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 function \( f(x) \), we need to consider its two pieces separately based on the defined intervals. 1. For \( x < 2 \), the function is given by \( f(x) = -2x + 3 \). This is a linear function with a slope of -2, which means it will decrease as \( x \) increases. The y-intercept occurs at \( (0, 3) \), and when \( x = 2 \), we'll find \( f(2) = -2(2) + 3 = -1 \). However, since \( x < 2 \), we will not include this point in the graph. 2. For \( x \geq 2 \), the function is defined as \( f(x) = x - 3 \). This is also a linear function but with a slope of 1, which means it will increase as \( x \) increases. When \( x = 2 \), \( f(2) = 2 - 3 = -1 \), and this is included in the graph. Now, plotting these two pieces on a coordinate axis, the line for \( x < 2 \) is decreasing and approaches the point \( (2, -1) \) but does not include it. The line for \( x \geq 2 \) begins at \( (2, -1) \) and increases from there. Next, we check for continuity at the point \( x = 2 \): - We compute the left-hand limit as \( x \) approaches 2: \[\lim_{x \to 2^-} f(x) = -2(2) + 3 = -1\] - The right-hand limit as \( x \) approaches 2 is: \[\lim_{x \to 2^+} f(x) = 2 - 3 = -1\] - Finally, we have \( f(2) = -1\). Since the left-hand limit, right-hand limit, and the value of the function all agree at \( x = 2 \), we conclude that \( f(x) \) is continuous everywhere, including at the junction point \( x = 2 \). In summary: - The graph consists of a decreasing line segment for \( x < 2 \) and an increasing line segment for \( x \geq 2 \). - The function \( f \) is continuous.
