Question

Prove that the following define continuous functions from \(D\) to R: (i) \(D=\\{x \in \mathbb{R}: x \geq 0\\}\) and \(f(x)=\sqrt{x}\) (ii) \(D=\mathbb{R}\) and \(f(s)=a_{n} s^{n}+a_{n-1} s^{n-1}+\cdots+a_{0}\), where \(a_{0}, a_{1}, \ldots a_{n}\) are fixed real numbers.

Short answer

In summary, we have proved the continuity of both the square root function and the polynomial function of degree \(n\). For the square root function, \(f(x) = \sqrt{x}\) with a domain of \(D = \{x \in \mathbb{R}: x\geq0\}\), it was shown that for every point \(x_0 \in D\) and every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(|x - x_0| < \delta\), then \(|f(x) - f(x_0)| < \epsilon\). For the polynomial function of degree \(n\), \(f(s) = a_{n}s^{n} + a_{n-1} s^{n-1}+\cdots+a_{0}\) with a domain of all real numbers, we used the Mean Value Theorem to show that for every point \(s_0 \in \mathbb{R}\), given any \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(|s - s_0| < \delta\), then \(|f(s) - f(s_0)| < \epsilon\). Thus, both functions have been proved to be continuous in their respective domains.

Step-by-step solution

Prove continuity of the square root function

To prove that the function \(f(x) = \sqrt{x}\) is continuous for \(D = \{x \in \mathbb{R}: x\geq0\}\), we'll use the definition of continuity. Fix an arbitrary point \(x_0 \in D\). We want to show that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(|x - x_0| < \delta\), then \(|f(x) - f(x_0)| < \epsilon\). So, let \(\epsilon > 0\) be given. Consider \(|f(x) - f(x_0)| = |\sqrt{x} - \sqrt{x_0}|\). Notice that we can multiply the numerator and denominator of the expression \(\frac{\sqrt{x} - \sqrt{x_0}}{(\sqrt{x} + \sqrt{x_0})}\) by the conjugate \((\sqrt{x} + \sqrt{x_0})\) to simplify it. This results in: $$|\sqrt{x} - \sqrt{x_0}| = \frac{|x - x_0|}{|(\sqrt{x} + \sqrt{x_0})|}.$$ Now, if we can find a \(\delta\) such that \(|x - x_0| < \delta\) implies \(\frac{|x - x_0|}{|(\sqrt{x} + \sqrt{x_0})|} < \epsilon\), then our proof is complete. For this to happen, we need: $$|x - x_0| < \delta \implies |x - x_0| < \epsilon(\sqrt{x} + \sqrt{x_0}).$$ If we choose \(\delta = \epsilon(\sqrt{x} + \sqrt{x_0})\), then we have: $$|x - x_0| < \delta = \epsilon(\sqrt{x} + \sqrt{x_0}) \implies |f(x) - f(x_0)| < \epsilon.$$ This proves that the function \(f(x) = \sqrt{x}\) is continuous for all \(x \in D\).

Prove the continuity of the polynomial function

In this step, we will prove that a polynomial function of degree \(n\) is continuous for all \(s \in \mathbb{R}\). Using the definition of continuity, we need to show that for every \(s_0 \in \mathbb{R}\) and every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(|s - s_0| < \delta\), then \(|f(s) - f(s_0)| < \epsilon\). So, let \(\epsilon > 0\) be given. \(f(s)-f(s_0)=a_{n}(s^n-s_0^n)+a_{n-1}(s^{n-1}-s_0^{n-1})+\cdots+a_{1}(s-s_0)\). This expression is a sum of products of the form \(s_j^{k}-s_0^{k}\), where \(j=1,2,\dots,n\), and \(k\) ranges from \(1\) to \(n\), by real constants \(a_j\). Let the function \(h(s)=f(s)-f(s_0)\). We use the Mean Value Theorem on \(h(s)\) on the interval \([s, s_0]\). According to the Mean Value theorem, \(\exists c\in [s, s_0]\), such that \(h'(c)=\frac{h(s)-h(s_0)}{s-s_0}\). Since we want \(|h(s)-h(s_0)|<\epsilon, \Rightarrow \left|\frac{h(s)-h(s_0)}{s-s_0}\right|<\frac{\epsilon}{|s-s_0|} \Rightarrow |h'(c)|<\frac{\epsilon}{|s-s_0|}\). Since \(h'(s) = na_n s^{n-1} + (n-1)a_{n-1} s^{n-2} + \cdots + a_1\), by choosing some \(\delta\leq\frac{\epsilon}{n\max_j|a_j|}\), we can ensure that \(|h(s)-h(s_0)|<\epsilon\). Thus, the polynomial function of degree \(n\) is continuous for all \(s\in \mathbb{R}\).

