Chapter 6: Problem 60
Draw the graphs of \(f\) and \(f^{\prime},\) where \(f(x)=1 /\left(1+e^{1 / x}\right)\). Then determine each of the following: (a) \(\lim _{x \rightarrow 0^{+}} f(x)\) (b) \(\lim _{x \rightarrow 0^{-}} f(x)\) (c) \(\lim _{x \rightarrow \pm \infty} f(x)\) (d) \(\lim _{x \rightarrow 0} f^{\prime}(x)\) (e) The maximum and minimum values of \(f\) (if they exist).
Short Answer
Step by step solution
Analyzing the Function f(x)
Calculating Limits as x Approaches 0 from the Right
Calculating Limits as x Approaches 0 from the Left
Calculating Limits as x Approaches Infinity
Finding the Derivative f'(x)
Evaluating Limit of the Derivative at x Approaches 0
Finding Maximum and Minimum Values of f(x)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits
- For \(x \to 0^+\): As \( \frac{1}{x} \to +\infty\), \( e^{1/x} \to +\infty\), turning \(f(x)\) into \(\frac{1}{1 + \infty} = 0\).
- For \(x \to 0^-\): As \( \frac{1}{x} \to -\infty\), \( e^{1/x} \to 0\), making \(f(x)\) approach \(\frac{1}{1 + 0} = 1\).
Continuity
- A function is continuous at a point if the limit of the function as it approaches the point from both directions equals the function's value at that point.
- For \(f(x)\) at \( x = 0\), the right-hand limit is 0 and the left-hand limit is 1. Since these limits differ, \(f(x)\) is not continuous at \( x = 0 \).
Derivative
Calculating the Derivative
Set \(u = 1 + e^{1/x}\). By applying the rules:- The derivative of \(f(x)\) is \( f'(x) = -\frac{1}{u^2} \times u'(x)\).
- For \(u'(x)\), we use \( u'(x) = -\frac{1}{x^2} e^{1/x} \).
- Thus, \( f'(x) = \frac{e^{1/x}}{(1 + e^{1/x})^2 x^2} \).
Graphing Functions
- As \(x \to 0^+\), \(f(x)\) approaches 0, indicated by the graph nearing the x-axis from the right.
- Conversely, approaching from left (* \(x \to 0^-\)) brings \(f(x)\) to 1, making the graph meet the line \(y = 1\).
- As \(x\) extends towards infinity or negative infinity, \(f(x)\) stabilizes at 0.5, seen as an asymptote parallel to \(y = \frac{1}{2}\).