Piecewise Functions - Accuplacer Aaf

Let $f(x) = \begin{cases} 2x+1 & \text{if } x < 0 \\ 2 & \text{if } 0 \le x < 2 \\ x-1 & \text{if } x \ge 2 \end{cases}$ . How many distinct solutions does the equation $f(x) = x$ have?
Answer: 1 We solve $f(x)=x$ for each piece. Case 1 ( $x<0$ ): $2x+1=x \implies x=-1$ . This is valid since $-1<0$ . Case 2 ( $0 \le x < 2$ ): $2=x \implies x=2$ . This is invalid since we need $x<2$ for this piece. Case 3 ( $x \ge 2$ ): $x-1=x \implies -1=0$ , which is impossible. Therefore, there is only one solution: $x=-1$ .
Let $f(x) = \begin{cases} x + 3 & \text{if } x < 1 \\ 5 & \text{if } x \ge 1 \end{cases}$ and $g(x) = x^2+1$ . At which positive $x$ -value do the graphs of $f(x)$ and $g(x)$ intersect?
Answer: 2 We must check for intersections on both intervals defined by $f(x)$ . Case 1: $x < 1$ . Set $x + 3 = x^2 + 1$ . This gives $x^2 - x - 2 = 0$ , which factors to $(x-2)(x+1) = 0$ . The solutions are $x = 2$ and $x = -1$ . Only $x=-1$ satisfies the condition $x < 1$ . Case 2: $x \ge 1$ . Set $5 = x^2 + 1$ . This gives $x^2 = 4$ , so $x = 2$ or $x = -2$ . Only $x = 2$ satisfies the condition $x \ge 1$ . The two intersection points are at $x=-1$ and $x=2$ . The question asks for the positive $x$ -value, which is 2.
A water tank's volume changes according to: $V(h) = \begin{cases} 50h & \text{if } 0 \leq h \leq 8 \\ 400 + 75(h-8) & \text{if } 8 < h \leq 12 \\ 700 & \text{if } h > 12 \end{cases}$ where $h$ is height in feet. At what height does the tank's volume first reach 550 cubic feet?
Answer: 10 feet Check each piece to find where V(h) = 550. First piece (0 ≤ h ≤ 8): 50h = 550 gives h = 11, but 11 > 8, so this piece doesn't reach 550. Second piece (8 < h ≤ 12): 400 + 75(h-8) = 550, so 75(h-8) = 150, thus h-8 = 2, so h = 10. Since 8 < 10 ≤ 12, this is valid. Third piece gives V(h) = 700 > 550. Therefore, the tank first reaches 550 cubic feet at h = 10 feet. Choice B incorrectly solves 400 + 75(h-8) = 550 as h = 10.5. Choice C uses wrong arithmetic. Choice D incorrectly assumes the volume cannot be determined.
The function $f(x)$ is defined as $f(x) = \begin{cases} x^2+a & \text{if } x < 1 \\ bx+2 & \text{if } x \ge 1 \end{cases}$ . The function is continuous at $x=1$ and $f(3) = 23$ . What is the value of $a$ ?
Answer: 8 First, use $f(3) = 23$ . Since $3 \ge 1$ , we use the second piece: $b(3) + 2 = 23$ , giving $3b = 21$ , so $b = 7$ . For continuity at $x=1$ , the pieces must be equal: $x^2+a = bx+2$ at $x=1$ . Substituting $x=1$ and $b=7$ : $1^2+a = 7(1)+2$ , which gives $1+a = 9$ , so $a = 8$ .
A function $g(x)$ is defined as: $g(x) = \begin{cases} x^2 - 4 & \text{if } x < 0 \\ 2x + 1 & \text{if } 0 \leq x < 3 \\ -x + 10 & \text{if } x \geq 3 \end{cases}$ What is the value of $g(3) - g(-2) + g(1.5)$ ?
Answer: 11 When you encounter a piecewise function, you need to determine which piece (or rule) applies to each input value based on the given conditions. Let's evaluate each term in $g(3) - g(-2) + g(1.5)$ : For $g(3)$ : Since $x = 3$ and $3 \geq 3$ , we use the third piece: $g(3) = -3 + 10 = 7$ For $g(-2)$ : Since $x = -2$ and $-2 < 0$ , we use the first piece: $g(-2) = (-2)^2 - 4 = 4 - 4 = 0$ For $g(1.5)$ : Since $x = 1.5$ and $0 \leq 1.5 < 3$ , we use the second piece: $g(1.5) = 2(1.5) + 1 = 3 + 1 = 4$ Therefore: $g(3) - g(-2) + g(1.5) = 7 - 0 + 4 = 11$ Choice A (7) represents just the value of $g(3)$ alone, missing the other terms. Choice B (4) could result from calculating only $g(1.5)$ or from arithmetic errors in combining the terms. Choice C (8) might occur if you miscalculated one of the function values or made an error in the final arithmetic. Choice D (11) correctly accounts for all three function evaluations and their proper combination. Study tip: Always pay careful attention to the boundary conditions in piecewise functions. Notice that $x = 3$ uses the third piece because of the "greater than or equal to" condition ( $x \geq 3$ ), while $x = 0$ would use the second piece. Double-check which inequality symbols include the boundary point.
