Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2].
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To show that the equation x^3 - 15x + c = 0 has at most one root in the interval [-2,2], let's use the concept of the intermediate value theorem.
The intermediate value theorem states that if a function is continuous on a closed interval [a, b] and takes on positive and negative values at the endpoints a and b, then it must have at least one root in between.
In this case, let's consider the interval [-2,2]. We need to determine the signs of the function at the endpoints.
1. At x = -2:
Plug in x = -2 into the equation: (-2)^3 - 15(-2) + c = -8 + 30 + c = 22 + c.
This tells us that the value of the function at x = -2 is 22 + c.
2. At x = 2:
Plug in x = 2 into the equation: (2)^3 - 15(2) + c = 8 - 30 + c = -22 + c.
This tells us that the value of the function at x = 2 is -22 + c.
Now, considering these two endpoints, we have two possible cases:
Case 1: If 22 + c < 0 and -22 + c > 0.
This means the function takes on negative values at x = -2 and positive values at x = 2. Therefore, by the intermediate value theorem, there must be at least one root in the interval (-2, 2).
Case 2: If 22 + c > 0 and -22 + c < 0.
This means the function takes on positive values at x = -2 and negative values at x = 2. Again, by the intermediate value theorem, there must be at least one root in the interval (-2, 2).
However, we notice that in both cases, the function changes signs between x = -2 and x = 2. Since the function is a cubic (degree 3), it can change signs at most twice. This implies that there can be at most two roots in the entire real number line. But since we have already confirmed the existence of at least one root in the interval (-2, 2), there can be at most one root in this interval.
Therefore, the equation x^3 - 15x + c = 0 has at most one root in the interval [-2,2].