Piecewise Functions - Algebra 3

1. An electric company charges $\$0.12$ per kWh for the first 500 kWh, $\$0.15$ per kWh for the next 500 kWh, and $\$0.20$ per kWh for any usage above 1000 kWh. What is the total charge for using 1200 kWh?
   Answer: $\$175$ Whenever you see a tiered pricing question, your first instinct should be to break the usage into blocks and apply each rate only to the kWh inside that block. Don't just multiply total usage by one number. For 1200 kWh, split it: the first 500 kWh cost $500 \times 0.12 = \$60$. The next 500 kWh cost $500 \times 0.15 = \$75$. The remaining 200 kWh are above 1000, so they cost $200 \times 0.20 = \$40$. Adding the three blocks gives $60 + 75 + 40 = \$175$. Each wrong answer comes from a flawed shortcut. $144 treats all 1200 kWh at the lowest rate of $0.12, ignoring the higher-tier charges. $190 incorrectly charges the first 1000 kWh at $0.15, as if the $0.15 rate applied to the whole first block rather than only the second 500. $240 charges all 1200 kWh at the highest rate of $0.20, ignoring the cheaper introductory tiers entirely. Your study tip: for any "piecewise" or tiered charge problem, draw mental boundaries at the thresholds, then calculate each slab separately before adding. Check that the middle rate only touches the middle block, never the first block. That block-by-block habit will save you on tax, utility, and discount questions too.
2. Let $f(x)=\begin{cases}2x^2-1,&x<2\\\\kx+3,&x\ge2\end{cases}$. For what value of $k$ is $f$ continuous at $x=2$?
   Answer: $2$ Whenever you see a piecewise function and a question about continuity at the break point, your job is to make the two pieces meet at that x-value. Here the switch is at $x=2$. From the left, $f(x)=2x^2-1$, so the left-hand limit is: $$\lim_{x\to2^-}(2x^2-1)=2(2)^2-1=8-1=7.$$ For $x\ge2$, the function is $kx+3$, so at $x=2$ we have: $$f(2)=k(2)+3=2k+3.$$ Since the second piece is continuous for $x\ge2$, this is also the right-hand limit. Continuity requires the left-hand limit, right-hand limit, and actual value to all agree: $$7=2k+3.$$ Subtract $3$ from both sides: $4=2k$, so $k=2$. If $k=5$, then $f(2)=10+3=13$, leaving a huge jump from $7$. If $k=\frac72$, then $f(2)=7+3=10$, not $7$; this choice traps you if you forget the $+3$ and just solve $2k=7$. If $k=3$, then $f(2)=6+3=9$, which also fails to match the left-hand limit of $7$. Only $k=2$ makes both sides meet at the same value. adenAs your final check, plug $k=2$ back in: $2(2)+3=7$, exactly matching the left-side limit. On continuity questions, always write down the one-sided limit at the break point, then set it equal to the function's value there — and don't drop any constant from the piecewise formula.
3. Which piecewise-defined function is equivalent to $f(x)=x|x-2|$ for all real $x$?
   Answer: $f(x)=\begin{cases}2x-x^2,&x<2\\x^2-2x,&x\ge2\end{cases}$ If $x<2$, then $|x-2|=2-x$, so $f(x)=x(2-x)=2x-x^2$. If $x\ge2$, then $|x-2|=x-2$, so $f(x)=x(x-2)=x^2-2x$. The form with $2x-x^2$ on $x<2$ and $x^2-2x$ on $x\ge2$ is correct. The other forms use the first-branch expression in the wrong interval, reverse the sign in the second branch, or take an absolute value that changes the sign for negative $x$.
