Piecewise Functions: Evaluate and Graph - College Algebra

1. Consider the piecewise function $k(x) = \begin{cases} |x + 1| & \text{if } x \leq 2 \\ 3x - 5 & \text{if } x > 2 \end{cases}$. What is the range of $k(x)$ for $x \in [-3, 4]$?
   Answer: $[0, 7]$ For $x \in [-3, 2]$, we use $k(x) = |x + 1|$. The vertex of $|x + 1|$ is at $x = -1$, where $k(-1) = 0$. At the endpoints: $k(-3) = |-3 + 1| = 2$ and $k(2) = |2 + 1| = 3$. Since the absolute value function decreases from $x = -3$ to $x = -1$ then increases from $x = -1$ to $x = 2$, the range on this interval is $[0, 3]$. For $x \in (2, 4]$, we use $k(x) = 3x - 5$. This is increasing, so $k(x)$ approaches $3(2) - 5 = 1$ as $x \to 2^+$ and reaches $k(4) = 3(4) - 5 = 7$. The range on this interval is $(1, 7]$. Combining: the overall range is $[0, 3] \cup (1, 7] = [0, 7]$ since these intervals overlap.
2. The piecewise function $s(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x \neq 3 \\ k & \text{if } x = 3 \end{cases}$ can be made continuous by choosing the appropriate value of $k$. However, if $k = 8$, what type of discontinuity occurs at $x = 3$?
   Answer: A removable discontinuity because the limit exists but differs from the function value First, we find the limit as $x \to 3$. For $x \neq 3$: $\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3$. Therefore, $\lim_{x \to 3} s(x) = \lim_{x \to 3} (x + 3) = 6$. Since $s(3) = k = 8$ and $\lim_{x \to 3} s(x) = 6 \neq 8$, there is a discontinuity. It's removable because the limit exists (equals 6) but doesn't equal the function value (8). The discontinuity could be removed by setting $k = 6$.
3. A piecewise function is defined as $g(x) = \begin{cases} 2x + 1 & \text{if } x \leq -1 \\ x^2 & \text{if } -1 < x < 3 \\ 5 - x & \text{if } x \geq 3 \end{cases}$. For which value(s) of $x$ does $g(x) = 4$?
   Answer: $x = 2$ only We need to solve $g(x) = 4$ in each piece. For $x \leq -1$: $2x + 1 = 4$, so $x = 1.5$. But $1.5 > -1$, so this is not valid. For $-1 < x < 3$: $x^2 = 4$, so $x = \pm 2$. Since we need $-1 < x < 3$, only $x = 2$ works ($x = -2$ doesn't satisfy $-2 > -1$). For $x \geq 3$: $5 - x = 4$, so $x = 1$. But $1 < 3$, so this is not valid. Therefore, only $x = 2$ satisfies $g(x) = 4$.
4. A parking garage charges according to the function $C(t) = \begin{cases} 3 & \text{if } 0 < t \leq 2 \\ 3 + 2(t - 2) & \text{if } 2 < t \leq 6 \\ 11 & \text{if } t > 6 \end{cases}$, where $C(t)$ is the cost in dollars and $t$ is time in hours. If someone pays exactly $7, what is the maximum time they could have parked?
   Answer: $4$ hours When you encounter piecewise functions like this parking garage pricing model, you need to determine which piece of the function applies to your situation, then work within that constraint. Since someone paid exactly $7, you need to find where $C(t) = 7$ and determine the maximum possible time. Let's check each piece systematically. For $0 < t \leq 2$: The cost is flat at $3, so this piece can't give us $7. For $2 < t \leq 6$: Here $C(t) = 3 + 2(t - 2)$. Setting this equal to 7: $7 = 3 + 2(t - 2)$ $4 = 2(t - 2)$ $2 = t - 2$ $t = 4$ Since $4$ is in the valid range $2 < t \leq 6$, this works. For $t > 6$: The cost is flat at $11, which exceeds $7, so this piece is impossible. Therefore, the maximum time for $7 is exactly 4 hours. Looking at the wrong answers: Choice A ($5$ hours) would actually cost $3 + 2(5-2) = 9$ dollars, which is too expensive. Choice B ($4.5$ hours) would cost $3 + 2(4.5-2) = 8$ dollars, also too much. Choice C ($3.5$ hours) would cost $3 + 2(3.5-2) = 6$ dollars, which is less than $7 but not the maximum time possible. Study tip: With piecewise functions, always identify which piece applies to your constraint first, then solve within that domain. Don't forget to verify your answer falls within the piece's specified range.
5. Consider $p(x) = \begin{cases} x^2 - 1 & \text{if } x \neq 3 \\ 5 & \text{if } x = 3 \end{cases}$. Which statement correctly describes the behavior of this function?
   Answer: The function has a removable discontinuity at $x = 3$ since $\lim_{x \to 3} p(x) \neq p(3)$ When you encounter a piecewise function like this, you need to analyze continuity by checking three conditions at any point of interest: the function value exists, the limit exists, and they're equal. Let's examine what happens at $x = 3$. First, find $\lim_{x \to 3} p(x)$. Since the limit only cares about values near 3 (not at 3), you use the first piece: $\lim_{x \to 3} (x^2 - 1) = 3^2 - 1 = 8$. Next, find $p(3)$. The function explicitly defines $p(3) = 5$. Since $\lim_{x \to 3} p(x) = 8$ but $p(3) = 5$, the limit and function value don't match. This creates a removable discontinuity—you could "fix" it by redefining $p(3) = 8$. Choice A is wrong because both left and right limits equal 8 (they approach the same value from both sides). Choice B misses the point—having $p(3)$ defined isn't enough for continuity; it must equal the limit. Choice D incorrectly assumes that having both the limit and function value exist guarantees continuity, but they must be equal, not just both exist. Choice C correctly identifies this as a removable discontinuity because the limit exists but doesn't equal the function value. Study tip: For piecewise functions, always check continuity at the boundary points by computing the limit using the appropriate piece, then comparing it to the actual function value at that point.

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