Math Problem Statement

prove that f(X) = x^2 is continuous on R

Solution

To prove that f(x)=x2f(x) = x^2 is continuous on R\mathbb{R}, we will use the ε-δ definition of continuity. According to this definition, a function f(x)f(x) is continuous at a point x=ax = a if, for every ϵ>0\epsilon > 0, there exists a δ>0\delta > 0 such that whenever xa<δ|x - a| < \delta, it follows that f(x)f(a)<ϵ|f(x) - f(a)| < \epsilon.

We need to show that for any aRa \in \mathbb{R}, the function f(x)=x2f(x) = x^2 satisfies this definition.

  1. Let f(x)=x2f(x) = x^2 and choose any point aRa \in \mathbb{R}.

  2. We want to show that for every ϵ>0\epsilon > 0, there exists a δ>0\delta > 0 such that xa<δ|x - a| < \delta implies f(x)f(a)<ϵ|f(x) - f(a)| < \epsilon.

  3. Compute f(x)f(a)|f(x) - f(a)|: f(x)f(a)=x2a2|f(x) - f(a)| = |x^2 - a^2| We can factor this expression as: x2a2=(xa)(x+a)|x^2 - a^2| = |(x - a)(x + a)| Thus, f(x)f(a)=xax+a|f(x) - f(a)| = |x - a| \cdot |x + a|

  4. Control x+a|x + a|: To ensure that f(x)f(a)<ϵ|f(x) - f(a)| < \epsilon, we need to find an appropriate δ\delta in terms of ϵ\epsilon. Notice that if we can bound x+a|x + a|, we can find a relationship between xa|x - a| and ϵ\epsilon.

    Assume that xa<1|x - a| < 1, which implies xx is in the interval (a1,a+1)(a - 1, a + 1). Then: x+aa+1+a=2a+1|x + a| \leq |a + 1| + |a| = 2|a| + 1 So, we have the bound x+a2a+1|x + a| \leq 2|a| + 1 when xa<1|x - a| < 1.

  5. Choose δ\delta: Now, to ensure f(x)f(a)<ϵ|f(x) - f(a)| < \epsilon, we need: xax+a<ϵ|x - a| \cdot |x + a| < \epsilon Since x+a2a+1|x + a| \leq 2|a| + 1 when xa<1|x - a| < 1, it suffices to choose δ\delta such that: δ(2a+1)<ϵ\delta (2|a| + 1) < \epsilon Solving for δ\delta, we get: δ=ϵ2a+1\delta = \frac{\epsilon}{2|a| + 1}

  6. Conclusion: Given any ϵ>0\epsilon > 0, choose δ=min(1,ϵ2a+1)\delta = \min \left(1, \frac{\epsilon}{2|a| + 1}\right). Then, for xa<δ|x - a| < \delta, we have f(x)f(a)<ϵ|f(x) - f(a)| < \epsilon. Therefore, f(x)=x2f(x) = x^2 is continuous at x=ax = a.

Since aa was arbitrary, f(x)=x2f(x) = x^2 is continuous on R\mathbb{R}.

Would you like further details on any part of the proof?


Here are some related questions to expand on this topic:

  1. How does the continuity of polynomial functions extend to all of R\mathbb{R}?
  2. What would be the proof for the continuity of f(x)=x3f(x) = x^3 on R\mathbb{R}?
  3. How does the ε-δ definition of continuity compare with limits-based definitions?
  4. Can the continuity of f(x)=x2f(x) = x^2 be generalized to functions of higher degrees?
  5. How would you prove continuity of f(x)=x2f(x) = x^2 using sequences?

Tip: For polynomial functions, continuity can often be more easily shown by leveraging that they are compositions and sums of continuous functions, all continuous over R\mathbb{R}.

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Math Problem Analysis

Mathematical Concepts

Continuity
Real Analysis
Limits
Functions

Formulas

ε-δ definition of continuity

Theorems

Continuity of polynomial functions
ε-δ definition of continuity

Suitable Grade Level

Undergraduate - Calculus I or II