A parking garage uses this rate structure: $R(t) = \begin{cases} 5 & \text{if } 0 < t \leq 2 \\ 5 + 2\lceil t-2 \rceil & \text{if } t > 2 \end{cases}$ where $t$ is hours parked and $\lceil \cdot \rceil$ is the ceiling function. If someone parks for exactly 4.3 hours, how much more do they pay compared to parking for exactly 2.8 hours?
Answer: $2.00 For 4.3 hours: Since t > 2, use R(t) = 5 + 2⌈4.3-2⌉ = 5 + 2⌈2.3⌉ = 5 + 2(3) = 5 + 6 = $11. For 2.8 hours: Since t > 2, use R(t) = 5 + 2⌈2.8-2⌉ = 5 + 2⌈0.8⌉ = 5 + 2(1) = 5 + 2 = $7. The difference is $11 - $7 = $2. Choice B incorrectly calculates ceiling functions as floor functions. Choice C uses the raw differences without the ceiling function. Choice D makes an error in applying the piecewise conditions.
What is the range of the function $f(x) = \begin{cases} x & \text{if } x < 1 \\ 3 & \text{if } x = 1 \\ (x-1)^2 + 1 & \text{if } x > 1 \end{cases}$ ?
Answer: $(-\infty, 1) \cup (1, \infty) \cup \{3\}$ We analyze the range of each piece. For $x < 1$ , $f(x) = x$ covers all values in the interval $(-\infty, 1)$ . For $x = 1$ , $f(x) = 3$ , so the single value $3$ is in the range. For $x > 1$ , $f(x) = (x-1)^2 + 1$ is a parabola with its vertex at $(1, 1)$ . Since $x > 1$ , this piece covers all values in the interval $(1, \infty)$ . Combining these parts, the total range is $(-\infty, 1) \cup (1, \infty) \cup \{3\}$ .
A delivery service charges based on package weight using this piecewise function: $f(w) = \begin{cases} 8.50 & \text{if } 0 < w \leq 2 \\ 8.50 + 3.25(w-2) & \text{if } 2 < w \leq 10 \\ 34.50 + 4.75(w-10) & \text{if } w > 10 \end{cases}$ where $w$ is weight in pounds. If a customer ships three packages weighing 1.5 lbs, 6 lbs, and 12 lbs respectively, what is the total shipping cost?
Answer: $52.00 For the 1.5 lb package: Since 0 < 1.5 ≤ 2, use f(w) = 8.50, so cost = $8.50. For the 6 lb package: Since 2 < 6 ≤ 10, use f(w) = 8.50 + 3.25(6-2) = 8.50 + 3.25(4) = 8.50 + 13.00 = $21.50. For the 12 lb package: Since 12 > 10, use f(w) = 34.50 + 4.75(12-10) = 34.50 + 4.75(2) = 34.50 + 9.50 = $22.00. Total = 8.50 + 21.50 + 22.00 = $52.00. Choice A incorrectly uses the first piece for all packages. Choice B incorrectly calculates the third piece as 34.50 + 4.75(12) instead of 4.75(12-10). Choice D makes an arithmetic error in the second piece calculation.
A specific job had a total fee of $1100. How many hours did the job take?
Answer: 18 First, check if the fee could be generated by the first rule. Set $50h = 1100$ , which yields $h = 22$ . This is not valid because the rule only applies for $h \le 10$ . So, we must use the second rule. Set $500 + 75(h - 10) = 1100$ . Subtract 500 from both sides: $75(h - 10) = 600$ . Divide by 75: $h - 10 = 8$ . Add 10 to both sides: $h = 18$ . This is a valid solution because $18 > 10$ .
Let $f(x) = \begin{cases} -x-1 & \text{if } x < -1 \\ x+1 & \text{if } x \ge -1 \end{cases}$ and $g(x) = 2$ . What is the value of $f(f(-2)) + g(3)$ ?
Answer: 4 This is a multi-step evaluation. First, find the value of the inner function $f(-2)$ . Since $-2 < -1$ , we use the first piece: $f(-2) = -(-2) - 1 = 2 - 1 = 1$ . Now we must find $f(1)$ . Since $1 \ge -1$ , we use the second piece: $f(1) = 1 + 1 = 2$ . So, $f(f(-2)) = 2$ . Next, evaluate $g(3)$ . The function $g(x)$ is a constant function, so $g(3) = 2$ . Finally, add the results: $f(f(-2)) + g(3) = 2 + 2 = 4$ .
