Show that the equation x^3 - 15x + c = o has exactly one real root.
All I know is that it has something to do with the Mean Value Theorem/Rolle's Theorem.
You must have a typo.
There are all kinds of values of c for which the above equation has 3 roots.
e.g. if c=0
x^3 - 15x = 0
x(x^2 - 15) = 0
x = 0 or x = ± √15
are you sure it wasn't x^3 + 15x + c = 0 ?
I will assume it was
It's been 50 years but if I recall, Rolle's theorem says that if there are two x-intercepts , then the derivative must be zero between those two points
so let f(x) = x^3 + 15x + c
f '(x) = 3x^2 +15 = 0
notice this is always positive, since x^2 = -5 has no real solution,
so there are no turning points, and the curve cuts only once.
You do know that every cubic must cut he x-axis at least once?
The missing info it the interval [-2,2]. In this interval there could be just one root.
To show that the equation \(x^3 - 15x + c = 0\) has exactly one real root, we can indeed use the Mean Value Theorem or Rolle's Theorem, as you mentioned.
Let's consider the function \(f(x) = x^3 - 15x + c\). The Mean Value Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \(c\) in \((a, b)\) such that the derivative of the function at that point is equal to the average rate of change of the function over the interval \([a, b]\).
In this case, we will consider an interval containing the real roots of the equation. Let's choose the interval \([-r, r]\), where \(r\) is a positive number that is greater than or equal to the largest possible real root of the equation. Since the function \(f(x) = x^3 - 15x + c\) is a polynomial, it is continuous and differentiable on the entire real line.
Now, we need to show that there is exactly one real root in the interval \([-r, r]\). If there are no real roots, it means that the equation has either no solutions or only complex solutions. If there is more than one real root, it means that there is at least one point within the interval \([-r, r]\) where the derivative of the function is equal to zero.
Using Rolle's Theorem, which is a special case of the Mean Value Theorem, we know that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), and the function values at both endpoints are equal, then there exists at least one point \(c\) in \((a, b)\) where the derivative of the function is equal to zero.
In this case, we will assume that there are three distinct real roots. Let's label them as \(x_1\), \(x_2\), and \(x_3\) such that \(x_1 < x_2 < x_3\). According to Rolle's Theorem, since \(f(x)\) has a value of zero at \(x_1\) and \(x_3\), there must exist at least one point \(c\) between \(x_1\) and \(x_3\) (i.e., \(x_1 < c < x_3\)) where the derivative of the function is equal to zero.
However, by the factor theorem, if \(x - x_1\) is a factor of \(f(x)\), then \(x_1\) must be a root of \(f(x)\). Therefore, if \(f(x)\) has three distinct real roots, it would mean that there are at least two factors of the form \((x - x_1)\), which would contradict the assumption of a unique real root. Hence, the equation \(x^3 - 15x + c = 0\) can have at most one real root.
To conclude that the equation has exactly one real root, we need to ensure that there is at least one real root. This can be done by showing that the function \(f(x)\) crosses the x-axis at least once. Since \(f(x)\) is a continuous polynomial of odd degree (degree 3 in this case), it is guaranteed that there is at least one real root.
Therefore, combining the fact that \(f(x)\) has at most one real root and that it has at least one real root, we can conclude that the equation \(x^3 - 15x + c = 0\) has exactly one real root.