Key concepts

Square Root Continuity

When students tackle calculus problems, understanding the continuity of various functions is crucial. The square root function, denoted as \(f(x) = \sqrt{x}\), is one such function students often encounter. To grasp the continuity of the square root function, it's important to know that it's defined for all non-negative real numbers, \(D = \{x \in \mathbb{R}: x \geq 0\}\).

Here's a simplified view of the continuity proof for the square root function. To begin, we pick any non-negative number, \(x_0\). We're tasked with proving that as \(x\) gets infinitesimally close to \(x_0\), the function's value at \(x\), \(f(x)\), will get arbitrarily close to \(f(x_0)\). In technical terms, for every \(\epsilon > 0\), we need to find a \(\delta > 0\) such that if \(\left| x - x_0 \right| < \delta\), then \(\left| f(x) - f(x_0) \right| < \epsilon\).

Utilizing the property that the difference between two square roots can be expressed as a single fraction by multiplying and dividing by their conjugate yields a handy formula that demonstrates continuity for all \(x \geq 0\). This understanding ensures that students can justify why the square root function doesn't suddenly leap to unexpected values and behaves 'nicely' in its domain.

Polynomial Function Continuity

Polynomials are another fundamental class of functions in mathematics, noted for their simplicity and continuity. A polynomial function of degree \(n\) takes the form \(f(s) = a_n s^n + a_{n-1} s^{n-1} + \cdots + a_0\), where \(a_0, a_1, \ldots, a_n\) are real constants.

To affirm the continuity of polynomial functions across all real numbers, we employ a similar continuity test as with the square root function, ensuring that the values of the function at points closely surrounding any given real number form a tight cluster around the function value at that number. This is confirmed by the fact that manipulating a polynomial's terms \(s^k\) does not allow for any sudden jumps or breaks in the function's graph.

This consistent behavior means that polynomials are continuous everywhere, a fact that brings a level of predictability when working with such functions. Whether dealing with linear, quadratic, cubic, or even higher degree polynomials, the continuous nature remains steadfast, allowing for smooth graphs without breaks or holes.

Proof of Continuity

Proofs of continuity are foundational exercises in calculus that ensure students comprehend how a function behaves near a specific point. A function is considered continuous at a point if three conditions are met: the function is defined at the point, the limit of the function exists as it approaches the point, and the limit equals the actual value of the function at that point.

To prove continuity, a common approach involves using limits. In essence, one must show that for every arbitrarily small distance \(\epsilon\) from the function value \(f(x_0)\) at a certain point \(x_0\), there exists a corresponding zone around \(x_0\), defined by \(\delta\), such that any \(x\) within this \(\delta-zone\) results in a \(f(x)\) value inside the \(\epsilon-tolerance\). If such a \(\delta\) can always be found no matter how small \(\epsilon\) is chosen, the function is continuous at \(x_0\).

While this concept is simple in theory, applying it to diverse functions requires careful analysis of the function's behavior, exploring its limits, and sometimes needing advanced calculus tools such as derivatives, which lead us to the next topic: the Mean Value Theorem.

Mean Value Theorem

The Mean Value Theorem (MVT) is a cornerstone of calculus that captures the intuitive idea that on any smooth journey from one point to another, there must be at least one moment where you are traveling at the average speed of the entire trip. In mathematical language, if you have a continuous function \(f\) on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there is at least one point \(c\) in \((a, b)\) where \(f'(c)\), the derivative of \(f\) at \(c\), equals the function's average rate of change over the interval from \(a\) to \(b\).

Understanding the MVT aids in grasping the behavior of differentiable functions and proving various properties, including the continuity of polynomials as seen in earlier sections. It serves as the linchpin for further results in differential calculus and assures that the behavior of functions within an interval can be captured by looking at just one point, at least temporarily. This elegant theorem not only secures a critical aspect of function behavior but also paves the way for further exploration in the realms of calculus.