The value of a collectible is modeled by the function $V(t)$ , where $t$ is the number of years after 2010. $V(t) = \begin{cases} 500 + 100t & \text{if } 0 \le t \le 5 \\ 1000(1.04)^{t-5} & \text{if } t > 5 \end{cases}$ . What is the approximate value of the collectible in the year 2020?
Answer: $1217 The year 2020 corresponds to $t = 2020 - 2010 = 10$ . Since $10 > 5$ , we use the second piece: $V(t) = 1000(1.04)^{t-5}$ . Substitute $t = 10$ : $V(10) = 1000(1.04)^{10-5} = 1000(1.04)^5$ . Calculating $(1.04)^5 \approx 1.217$ , so $V(10) \approx 1000(1.217) = \$1217$ . Choice C represents using $t$ instead of $t-5$ in the exponent, choice D uses the linear piece incorrectly, and choice A is the function value at $t=5$ .
A manufacturing company's daily production cost follows: $P(u) = \begin{cases} 1000 + 50u & \text{if } 0 \leq u \leq 40 \\ 3000 + 25u & \text{if } 40 < u \leq 80 \\ 5000 + 10u & \text{if } u > 80 \end{cases}$ where $u$ is units produced. The company wants to minimize cost per unit. Which production level achieves this?
Answer: 100 units When you encounter a piecewise cost function and need to minimize cost per unit, you must calculate the cost per unit (total cost divided by units) at critical points where the function changes behavior. Let's evaluate cost per unit at each boundary and test point. For any production level $u$ , cost per unit equals $\frac{P(u)}{u}$ . At 40 units: $\frac{P(40)}{40} = \frac{1000 + 50(40)}{40} = \frac{3000}{40} = 75$ per unit At 80 units: $\frac{P(80)}{80} = \frac{3000 + 25(80)}{80} = \frac{5000}{80} = 62.5$ per unit At 100 units: $\frac{P(100)}{100} = \frac{5000 + 10(100)}{100} = \frac{6000}{100} = 60$ per unit For choice C (60 units), we use the second piece: $\frac{P(60)}{60} = \frac{3000 + 25(60)}{60} = \frac{4500}{60} = 75$ per unit Choice A (40 units) gives a cost per unit of $75, which is higher than other options. Choice B (80 units) yields $62.50 per unit, better than A but not optimal. Choice C (60 units) also gives $75 per unit, matching A's inefficiency. Choice D (100 units) achieves the lowest cost per unit at $60. The pattern here shows that as production increases beyond 80 units, the fixed cost component becomes spread over more units while the variable cost per unit drops to just $10, creating the most efficient production level. Study tip: With piecewise cost functions, always check the cost per unit at boundary points and test higher production levels—economies of scale often make higher production more efficient per unit.
Let $f(x) = \begin{cases} \sqrt{1-x} & \text{if } x \le 1 \\ -(x-1)^2 & \text{if } x > 1 \end{cases}$ . What is the domain of $f(x)$ ?
Answer: $(-\infty, \infty)$ The domain of a piecewise function is the union of the domains of its pieces. The first piece, $\sqrt{1-x}$ , is defined for $1-x \ge 0$ , which means $x \le 1$ . The function's definition restricts this piece to $x \le 1$ , which is consistent. The second piece, $-(x-1)^2$ , is a polynomial and is defined for all real numbers. The function's definition restricts this piece to $x > 1$ . The total domain is the union of these two intervals: $(-\infty, 1] \cup (1, \infty)$ . This union covers all real numbers, so the domain is $(-\infty, \infty)$ . Choice D is a description of the parts of the domain, but their union is all real numbers.
Consider the function $g(x) = \begin{cases} 2x + 7 & \text{if } x \le -1 \\ x^2 + 4 & \text{if } x > -1 \end{cases}$ . What is the average rate of change of $g(x)$ over the interval $[-3, 2]$ ?
Answer: $\frac{7}{5}$ The average rate of change is given by the formula $\frac{g(b) - g(a)}{b - a}$ . Here, $a = -3$ and $b = 2$ . First, find $g(-3)$ . Since $-3 \le -1$ , use the first piece: $g(-3) = 2(-3) + 7 = 1$ . Next, find $g(2)$ . Since $2 > -1$ , use the second piece: $g(2) = 2^2 + 4 = 8$ . Now, calculate the average rate of change: $\frac{g(2) - g(-3)}{2 - (-3)} = \frac{8 - 1}{2 + 3} = \frac{7}{5}$ .
