Problem 39 - Calculus: Early Transcendentals

Question

Show that \( f \) is continuous on \( (-\infty, \infty ) \). \( f(x) = \left\\{ \begin{array}{ll} 1 - x^2 & \mbox{if \) x \le 1 \(}\\\ \ln x & \mbox{if \) x > 1 \(} \end{array} \right.\)

Step-by-Step Solution

Answer

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

Step 1: Understanding Continuity

A function \( f \) is continuous at a point \( c \) if \( \lim_{x \to c} f(x) = f(c) \). To show that \( f \) is continuous over \(( -\infty, \infty )\), we need to verify continuity at each point in both piecewise parts and specifically check the point \( x = 1 \) where the definition of the function changes.

Step 2: Continuity for \( x \le 1 \)

For \( x \le 1 \), \( f(x) = 1 - x^2 \), which is a polynomial. Polynomials are continuous everywhere on their domain. Therefore, \( f(x) \) is continuous for all \( x \le 1 \).

Step 3: Continuity for \( x > 1 \)

For \( x > 1 \), \( f(x) = \ln x \). The natural logarithm function \( \ln x \) is continuous for all positive \( x \). Thus, \( f(x) \) is continuous for \( x > 1 \).

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

We must check the limit from the left and right, ensuring they equal \( f(1) \).The left-hand limit as \( x \to 1^- \) for \( f(x) = 1 - x^2 \):\[ \lim_{x \to 1^-} (1 - x^2) = 1 - (1)^2 = 0 \].The right-hand limit as \( x \to 1^+ \) for \( f(x) = \ln x \):\[ \lim_{x \to 1^+} \ln x = \ln 1 = 0 \].Both limits equal \( f(1) = 0 \), so \( f \) is continuous at \( x = 1 \).

Step 5: Conclusion

Since \( f(x) \) is continuous on \( x \le 1 \), \( x > 1 \), and the limits from both sides at \( x = 1 \) are equal, \( f \) is continuous for all \( x \) over \((-\infty, \infty)\).

Key Concepts

Understanding Limits

When we talk about limits, we are discussing what value a function approaches as the input approaches a certain number. Limits are crucial for determining the behavior of a function at specific points, especially where a function's definition changes, like at piecewise boundaries. It helps us analyze whether there is a smooth transition of values without any jumps or gaps.

• The notation \( \lim_{x \to c} f(x) = L \) means as \( x \) gets closer to \( c \), \( f(x) \) approaches \( L \).
• For a piecewise function, checking the limits from both sides of a boundary ensures continuity across that point.
• In this exercise, we specifically check the left-hand and right-hand limits at \( x = 1 \) to ensure the value matches the function's result at that point.

Without equal limits from both sides, the function would be discontinuous at that point.

Polynomials and Continuity

Polynomial functions are mathematical expressions involving sums and products of variables raised to whole number powers, such as \( 1 - x^2 \). These functions are essential because they are inherently continuous over all real numbers. This means we can graph them without lifting our pencil.

• Since polynomials are continuous everywhere in their domain, for piecewise functions involving polynomials, the challenge is just their boundaries with other functions.
• The polynomial part in our function, \( 1-x^2 \), ensures continuity for all \( x \le 1 \), without any additional calculations needed.
• This intrinsic continuity simplifies the analysis, focusing only on points where the piecewise condition changes.

Understanding the polynomial's properties can quickly confirm parts of a function's continuity.

Exploring the Natural Logarithm Function

The natural logarithm function, denoted as \( \ln x \), is another classic mathematical function. It is only defined for positive \( x \), but within this domain, it is continuous. This means if you were plotting \( \ln x \) on a graph, the line would never break.

• The natural logarithm is particularly important because of its applications in growth processes, calculus, and complex numbers.
• For \( \ln x \), continuity implies that the function smoothly approaches values without abrupt shifts or breaks, making it predictable in its realm.
• In this piecewise function, the continuity of \( \ln x \) ensures that for \( x > 1 \), the function transitions seamlessly.

The smooth progression of \( \ln x \) assures us of its role in a continuous piecewise function.

Ensuring Continuity at a Point

To show that a function is continuous at a specific point, such as \( x = 1 \) in this exercise, we need to verify a few conditions using limits.

• First, evaluate the left-hand limit, where \( x \) approaches the point from smaller values. Here, it uses \( f(x) = 1-x^2 \), leading to \( \lim_{x \to 1^-} (1 - x^2) = 0 \).
• Next, evaluate the right-hand limit, where \( x \) approaches the point from larger values. Here, it uses \( f(x) = \ln x \), leading to \( \lim_{x \to 1^+} \ln x = 0 \).
• Finally, check the function's value at that exact point: \( f(1) = 0 \).

All these values must match for the function to be continuous at \( x = 1 \). Without equality, the function has a discontinuity at that point, making it vital to check any piecewise transitions thoroughly.