show that the equation x^3-15x+c=0 has at most one root on the interval [-2,2.
f(-2) = -8+30+c = 22+c
f(2) = 8-30+c = -22+c
df/dx = 3x^2-15 = 3(x^2-5)
df/dx=0 at x=±√5
so, f(x) has no turning points on [-2,2]
so, either f stays negative or positive on [-2,2], or it crosses somewhere. But, it can cross only once, since it does not turn around anywhere.
To show that the equation has at most one root on the interval [-2,2], we can consider the behavior of the function f(x) = x^3 - 15x + c on this interval.
Step 1: Finding the critical points
First, let's find the critical points by taking the derivative of f(x) with respect to x and setting it equal to zero.
f'(x) = 3x^2 - 15
Setting f'(x) = 0, we solve for x:
3x^2 - 15 = 0
3x^2 = 15
x^2 = 15/3
x^2 = 5
x = ±√5
So, we have two critical points x = -√5 and x = √5.
Step 2: Determining the behavior of f(x)
Next, let's analyze the behavior of the function f(x) on the interval [-2,2] by evaluating f(x) at the critical points.
For x = -√5,
f(-√5) = (-√5)^3 - 15(-√5) + c
= -5√5 - 15√5 + c
= (-20√5 + c)
For x = √5,
f(√5) = (√5)^3 - 15(√5) + c
= 5√5 - 15√5 + c
= (-10√5 + c)
Since f(x) is a continuous function and the function values for x = -√5 and x = √5 are given by -20√5 + c and -10√5 + c respectively, we can deduce the following:
- If -20√5 + c > 0 and -10√5 + c < 0, then f(x) changes sign from negative to positive between x = -√5 and x = √5. This means there is exactly one root between x = -√5 and x = √5.
- If -20√5 + c < 0 and -10√5 + c > 0, then f(x) changes sign from positive to negative between x = -√5 and x = √5. This means there is exactly one root between x = -√5 and x = √5.
- If -20√5 + c and -10√5 + c have the same sign, then f(x) does not change sign between x = -√5 and x = √5. This means there are no roots or more than one root on the interval [-2,2].
Therefore, in order to show that the equation x^3 - 15x + c = 0 has at most one root on the interval [-2,2], we need to ensure that -20√5 + c and -10√5 + c have opposite signs.
To show that the equation x^3 - 15x + c = 0 has at most one root on the interval [-2, 2], we can use the Intermediate Value Theorem and the fact that the function is continuous.
Here's how you can proceed:
1. Begin by finding the critical points of the function. Take its derivative, f'(x), and set it equal to zero to find the critical points.
- In this case, the derivative of f(x) = x^3 - 15x + c would be f'(x) = 3x^2 - 15.
2. Solve the equation f'(x) = 3x^2 - 15 = 0 to find the critical points.
- By factoring the expression, we get 3(x^2 - 5) = 0. Setting each factor equal to zero, we get x = ±√5.
3. Now, check the value of the function at the endpoints of the interval [-2, 2].
- Evaluate f(-2) and f(2) by substituting these values into the equation: (-2)^3 - 15(-2) + c = -8 + 30 + c = 22 + c.
- Similarly, f(2) = 8 - 30 + c = -22 + c.
4. Use the Intermediate Value Theorem to show that if f(-2) and f(2) have different signs, then there must be at least one root on the interval [-2, 2].
- In this case, if f(-2) = 22 + c > 0 and f(2) = -22 + c < 0, then there must be at least one root between -2 and 2.
However, to prove that there is at most one root, we need to show that f(x) does not change signs on the interval.
5. Calculate f'(x) for values less than √5 and greater than √5, such as x = √5 + 1 and x = √5 - 1.
- Evaluate f'(√5 + 1) and f'(√5 - 1) to determine the sign of f'(x) in these regions.
6. Based on the sign of f'(x), determine if the function is increasing or decreasing in each interval.
- If f'(x) > 0 for x < √5 and f'(x) < 0 for x > √5, then the function is increasing up to √5 and decreasing after √5.
7. Using this information, determine the sign of f(x) for x < √5 and x > √5.
- If the function is increasing up to √5 and then decreasing after √5, determine the sign of f(x) for x < √5 and x > √5.
8. Combine the information from steps 4 and 7 to draw a conclusion about the number of roots on the interval [-2, 2].
- If f(x) has different signs at the endpoints, -2 and 2, and maintains the same sign throughout the domain, then there can be at most one root in the interval [-2, 2].
By following these steps, you can show that the equation x^3 - 15x + c = 0 has at most one root on the interval [-2, 2].