Chapter 2: Problem 70
Exercises \(69-74:\) Complete the following for \(f(x)\) (a) Determine the domain of \(f\) (b) Evaluate \(f(-2), f(0),\) and \(f(3)\) (c) Graph \(f\) (d) Is \(f\) continuous on its domain? $$ f(x)=\left\\{\begin{array}{ll} 2 x+1 & \text { if }-3 \leq x<0 \\ x-1 & \text { if } \quad 0 \leq x \leq 3 \end{array}\right. $$
Short Answer
Step by step solution
Determine the Domain
Evaluate f(-2)
Evaluate f(0)
Evaluate f(3)
Graph the Function
Analyze Continuity
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of a Function
For this specific function, we have two pieces with different conditions:
- The expression \(2x + 1\) is valid for \(-3 \leq x < 0\).
- The expression \(x - 1\) is valid for \(0 \leq x \leq 3\).
Function Evaluation
In our exercise, we evaluated the function at specific points:
- For \(x = -2\), since it's in the interval \(-3 \leq x < 0\), we use the formula \(2x + 1\). Calculating,the result is \(f(-2) = -3\).
- For \(x = 0\), it falls in the interval \(0 \leq x \leq 3\), so we use \(x - 1\). Hence, \(f(0) = -1\).
- For \(x = 3\), still within \(0 \leq x \leq 3\), use \(x - 1\), giving \(f(3) = 2\).
Graphing Functions
To graph the given function, follow these steps:
- First, graph \(y = 2x + 1\) for \(-3 \leq x < 0\). This includes points from \((-3, -5)\) to almost \((0, 1)\), using an open circle at \((0, 1)\) since \(0\) is not included.
- Next, graph \(y = x - 1\) for \(0 \leq x \leq 3\), connecting points \((0, -1)\) to \((3, 2)\) with solid lines.
Continuity of Functions
In our example, the function is analyzed for continuity at \(x = 0\). Here's what we observe:
- The left limit as \(x\) approaches zero (from \(-\)) using \(2x + 1\) is \(1\).
- The right limit as \(x\) approaches zero (from \(+\)) and the function value using \(x - 1\) is \(-1\).