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Problem 59

# Let $$\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x}^{2}+3 \mathrm{y}$$ and let $$\varepsilon>0$$ be given. Prove that $$\mathrm{F}$$ is continuous in the whole plane by finding $$\delta>0$$ such that for $\left|(x, y)-\left(x_{0}, y_{0}\right)\right|<\delta$ $$\left|F(x, y)-F\left(x_{0}, y_{0}\right)\right|<\varepsilon$$ where $x_{0}, y_{0}$ is an arbitrary point in the plane.

Expert verified
For the given function $$F(x, y)=x^2+3y$$, we can find a bound $$\delta>0$$ such that if the distance between $$(x, y)$$ and $$(x_0, y_0)$$ is less than $$\delta$$, then the distance between $$F(x, y)$$ and $$F(x_0, y_0)$$ is less than $$\varepsilon$$. We have found that $$\delta = \frac{\varepsilon}{2 (|x_0| + 1) + 7}$$ satisfies this condition. Therefore, the function $$F(x, y)=x^2+3y$$ is continuous in the whole plane.
See the step by step solution

## Step 1: Analyze the difference between function values

For the given function, $$\mathrm{F}(x, y)=x^2+3y$$, we will first analyze the difference between the function values at $$(x, y)$$ and $$(x_0, y_0)$$: $|F(x, y) - F(x_0, y_0)| = |(x^2 - x_0^2) + 3(y - y_0)|$

## Step 2: Apply triangle inequality

By using triangle inequality, we can say that $$(x^2 - x_0^2) + 3(y - y_0)| \leq |x^2 - x_0^2| + 3|y - y_0|$$. Now, using the identity, $$||a| - |b|| \leq |a - b| \leq |a| + |b|$$, we get, $|x^2 - x_0^2| \leq |x + x_0||x - x_0|$

## Step 3: Bound the difference in function values

Combining the bounds from the previous steps, we get: $|F(x, y) - F(x_0, y_0)| \leq |x + x_0||x - x_0| + 3|y - y_0|$ Notice that we have the terms $$|x - x_0|$$ and $$|y - y_0|$$, which can be related to the distance between $$(x, y)$$ and $$(x_0, y_0)$$. So, let's calculate this distance and bound it by some $$\delta$$.

## Step 4: Calculate the distance between (x, y) and (x_0, y_0)

The distance between $$(x, y)$$ and $$(x_0, y_0)$$ is given by: $|(x, y) - (x_0, y_0)| = \sqrt{(x - x_0)^2 + (y - y_0)^2}$ Now, we need to find a bound for this distance so that it influences the inequality involving the function values.

## Step 5: Find the bound to relate the distance and function values

Let's find $$\delta$$ such that $\delta = \sqrt{(x - x_0)^2 + (y - y_0)^2}$ thus $|(x, y) - (x_0, y_0)| < \delta \Rightarrow |x - x_0|^2 + |y - y_0|^2 < \delta^2$ and also, $|x - x_0| < \delta, |y - y_0| < \delta$ Now, choose $$\delta = \frac{\varepsilon}{2 (|x_0| + 1) + 7}$$. Observe that: $|x + x_0||x - x_0| + 3|y - y_0|$ $\leq (|x_0| + |x - x_0|)|x - x_0| + 3|y - y_0|$ $\leq (|x_0| + \delta)|\delta| + 3\delta$ $\leq (|x_0| + 1)\delta + 3\delta$ $\leq |x_0| \delta + \delta + 3\delta$ $\leq (2 |x_0| + 4)\delta$ $\leq \varepsilon$ Therefore, we have shown that there exists a $$\delta > 0$$ such that for any $$(x, y)$$ with $$|(x, y)-(x_0, y_0)| < \delta$$, $$|F(x, y)-F(x_0, y_0)| < \varepsilon$$. This means the function $$\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x}^{2}+3 \mathrm{y}$$ is continuous in the whole plane.

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