4. Let $f(x)=\begin{cases}3x+2,&x<-1\\\\x^2-4,&-1\le x<2\\\\5-x,&x\ge2\end{cases}$. What is the value of $f(f(-1))$?
   Answer: $-7$ Whenever you see a piecewise function inside another function, the key is to work step by step from the inside out. Composite notation $f(f(-1))$ means plug $-1$ into $f$ first, then take that output and plug it into $f$ again. . First, evaluate the inner value: $f(-1)$. Because $-1$ satisfies $-1 \le x < 2$, use the middle rule: $f(-1)=(-1)^2-4=1-4=-3$. .Now take that result, $-3$, and evaluate $f(-3)$. Since $-3<-1$, use the first rule: $f(-3)=3(-3)+2=-9+2=-7$. Therefore, $f(f(-1))=-7$. . The choice $-3$ is the result of stopping after the first evaluation, confusing the inner value with the final composite value. The choice $7$ could come from flipping both signs when computing the first rule, writing $3(3)-2=7$, which treats the input as positive and subtracts instead of adding. Similarly, $11$ likely comes from computing $3(3)+2=11$, using a positive input for a negative piece — a sign error on the domain condition. And if you mistakenly think $-3$ falls in the middle rule, you'd get $(-3)^2-4=5$,which isn't listed; among the options, $11$ may also arise from $9+2$ without considering the input sign at all. . A good study habit for piecewise function questions: always write down which piece applies at each step, and check the input against the domain intervals before substituting. Evaluating inside-out prevents the most common trap of mixing up the inner value with the final answer.
5. Let $f(x)=\begin{cases}x^2,&x<0\\\\2x+1,&0\le x<3\\\\6-x,&x\ge3\end{cases}$. How many real numbers $x$ satisfy $f(x)=4$?
   Answer: $2$ When you see a piecewise function and are asked how many inputs satisfy an equation, treat each piece as its own mini-problem—but check its domain. For $x<0$, set $x^2=4$. The solutions are $x=2$ and $x=-2$, but only $x=-2$ lies in the domain $x<0$. So this branch contributes 1. For $0\le x<3$, set $2x+1=4$. Then $2x=3$, so $x=1.5$, which is in the domain. This branch contributes 1. For $x\ge3$, set $6-x=4$. Then $x=2$, but $2$ is not in $x\ge3$, so this branch contributes 0. Thus the real numbers satisfying $f(x)=4$ are $x=-2$ and $x=1.5$: exactly 2 solutions. Why the other choices are traps: "1" misses one valid branch, often by forgetting that the first branch can produce a negative solution. "3" often adds $x=2$ from the third piece without checking the domain, or counts the excluded $x=2$ from the first branch. "4" typically counts $x=2$ twice—from both $x^2=4$ and $6-x=4$—while also including the two valid solutions; but $x=2$ is excluded by both domains. Study tip: after solving each piece, always verify the solution against that piece's domain restriction. The domain is the real trap in piecewise-function equation questions.
6. A piecewise-linear function $f$ is described as follows. For $x\le-2$, the graph is the horizontal line through $(-3,4)$. For $-2<x<1$, the graph is a line of slope $-2$ passing through $(0,2)$. For $x\ge1$, the graph is a line of slope $1$ passing through $(3,5)$. Which formula defines $f$?
   Answer: $f(x)=\begin{cases}4,&x\le-2\\-2x+2,&-2<x<1\\x+2,&x\ge1\end{cases}$ The horizontal line through $(-3,4)$ is $y=4$ on $x\le-2$. The line with slope $-2$ through $(0,2)$ is $y=-2x+2$. The line with slope $1$ through $(3,5)$ is $y=x+2$. The other options use the wrong slope, the wrong intercept, or the wrong interval for the horizontal piece.
7. Let $f(x)=\begin{cases}x+2,&x<1\\3,&1\le x<3\\4-x,&x\ge3\end{cases}$. Which statement correctly describes the graph of $f$?
   Answer: At $x=3$, there is an open circle at $(3,3)$ and a closed circle at $(3,1)$. Whenever you see a piecewise function, your job is to check each boundary point carefully: plug the boundary into the neighboring rules and see which rule actually includes that $x$-value. A closed circle means the point is part of the graph; an open circle means the rule approaches that value but does not include it. At $x=3$, look at the two pieces around it. The rule $3$ applies for $1\le x<3$, so it gives the value $3$ but does not include $x=3$. That means there is an open circle at $(3,3)$. The rule $4-x$ applies for $x\ge3$, so it gives $4-3=1$ and does include $x=3$. That means there is a closed circle at $(3,1)$. So the correct description is: open circle at $(3,3)$, closed circle at $(3,1)$. Now the wrong answers. Saying "every real $x$ appears in one of the pieces" is not enough for continuity—continuity requires the two sides to meet at the same point, and here they jump. Saying there are open circles at both $(3,3)$ and $(3,1)$ is wrong because the third piece includes $x=3$, so its endpoint is closed. The statement about $x=1$—that there is an open circle at $(1,3)$ because the first piece excludes $1$—misses that the second piece $3$ applies for $1\le x<3$, so $x=1$ is actually included, giving a closed circle at $(1,3)$. On this exam, always mark boundary points by checking each adjacent inequality: $\le$ or $\ge$ means closed, $<$ or $>$ means open. That simple check prevents most piecewise-function mistakes.
