Final answer:
To prove that the function F(x) = x² is uniformly continuous on [−7,7], we need to show that for any ε > 0, there exists a δ > 0 such that whenever |x - y| < δ, then |F(x) - F(y)| < ε for all x, y in [−7,7]. We can choose δ = ε/14 and show that |x² - y²| < ε.
Step-by-step explanation:
To prove that the function F(x) = x² is uniformly continuous on [−7,7], we need to show that for any ε > 0, there exists a δ > 0 such that whenever |x - y| < δ, then |F(x) - F(y)| < ε for all x, y in [−7,7].
Let ε > 0 be given. We choose δ = ε/14. Now let x, y be any two points in [−7,7] such that |x - y| < δ. Since |x - y| < δ = ε/14, multiplying both sides by |x + y| gives us |x² - y²| < (ε/14)(|x + y|). Since |x + y| ≤ 14 for all x, y in [−7,7], we have |x² - y²| < ε.
Therefore, we have shown that for any ε > 0, there exists a δ > 0 such that whenever |x - y| < δ, then |F(x) - F(y)| < ε for all x, y in [−7,7]. Hence, the function F(x) = x² is uniformly continuous on [−7,7].