressions 22 + c or -22 + c will be positive, while the other will be negative. Thus, the function changes signs between x = -2 and x = 2. By the IVT, we can conclude that the function has at least one root in this interval.

4. To summarize, in Cases 1 and 2, the function has no root in the interval [-2, 2]. In Case 3, the function has at least one root in the interval [-2, 2]. Therefore, the equation x^3 - 15x + c = 0 has at most one root in the interval [-2, 2].

It's worth noting that this approach does not provide the exact number of roots or their values, but it does establish an upper bound on the number of roots within the given interval.