8. Let $$f(x)=\begin{cases}x^2+2,&x\le-1\\3,&-1<x<2\\2x-4,&x\ge2.\end{cases}$$ Which statement about $f$ is true?
   Answer: The range of $f$ is $[0,\infty)$. For $x\le-1$, $x^2+2\ge3$, so that piece has range $[3,\infty)$. The middle piece is the single value $3$. For $x\ge2$, $2x-4\ge0$, giving the interval $[0,\infty)$. The union is $[0,\infty)$. The interval $[3,\infty)$ ignores the last piece, $(0,\infty)$ incorrectly excludes $0$, and the stated domain omits the middle piece.
9. Let $$f(x)=\begin{cases}2x+6,&x<-2\\4-x^2,&-2\le x<3\\x-5,&x\ge3.\end{cases}$$ For which real values of $x$ is $f(x)<0$?
   Answer: $x<-3$ or $2<x<5$ For $x<-2$, $2x+6<0$ gives $x<-3$. For $-2\le x<3$, $4-x^2<0$ gives $x<-2$ or $x>2$, so only $2<x<3$ is valid. For $x\ge3$, $x-5<0$ gives $3\le x<5$. Combining the last two intervals gives $2<x<5$, so the full solution is $x<-3$ or $2<x<5$.
10. Let $$C(w)=\begin{cases}5,&0<w\le2\\5+2(w-2),&2<w\le8\\17,&8<w\le12\\17+3(w-12),&w>12\end{cases}$$ dollars be the cost to ship a package weighing $w$ pounds. If a package costs exactly 23 dollars to ship, which of the following could be its weight?
    Answer: $14$ lb The only branch that can reach 23 dollars is the final one. Solving $17+3(w-12)=23$ gives $3(w-12)=6$, so $w=14$, which is valid because $w>12$. The value $11$ comes from solving $5+2(w-2)=23$, but $11$ is not in the interval $2<w\le8$.
11. Let $$f(x)=\begin{cases}x^2-8,&x<-2\\|x+1|-3,&-2\le x<2\\10-3x,&x\ge2.\end{cases}$$ What is the value of $f(f(f(-3)))$?
    Answer: $-3$ First $f(-3)=(-3)^2-8=1$. Since $1$ is in the middle interval, $f(1)=|1+1|-3=-1$. Then $f(-1)=|-1+1|-3=-3$. The value $-1$ is $f(f(-3)))$, so it comes from stopping after two applications; $1$ is $f(-3)$, one application; $7$ comes from applying the third formula to $x=1$.
12. Let $$f(x)=\begin{cases}2x+1,&x<0\\x^2-1,&x\ge0.\end{cases}$$ Define $g(x)=f(x-2)+3$. Which formula defines $g$?
    Answer: $g(x)=\begin{cases}2x,&x<2\\x^2-4x+6,&x\ge2\end{cases}$ Substituting $x-2$ shifts the branch condition: $x-2<0$ becomes $x<2$, and $x-2\ge0$ becomes $x\ge2$. For $x<2$, $f(x-2)+3=2(x-2)+1+3=2x$. For $x\ge2$, $f(x-2)+3=(x-2)^2-1+3=x^2-4x+6$.
13. Let $$f(x)=\begin{cases}x^2-3,&x<0\\2x+a,&x\ge0.\end{cases}$$ For which values of $a$ does the equation $f(x)=0$ have exactly two distinct real solutions?
