Show that the equation x^5+x+1 = 0 has exactly one real root. Name the theorems you use to prove it.
I.V.T.
*f(x) is continuous
*Lim x-> inf x^5+x+1 = inf >0
*Lim x-> -inf x^5+x+1 = -inf <0
Rolles
*f(c)=f(d)=0
*f(x) is coninuous
*f(x) is differentiable
f'(x) = 5x^4+1=0
As f'(x) does not equal zero, because of rolles theorem, the assumptioin f(x) has two roots is false.
Used Intermediate Value theorem and Rolles Theorm.
you mean f'(x) has no real roots I think.
It does cross the x axis at least once as you showed with intermediate value
however the slope is never zero so it never can cross the axis again. The other four roots are therefore complex numbers.
To prove that the equation x^5 + x + 1 = 0 has exactly one real root, we can use the Intermediate Value Theorem (IVT) and Rolle's Theorem. Here is how we can apply these theorems step by step:
1. Intermediate Value Theorem (IVT): The IVT states that if a function f(x) is continuous on a closed interval [a, b], and if y lies between f(a) and f(b), then there exists at least one c in the interval [a, b] such that f(c) = y.
First, we need to show that f(x) = x^5 + x + 1 is continuous. This is true because polynomials are continuous for all real numbers x.
2. Evaluate the function at the limits: Next, we consider the behavior of f(x) as x approaches positive infinity and negative infinity.
- As x approaches positive infinity, x^5 grows without bound, while x and 1 have negligible effects. Therefore, the limit of f(x) as x approaches positive infinity is positive infinity.
- As x approaches negative infinity, x^5 still grows without bound, but x and 1 have little effect. Therefore, the limit of f(x) as x approaches negative infinity is negative infinity.
By the IVT, if f(x) is continuous and the function takes on values exceeding 0 and below 0 for various x, there must exist at least one real root, since f(x) changes sign between the two limits.
3. Rolle's Theorem: Rolle's Theorem states that if f(x) is continuous on the closed interval [a, b], differentiable on the open interval (a, b), and if f(a) = f(b) = 0, then there exists at least one c in the interval (a, b) such that f'(c) = 0.
To apply Rolle's Theorem, we need to find the derivative of f(x). Taking the derivative of f(x) gives us f'(x) = 5x^4 + 1. We need to find the roots of f'(x) to check if there are any critical points.
4. Solve for critical points: Setting f'(x) = 0, we have 5x^4 + 1 = 0. However, since 5x^4 is always positive, there is no real value of x that satisfies this equation. Therefore, f'(x) ≠ 0 for any x value.
By Rolle's Theorem, if f(x) has two or more roots between a and b, then f'(x) has at least one root in the interval (a, b). However, since f'(x) has no roots, the assumption that f(x) has two roots is false.
Hence, using the Intermediate Value Theorem and Rolle's Theorem, we can conclude that the equation x^5 + x + 1 = 0 has exactly one real root.