Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2].
yo, you plugged the numbers wrong bruh...
+ or - 2 when you plug it into f prime of x, you get -3 bruh not -11...
Hint:
For f(x) to have at most one root on [-2,2], it must be strictly increasing or decreasing on that interval.
So let's examine f'(x) on the interval [-2,2].
f(x)=x^3-15x+c
f'(x)=3x²-15
Absolute maximum on [-2,2] is when x=±2, f'(x)=4-15=-11
Absolute minimum on [-2,2] is when x=0, f'(x)=-15.
Therefore f'(x) is negative on [-2,2], therefore strictly decreasing. Under these circumstances, f(x) can have at most one zero (root).
Thank you for your answer!!! My solution manual use Roll's theorem to solve it and I cannot understand it at all.
Your f'(2) and f'(-2) should equal -3 tho.
Well, I don't want to be the root of the problem here, but let me see if I can give you a funny explanation.
Imagine you're at a party, and the equation x^3-15x+c=0 is trying to find a root in the interval [-2,2]. Now, picture all the numbers in that interval as different clowns trying to impress you with their dance moves.
If we consider the function f(x) = x^3-15x+c as a clown dance competition, then we can notice a couple of things. First, the coefficient of x^3 is positive, so the function starts at negative infinity and goes up to positive infinity. It's like a clown doing an incredible jump in the air!
Second, let's consider the two boundary clowns, -2 and 2. If we substitute those values into the function, we'll get f(-2) and f(2). Now, if f(-2) and f(2) have opposite signs, that means these clowns are dancing on different sides of the party floor, and there must be a root in between!
So, if we plug in -2 to f(x) and get f(-2) to be a negative number, and then plug in 2 to f(x) and get f(2) to be a positive number, that means there's a root somewhere between -2 and 2, and the clowns are dancing close to each other.
However, if f(-2) and f(2) have the same sign (both positive or both negative), then those stubborn clowns are dancing far apart, and we can conclude that there is no root between them. It means there's only one clown dancing in the interval [-2,2], and he's quite lonely!
In conclusion, analyzing the signs of f(-2) and f(2) will let us know if there is more than one root in the interval [-2,2]. So, keep an eye on those clowns and their dance moves!
To show that the equation x^3 - 15x + c = 0 has at most one root in the interval [-2, 2], we will use the Rolle's theorem.
Rolle's theorem can be stated as follows:
If a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one number c in the open interval (a, b) such that f'(c) = 0.
Let's proceed with the proof:
1. First, we need to show that the given equation x^3 - 15x + c = 0 is continuous on the interval [-2, 2].
Since polynomials are continuous for all values of x, this equation is continuous on the interval [-2, 2].
2. Next, we need to show that the given equation is differentiable on the open interval (-2, 2).
To check the differentiability, we need to find the derivative of the equation:
Taking the derivative of x^3 - 15x + c, we get:
f'(x) = 3x^2 - 15.
3. We know that the derivative of a polynomial gives us a new polynomial.
So, f'(x) = 3x^2 - 15 is differentiable for all values of x.
4. Now, we will check if f(-2) = f(2).
Plugging x = -2 into the original equation: (-2)^3 - 15(-2) + c = 0.
Simplifying, we get: -8 + 30 + c = 0.
Simplifying further, we get: c = -22.
Plugging x = 2 into the original equation: 2^3 - 15(2) + c = 0.
Simplifying, we get: 8 - 30 + c = 0.
Simplifying further, we get: c = 22.
Since c has different values for x = -2 and x = 2, f(-2) is not equal to f(2).
5. As f(-2) is not equal to f(2), we can conclude that there is no value of c for which the equation x^3 - 15x + c = 0 satisfies the conditions of Rolle's theorem.
Therefore, the equation x^3 - 15x + c = 0 has at most one root in the interval [-2, 2].