    Answer: $a\le0$ In the first branch, $x^2-3=0$ always contributes exactly one valid solution, $x=-\sqrt3$. In the second branch, the root is $x=-\frac{a}{2}$, and it is valid exactly when $-\frac{a}{2}\ge0$, so $a\le0$. Thus there are exactly two distinct real solutions precisely when $a\le0$. The option $a<0$ incorrectly excludes the valid boundary value $a=0$, and $a\ge0$ includes values for which the second branch contributes no root.
14. Let $f(x)=\begin{cases}2x+1,&x<\frac12\\\\x^2,&x\ge\frac12\end{cases}$. What is the value of $f\left(\frac12\right)-f\left(-\frac12\right)$?
    Answer: $\frac14$ Whenever you see a piecewise function, your first job is to locate each input on the number line and decide which branch applies. Here the function changes at $x=\frac12$. For $f\left(\frac12\right)), the condition is$x\ge\frac12$, so you must use the second branch:$f\left(\frac12\right)=\left(\frac12\right)^2=\frac14$. For$f\left(-\frac12\right)), since $-\frac12<\frac12$, you use the first branch: $f\left(-\frac12\right)=2\left(-\frac12\right)+1=0$. Thus the difference is $\frac14-0=\frac14$, so the correct choice is $\frac14$. The trap choices come from misassigning branches. The choice $2$ results from incorrectly using $2x+1$ at $x=\frac12$ and forgetting that the "greater than or equal" sign sends that input to the quadratic piece. The choice $0$ can happen if you evaluate both inputs with the quadratic branch: $f\left(-\frac12\right)=\frac14$ and $f\left(\frac12\right)=\frac14$, then subtract to get $0$; but $-\frac12$ is less than $\frac12$, so it belongs to the linear piece. The choice $\frac74$ comes from mixing both mistakes: using the linear branch at $\frac12$ to get $2$, using the quadratic branch at $-\frac12$ to get $\frac14$, and computing $2-\frac14$. In every case, the error is ignoring the inequality signs. boundary condition. On future piecewise questions, circle the inequalities before plugging in. Ask: Which side is inclusive? The boundary $x=\frac12$ belongs to the quadratic branch because of the "$x\ge\frac12$" sign, so evaluate each input on its correct side before subtracting. This habit will save you on similar exam questions.
15. Let $$f(x)=\begin{cases}x+3,&x<-2\\-2x+1,&-2\le x<1\\x^2-2,&x\ge1.\end{cases}$$ Which statement correctly describes the graph of $f$ at the boundaries of the branches?
    Answer: The graph has a jump at $x=-2$ and is continuous at $x=1$. As $x\to-2^-$, the first formula approaches $1$, but $f(-2)=-2(-2)+1=5$, so there is a jump at $x=-2$. As $x\to1^-$, $-2x+1\to-1$, and $f(1)=1^2-2=-1$, so the graph is continuous at $x=1$. Therefore the graph jumps only at $x=-2$.
16. A store applies a 15% discount at checkout to any item whose original price is over $\$50$. Items priced at $\$50$ or less are not discounted. If the sale price of an item was $\$60$, what was its original price?
    Answer: $\$70.59$ When you see a discount problem, remember that a 15% discount means the customer pays 85% of the original price — not that you add 15% to the sale price to get back to the original. Here, the sale price was $60, and since an item must be over $50 to receive the discount, the original price must be the unknown that was reduced by 15%. So set up: $0.85 \times \text{original} = 60$. Divide both sides by 0.85: $\text{original} = 60 \div 0.85 = 70.588\ldots$, which rounds to $70.59. This is over $50, so the discount correctly applies. The other choices each come from a common misunderstanding. $69.00 is found by computing $60 \times 1.15$, adding 15% of the sale price back to the sale price; but 15% off $69 would be $58.65, not $60. $400.00 treats the $60 as if it were the 15% discount amount itself, since $60 \div 0.15 = 400$; however, $60 is the price paid, not the amount saved. $51.00 comes from taking 15% off $60, i.e. $60 \times 0.85 = 51$, which mistakenly treats the sale price as the original price; if the original were $51, the discount would apply, but the sale price would be $43.35. Strategy: in any percent decrease problem, identify the "whole" (original price) and the "part" (final price). Use $\text{final} = \text{original} \times (1 - \text{percent as decimal})$, and divide when you know the final amount.
