How do you show that an equation has one real solution?

Rolle’s Theorem states that if a function f:[a,b]→R is continuous on [a,b] and differentiable on (a,b) then if f(a)=f(b), there exists a point c∈(a,b) such that f′(c)=0. We assume that there is more than one real solution for this equation, namely f(a)=0=f(b).

How do you prove Rolle’s theorem?

Proof of Rolle’s Theorem

1. If f is a function continuous on [a,b] and differentiable on (a,b), with f(a)=f(b)=0, then there exists some c in (a,b) where f′(c)=0.
2. f(x)=0 for all x in [a,b].

How do you solve intermediate value theorem problems?

Solving Intermediate Value Theorem Problems

1. Define a function y=f(x).
2. Define a number (y-value) m.
3. Establish that f is continuous.
4. Choose an interval [a,b].
5. Establish that m is between f(a) and f(b).
6. Now invoke the conclusion of the Intermediate Value Theorem.

How do you prove something has only one root?

To prove that the equation has at least one real root, we will rewrite the equation as a function, then find a value of x that makes the function negative, and one that makes the function positive. . The function f is continuous because it is the sum or difference of a continuous inverse trig function and a polynomial.

How do you prove that f(x) = x2 is continuous at x0?

First off, the definition: f ( x) = x 2 is continuous at x 0 if, for any ϵ > 0, we can find a δ > 0 such that for any c with | x 0 − c | < δ, we have | f ( x 0) − f ( c) | < ϵ. f is continuous if it is continuous at every real number. In terms of how we write the proof, that means we’re given arbitrary x 0, ϵ, and we must find a δ that works.

How do you prove that the limit(x2) = 4 as x approaches 2?

How do you prove that the limit (x2) = 4 as x approaches 2 using the formal definition of a limit? Given a function f (x), we say that the limit as x approaches a of f (x) is L, denoted lim x→a f (x) = L, if for every ε > 0 there exists a δ > 0 such that 0 < |x −a| < δ implies that |f (x) − L| < ε .

Is x2 – 2x – 2 factorable over the reals?

We could say that x2 − 2x − 2 is not factorable over the rational numbers. That is it has no factors whose coefficients are rational numbers. However it is factorable if you allow irrational coefficients. The next level would be Complex coefficients. For example: The quadratic x2 − 2x + 2 is not factorable over the reals.

How many negative real zeros does f(x) have?

With 2 changes of sign, that means that f (x) has 2 or 0 negative Real zeros. Given any polynomial in one variable with integer coefficients: