Chapter 1: Problem 31
Give the intervals on which the given function is continuous. $$ g(s)=\ln s $$
Short Answer
Expert verified
The function is continuous for \( s > 0 \) or on the interval \((0, \infty)\).
Step by step solution
01
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 \( s > 0 \).
02
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.
03
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 \( (0, \infty) \).
04
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 \( (0, \infty) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.