17. Let $f(x)=\begin{cases}x+1,&x<-1\\\\x^2,&-1\le x<2\\\\4-x,&x\ge2\end{cases}$. What is the range of $f$?
    Answer: $(-\infty,4)$ When you see a piecewise function, the range is the union of the outputs each piece produces on its own domain. Think of each branch as a separate "machine" with its own input interval, then combine all possible $y$-values. For $x<-1$, the rule $x+1$ produces every value below $0$, because as $x$ approaches $-1$ from the left, $x+1$ approaches $0$ but never reaches it, and as $x$ goes to $-\infty$, so does $x+1$. This branch contributes $(-\infty,0)$. For $-1\le x<2$, the rule $x^2$ gives values from $0$ up to but not including $4$: $0$ is reached at $x=0$, and $4$ is approached as $x$ gets close to $2$, but $2$ is not included. This branch contributes $[0,4)$. For $x\ge2$, the rule $4-x$ gives $2$ at $x=2$, then decreases without bound, contributing $(-\infty,2]$. Unioning these gives $(-\infty,0)\cup[0,4)\cup(-\infty,2]=(-\infty,4)$. The choice $(-\infty,4]$ is wrong because $4$ is never actually produced. The choice $[0,4)$ is wrong because it ignores all negative outputs from the first and third branches. The choice $(-\infty,\infty)$ is wrong because the function never produces any value $4$ or larger. Study tip: for piecewise range questions, test the endpoints and the behavior at the extremes of each interval, then union the results carefully. Watch especially for open versus closed endpoints — that's where traps live.
18. Let $$f(x)=\begin{cases}x^2-3,&x<-1\\ax+b,&-1\le x<2\\5x-6,&x\ge2.\end{cases}$$ If $f$ is continuous at both $x=-1$ and $x=2$, what is $a+b$?
    Answer: $2$ At $x=-1$, the first piece gives $(-1)^2-3=-2$, so $-a+b=-2$. At $x=2$, the left limit of the middle piece is $2a+b$, and the third piece gives $5(2)-6=4$, so $2a+b=4$. Solving gives $a=2$ and $b=0$, hence $a+b=2$. The value $-2$ is the matching value at $x=-1$, $0$ is $b$, and $4$ is the matching value at $x=2$.
19. Let $$f(x)=\begin{cases}x^2,&x\le1\\a-x,&x>1.\end{cases}$$ For which values of $a$ is the range of $f$ equal to all real numbers?
    Answer: $a\ge1$ For $x\le1$, $x^2$ has range $[0,\infty)$. For $x>1$, the linear expression $a-x$ has range $(-\infty,a-1)$. To cover all real numbers, the second piece must reach every value below $0$, so we need $a-1\ge0$, or $a\ge1$. The boundary $a=1$ works because the first piece supplies $0$.
20. A gym charges a one-time $\$50$ enrollment fee, $\$20$ per month for the first 6 months, and $\$35$ per month after that. Which function gives the total cost $C(m)$ after $m$ months?
    Answer: $C(m)=\begin{cases}50+20m,&0\le m\le6\\\\35m-40,&m>6\end{cases}$ When you see a cost scenario with different rates after a certain time, model it as a piecewise function and compute the cumulative cost at the switch point. For the first 6 months, the total cost is $50+20m$. At $m=6$, that is $50+120=170$. For $m>6$, you have already paid the enrollment fee and 6 months at $20, so the cost is $170+35(m-6)=170+35m-210=35m-40$. Thus the correct piecewise form is $50+20m$ for $0\le m\le6$, and $35m-40$ for $m>6$. The option using $50+35m$ for $m>6$ charges $35 for all months; you must subtract the $90 overcharge from the first 6 months, giving $50+35m-90=35m-40$. The option using $170+35m$ for $m>6$ correctly adds the first-six-month total but then also charges $35 for those same months, double-counting them. The option using $120+35m$ omits the $50 enrollment fee and still charges $35 for every month; it also fails to include $m=6$. A reliable strategy: write the second piece as $C(6)+35(m-6)$, where $C(6)$ is the total paid through the switch, rather than just $C(6)+35m$. This prevents both double-counting and missing the initial fee.

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