Problem 35 - Calculus Early Transcendentals

Question

Show that \(f\) is continuous on \((-\infty, \infty)\) \(f(x)=\left\\{\begin{array}{ll}{x^{2}} & {\text { if } x<1} \\ {\sqrt{x}} & {\text { if } x \geqslant 1}\end{array}\right.\)

Step-by-Step Solution

Answer

The function \( f \) is continuous on \(( -\infty, \infty )\).

Step 1: Understand the Form of the Function

The function \( f(x) \) is a piecewise function with two cases: \( f(x) = x^2 \) for \( x < 1 \) and \( f(x) = \sqrt{x} \) for \( x \geq 1 \). This means that the function is defined using different expressions based on the value of \( x \).

Step 2: Check Continuity for \( x < 1 \)

For \( x < 1 \), the function \( f(x) = x^2 \) is a polynomial function, which is inherently continuous on its domain. Therefore, \( f(x) \) is continuous for all \( x < 1 \).

Step 3: Check Continuity for \( x \geq 1 \)

For \( x \geq 1 \), the function \( f(x) = \sqrt{x} \) is a square root function, which is continuous for \( x \geq 1 \). Therefore, \( f(x) \) is continuous for \( x \geq 1 \).

Step 4: Verify Continuity at \( x = 1 \)

To confirm continuity at \( x = 1 \), check if the limits from both sides exist and are equal to the function value at \( x = 1 \).Calculate \( \lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} x^2 = 1^2 = 1 \).Calculate \( \lim_{{x \to 1^+}} f(x) = \lim_{{x \to 1^+}} \sqrt{x} = \sqrt{1} = 1 \).Since the left-hand limit, right-hand limit, and \( f(1) = \sqrt{1} = 1 \) are all equal, \( f \) is continuous at \( x = 1 \).

Step 5: Conclude the Continuity of the Function

Since the function is continuous for \( x < 1 \), \( x \geq 1 \), and at the boundary point \( x = 1 \), we conclude that \( f(x) \) is continuous on the entire real line \(( -\infty, \infty )\).

Key Concepts

Piecewise Functions

Piecewise functions are functions that are defined by multiple sub-functions, each applying to a certain interval of the main function's domain. For our function, \(f(x)\), it switches between two expressions based on the value of \(x\).

Understanding piecewise functions involves:

• Identifying intervals and which expression applies to each interval
• Evaluating how these expressions change at the boundaries where these pieces meet

In the example given, the expression \(x^2\) applies for \(x < 1\), while \(\sqrt{x}\) takes over for \(x \geq 1\). Understanding this structure is crucial for analyzing continuity and ensuring that no breaks or jumps occur where the pieces connect.

Polynomial Functions

Polynomial functions, like \(x^2\) in our piecewise example, are functions represented by polynomials. These functions are inherently continuous on their entire domain. A polynomial function can be expressed as:

\[ p(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \]

Polynomials are smooth and continuous because they involve only operations of addition, subtraction, and multiplication of variables with constants. There are no gaps or jumps in their graphs.

This inherent continuity is particularly useful when analyzing segments of piecewise functions, allowing us to confirm continuity easily for intervals where polynomial expressions are used.

Square Root Functions

Square root functions, such as \(\sqrt{x}\), involve the square root operation and are defined for non-negative values of \(x\). They are continuous over their domain, which typically includes \([0, \infty)\). Understanding square root functions means recognizing:

• They start at \(x=0\) and extend infinitely
• There are no negative outputs
• The function slowly increases, flattening out as \(x\) becomes very large

In our example, the square root function covers the interval \(x \geq 1\), ensuring continuous behavior from that point onward. This characteristic ensures no disruptions in the function's graph, adding to the overall continuity of the piecewise function on its defined intervals.

Limits and Continuity

A fundamental concept in calculus is limits and continuity. Continuity at a point means that a function is uninterrupted or unbroken at that point. To determine continuity, you check:

• Existence of the function at the point
• Existence of the limit as \(x\) approaches the point from both sides
• Equality of the value of the function and the limits from both sides at that point

For our piecewise function \(f(x)\), we especially need to focus on \(x = 1\), where the expression changes. Here, both the left and right hand limits equal \(f(1)\) as calculated:

\( \lim_{{x \to 1^-}} x^2 = 1 \) and \( \lim_{{x \to 1^+}} \sqrt{x} = 1 \) confirm that all are equivalent, proving continuity at that point.

Understanding and applying these concepts ensures a function like \(f(x)\) is indeed continuous across its entire domain \((-\infty, \infty)\).