A company's shipping costs are as follows: a flat fee of $15 for any order weighing up to 5 pounds. For orders weighing more than 5 pounds, the cost is the $15 flat fee plus an additional $2 per pound for the weight exceeding 5 pounds. Which function $C(w)$ correctly models the shipping cost for an order weighing $w$ pounds, where $w > 0$ ?
Answer: $C(w) = \begin{cases} 15 & \text{if } 0 < w \le 5 \\ 2w+5 & \text{if } w > 5 \end{cases}$ For weights $w$ such that $0 < w \le 5$ , the cost is a flat $15. For weights $w > 5$ , the cost is the base $15 plus $2 for each pound over 5. The weight exceeding 5 pounds is $w - 5$ . So the cost for $w > 5$ is $15 + 2(w - 5)$ . Distributing the 2 gives $15 + 2w - 10$ , which simplifies to $2w + 5$ . Therefore, the correct piecewise function is $C(w) = \begin{cases} 15 & \text{if } 0 < w \le 5 \\ 2w+5 & \text{if } w > 5 \end{cases}$ .
Let the function $f(x)$ be defined by $f(x) = \begin{cases} x^2 - 1 & \text{if } x \le 2 \\ 2x + 1 & \text{if } x > 2 \end{cases}$ . What is the value of $2f(2) - f(3)$ ?
Answer: -1 First, evaluate $f(2)$ . Since $2 \le 2$ , we use the first piece: $f(2) = 2^2 - 1 = 4 - 1 = 3$ . Next, evaluate $f(3)$ . Since $3 > 2$ , we use the second piece: $f(3) = 2(3) + 1 = 6 + 1 = 7$ . Finally, substitute these values into the expression: $2f(2) - f(3) = 2(3) - 7 = 6 - 7 = -1$ .
For what value of the constant $c$ is the function $f(x) = \begin{cases} cx + 1 & \text{if } x \le 3 \\ x^2 - c & \text{if } x > 3 \end{cases}$ continuous at $x = 3$ ?
Answer: 2 For the function to be continuous at $x = 3$ , the values of the two pieces must be equal at $x = 3$ . We set the expression for $x \le 3$ equal to the expression for $x > 3$ and substitute $x = 3$ . This gives the equation $c(3) + 1 = (3)^2 - c$ . Simplifying, we get $3c + 1 = 9 - c$ . Adding $c$ to both sides gives $4c + 1 = 9$ . Subtracting 1 gives $4c = 8$ . Dividing by 4 yields $c = 2$ .
A cell phone plan charges according to: $C(m) = \begin{cases} 25 & \text{if } 0 \leq m \leq 500 \\ 25 + 0.10(m-500) & \text{if } 500 < m \leq 2000 \\ 175 + 0.05(m-2000) & \text{if } m > 2000 \end{cases}$ where $m$ is minutes used. What is the average cost per minute for someone who uses exactly 1800 minutes?
Answer: $0.089 per minute For m = 1800 minutes, since 500 < 1800 ≤ 2000, use the second piece: C(1800) = 25 + 0.10(1800-500) = 25 + 0.10(1300) = 25 + 130 = $155. Average cost per minute = $155/1800 = $0.08611... ≈ $0.089 per minute. Choice A incorrectly calculates using the third piece formula. Choice C uses an error in the arithmetic (155/1800). Choice D incorrectly uses just the overage rate of $0.10 without accounting for the base fee.
Let $f(x) = \begin{cases} 2x & \text{if } x \le 2 \\ x+2 & \text{if } x > 2 \end{cases}$ and $g(x) = x^2$ . What is the value of $f(g(-1))$ ?
Answer: 2 This is a composition of functions, $f(g(x))$ , evaluated at $x=-1$ . First, find the value of the inner function, $g(-1)$ . $g(-1) = (-1)^2 = 1$ . Now, substitute this result into the outer function, $f(x)$ . We need to find $f(1)$ . Since $1 \le 2$ , we use the first piece of the definition for $f(x)$ . $f(1) = 2(1) = 2$ .
Which of the following piecewise functions is equivalent to $f(x) = |x - 3| + 2x$ ?
Answer: $f(x) = \begin{cases} x + 3 & \text{if } x < 3 \\ 3x - 3 & \text{if } x \ge 3 \end{cases}$ The absolute value function $|x - 3|$ can be written as a piecewise function: it is $-(x - 3)$ or $3 - x$ when $x < 3$ , and it is $x - 3$ when $x \ge 3$ . We must add $2x$ to each piece. For $x < 3$ , $f(x) = (3 - x) + 2x = x + 3$ . For $x \ge 3$ , $f(x) = (x - 3) + 2x = 3x - 3$ . Combining these gives the correct piecewise definition.

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