Suppose that the function \( f \) is defined, for all real numbers, as follows. \[ f(x)=\left\{\begin{array}{cc}-2 x+3 & \text { if } x<2 \\ x-3 & \text { if } x \geq 2\end{array}\right. \] Graph the function \( f \). Then determine whether or not the function is continuous
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To graph the piecewise function \( f(x) \), we’ll break it down into two parts based on the defined intervals: 1. For \( x < 2 \), the function is given by \( f(x) = -2x + 3 \). This is a line with a slope of -2 that intersects the y-axis at (0, 3). For \( x < 2 \), you can plot points like \( f(0) = 3 \), \( f(1) = 1 \), and \( f(2) \) (though it won't be included in this part of the graph). 2. For \( x \geq 2 \), the function is defined as \( f(x) = x - 3 \). This is another line with a slope of 1 that intersects the y-axis at (0, -3). You can plot points like \( f(2) = -1 \), \( f(3) = 0 \), and so on. When you put these two parts together, you will see that the graph consists of two linear pieces: one decreasing line segment for \( x < 2 \) and another increasing line segment starting from \( x = 2 \). To check for continuity, we need to see if the function is continuous at the point where the definitions change, \( x = 2 \): 1. Evaluate \( f(2) = 2 - 3 = -1 \). 2. Check the limit from the left as \( x \) approaches 2: \( \lim_{x \to 2^-} f(x) = -2(2) + 3 = -1 \). 3. Check the limit from the right as \( x \) approaches 2: \( \lim_{x \to 2^+} f(x) = 2 - 3 = -1 \). Since the left limit, right limit, and the function value at \( x = 2 \) are all equal to -1, \( f(x) \) is continuous everywhere, including at \( x = 2 \). Therefore, the function is continuous.
