Question

Prove that f(x, y)=x²-6y is continuous at (0,-3) using the e-6 definition.

Answer 1

To prove that f(x, y) = x² - 6y is continuous at (0, -3) using the ε-δ definition, we need to show that for any ε > 0, there exists a δ > 0 such that whenever |x - 0| < δ and |y - (-3)| < δ, |f(x, y) - f(0, -3)| < ε. We can prove this by computing |f(x, y) - f(0, -3)| and choosing an appropriate δ based on ε.

Step-by-step explanation:

To prove that f(x, y) = x² - 6y is continuous at (0, -3) using the ε-δ definition, we need to show that for any ε > 0, there exists a δ > 0 such that whenever |x - 0| < δ and |y - (-3)| < δ, |f(x, y) - f(0, -3)| < ε.

Let's compute |f(x, y) - f(0, -3)|:

|f(x, y) - f(0, -3)| = |x² - 6y - (-3)² - 6(-3)| = |x² - 6y + 9 + 18| = |x² - 6y + 27|.

Since this is a quadratic function, we can factor it as follows:

|x² - 6y + 27| = |(x - 3)² - 9(y + 3)|.

Now, if we choose δ = ε/9, then whenever |x - 0| < δ and |y - (-3)| < δ, we have:

|f(x, y) - f(0, -3)| = |(x - 3)² - 9(y + 3)| < δ² + 9δ = (ε/9)² + 9(ε/9) = ε/9 + ε = 2ε/3 < ε.

Therefore, we have shown that for any ε > 0, there exists a δ > 0 such that whenever |x - 0| < δ and |y - (-3)| < δ, |f(x, y) - f(0, -3)| < ε. Hence, f(x, y) = x² - 6y is continuous at (0, -3).

Answer 2

To prove the continuity of f(x, y)=x²-6y at (0,-3), we use the ε-δ definition of continuity by showing that for any ε>0, there exists a δ>0 such that |f(x, y) - f(0, -3)|<ε for all points within δ distance from (0,-3).

Step-by-step explanation:

To prove that the function f(x, y)=x²-6y is continuous at the point (0,-3) using the ε-δ definition of continuity, we need to show that for any ε>0, there exists a δ>0 such that for all (x, y) within δ distance from (0,-3), |f(x, y) - f(0, -3)| < ε. The function value at (0, -3) is f(0, -3) = 0² - 6(-3) = 18.

Let's take any ε>0 and find a corresponding δ. For (x, y) near (0, -3), we have |(x² - 6y) - 18| = |x² - 6(y+3)|. If we choose δ to be min{1, ε/12}, then for |x|<δ and |y+3|<δ, the inequality |x²| ≤ δ² and |6(y+3)| ≤ 6δ will hold. Hence, |x² - 6(y+3)| ≤ |x²| + 6|y+3| ≤ δ² + 6δ ≤ ε, ensuring continuity at (0, -3).