Problem 31 - APEX Calculus

Question

Give the intervals on which the given function is continuous. $$ g(s)=\ln s $$

Step-by-Step Solution

Answer

The function is continuous for \( s > 0 \) or on the interval \((0, \infty)\).

Step 1: Identify the Function and Its Domain

The function given is the natural logarithm function, \( g(s) = \ln s \). The natural logarithm is defined only for positive real numbers. Thus, its domain is \( s > 0 \).

Step 2: Check for Points of Discontinuity

Functions are discontinuous at points where they are not defined or where their definition changes. Since \( g(s) = \ln s \) is not defined for \( s \leq 0 \), check values near 0 from the positive side to confirm continuity. \( g(s) \) is defined and differentiable for all \( s > 0 \), meaning it has no discontinuity points in this interval.

Step 3: Conclusion of Continuity Interval

Since \( g(s) = \ln s \) is continuous wherever it is defined, and it is defined for all \( s > 0 \), it is continuous on the interval \( (0, \infty) \).

Step 4: Verify by Graph Observation (Optional)

The graph of \( g(s) = \ln s \) does not contain any holes or jumps for \( s > 0 \). This further validates that the function is continuous on \( (0, \infty) \).

Key Concepts

Natural Logarithm

The natural logarithm, commonly denoted as \( \ln s \), is a fascinating function in mathematics. It is the inverse of the exponential function \( e^s \), where \( e \approx 2.718 \). This function is only defined for positive values of \( s \), meaning that you can only find the natural logarithm of a positive number. Notice that \( \ln 1 = 0 \), since \( e^0 = 1 \). The natural logarithm grows very slowly compared to polynomial or exponential functions, making it unique.

• The domain of \( \ln s \) is \( s > 0 \).
• It approaches negative infinity as \( s \) approaches 0 from the right.
• The function is continuous and differentiable for all its domain.

Understanding the natural logarithm involves acknowledging that it's undefined for non-positive numbers. This helps explain why its domain, where it behaves nicely, is limited to positive real numbers.

Domain of a Function

When examining any function, understanding its domain is crucial. The domain is essentially all the input values for which the function is defined. For lots of functions, this means the set of all real numbers. However, some functions, like the natural logarithm, have restrictions. For \( g(s) = \ln s \), the domain is restricted to values where \( s > 0 \), because the natural logarithm of zero or negative numbers doesn't exist. This is an important aspect since any value outside this range would make the function undefined.

• The domain dictates where the function graphs over the x-axis.
• Check for values that might cause division by zero, negative logarithms, or other undefined operations.

By ensuring you have identified the correct domain, you can avoid errors in calculations and properly graph the function. This means correctly understanding the limits where this function can be applied.

Discontinuity

Discontinuity in functions is where a function "breaks"—where it's not defined or does not "behave" normally. For \( g(s) = \ln s \), discontinuities occur where the function isn't defined. Since \( \ln s \) is only defined for \( s > 0 \), you might consider whether there is a discontinuity at \( s = 0 \).A function like \( \ln s \) does not include \( s \leq 0 \) in its domain, so there’s no real "jump" or "hole" across \( s > 0 \). This leaves it continuous for all \( s > 0 \). Thus, there is technically no discontinuity in the range of its domain.

• A point of discontinuity means the function isn't continuous at that spot.
• The absence of such points in \( (0, \infty) \) shows the function is nicely behaved there.

By understanding discontinuities, or their absence, you can better predict where a function may have unexpected behaviors. For \( g(s) = \ln s \), knowing it is continuous over \( (0, \infty) \) allows for confident calculations and analysis.