Chapter 2: Problem 39
\(37-39\) Find the numbers at which \(f\) is discontinuous. At which of these numbers is \(f\) continuous from the right, from the left, or neither? Sketch the graph of f. \(f(x)=\left\\{\begin{array}{ll}{x+2} & {\text { if } x<0} \\ {e^{x}} & {\text { if } 0 \leq x \leqslant 1} \\ {2-x} & {\text { if } x>1}\end{array}\right.\)
Short Answer
Step by step solution
Analyze Each Piece of the Function
Check Continuity at x = 0
Check Continuity at x = 1
Sketch the Graph
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuity
- The function is defined at that point.
- The left-hand and right-hand limits at that point exist.
- The left-hand and right-hand limits are equal to each other and to the function's value at that point.
Discontinuity
- At \(x = 0\), the function is discontinuous because the left-hand limit \((2)\) and right-hand limit \((1)\) are not equal. The function has a jump discontinuity here.
- At \(x = 1\), the function is also discontinuous because the left-hand limit \((e)\) and right-hand limit \((1)\) are not equal. There is also a jump discontinuity at this point.
Limits
- If the limits are not equal or don't match the function value, this indicates a discontinuity.
Left-hand Limit
- At \(x = 0\): The left-hand limit is calculated as \(\lim_{x \to 0^-} (x + 2) = 2\).
- At \(x = 1\): The left-hand limit is calculated as \(\lim_{x \to 1^-} e^x = e\).
Right-hand Limit
- At \(x = 0\): The right-hand limit is \(\lim_{x \to 0^+} e^x = 1\).
- At \(x = 1\): The right-hand limit is \(\lim_{x \to 1^+} (2 - x) = 1\).