Algebra 3 Quiz: Domain And Range
12 questions · exam conditions
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Domain And RangeQuestion 1 of 12

What are the domain and range of f(x)=ln⁡(x2−1)f(x)=\ln(x^2-1)?

Domain: (−∞,−1)∪(1,∞)(-\infty,-1)\cup(1,\infty); Range: (−∞,∞)(-\infty,\infty)
Domain: (−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty); Range: (−∞,∞)(-\infty,\infty)
Domain: (−∞,−1)∪(1,∞)(-\infty,-1)\cup(1,\infty); Range: (0,∞)(0,\infty)
Domain: (−1,1)(-1,1); Range: (−∞,∞)(-\infty,\infty)
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Algebra 3 Quiz

Algebra 3 Quiz: Domain And Range

Practice Domain And Range in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Domain And Range, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What are the domain and range of f(x)=ln⁡(x2−1)f(x)=\ln(x^2-1)?

  1. Domain: (−∞,−1)∪(1,∞)(-\infty,-1)\cup(1,\infty); Range: (−∞,∞)(-\infty,\infty) (correct answer)
  2. Domain: (−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty); Range: (−∞,∞)(-\infty,\infty)
  3. Domain: (−∞,−1)∪(1,∞)(-\infty,-1)\cup(1,\infty); Range: (0,∞)(0,\infty)
  4. Domain: (−1,1)(-1,1); Range: (−∞,∞)(-\infty,\infty)
Explanation: When you see a logarithmic function, the first thing to check is always the domain: the expression inside the logarithm must be strictly positive. Here, that means x2−1>0x^2-1>0. Solving x2>1x^2>1 gives x<−1x<-1 or x>1x>1, so the domain is (−∞,−1)∪(1,∞)(-\infty,-1)\cup(1,\infty). Next, ask what values the inside can produce: as xx moves through that domain, x2−1x^2-1 can take every positive number, from just above 00 all the way to ∞\infty. Since ln⁡(t)\ln(t) can output any real number when t>0t>0, the range is all real numbers, (−∞,∞)(-\infty,\infty). That matches the choice with domain (−∞,−1)∪(1,∞)(-\infty,-1)\cup(1,\infty) and range (−∞,∞)(-\infty,\infty). The choice using closed brackets [−1,1][-1,1] at the endpoints is tempting, but x=±1x=\pm1 makes x2−1=0x^2-1=0, and ln⁡(0)\ln(0) is undefined. The choice with range (0,∞)(0,\infty) confuses the range of the logarithm with the fact that its input is positive — logarithms themselves can be negative, zero, or positive. The choice with domain (−1,1)(-1,1) is exactly backwards: that interval is where x2−1<0x^2-1<0, which makes the logarithm undefined, not defined. A solid habit: for any logarithm, first solve "inside >0>0" for the domain, then decide whether the inside expression covers all positive values to determine the range. Natural log always has range (−∞,∞)(-\infty,\infty) when its input can be any positive number.

Question 2

Let r(x)=2x2+3x2−4r(x)=\frac{2x^2+3}{x^2-4}. Which option gives the domain and range of this function in interval notation?

  1. Domain: (−∞,−2)∪(2,∞)(-\infty,-2)\cup(2,\infty); Range: (−∞,−34]∪(2,∞)(-\infty,-\frac{3}{4}]\cup(2,\infty)
  2. Domain: (−∞,−2)∪(−2,2)∪(2,∞)(-\infty,-2)\cup(-2,2)\cup(2,\infty); Range: (−∞,2)∪(2,∞)(-\infty,2)\cup(2,\infty)
  3. Domain: (−∞,−2)∪(−2,2)∪(2,∞)(-\infty,-2)\cup(-2,2)\cup(2,\infty); Range: (−∞,−34]∪[2,∞)(-\infty,-\frac{3}{4}]\cup[2,\infty)
  4. Domain: (−∞,−2)∪(−2,2)∪(2,∞)(-\infty,-2)\cup(-2,2)\cup(2,\infty); Range: (−∞,−34]∪(2,∞)(-\infty,-\frac{3}{4}]\cup(2,\infty) (correct answer)
Explanation: Whenever you see a rational function and need its range, solve for xx in terms of yy rather than guessing. The domain is found first: x2−4=0x^2-4=0 at x=±2x=\pm2, so the domain is (−∞,−2)∪(−2,2)∪(2,∞).(-\infty,-2)\cup(-2,2)\cup(2,\infty). A domain like (−∞,−2)∪(2,∞)(-\infty,-2)\cup(2,\infty) drops the entire valid interval (−2,2)(-2,2); for example, x=0x=0 gives r(0)=−34r(0)=-\frac34, so it cannot be excluded. For range, set y=2x2+3x2−4y=\frac{2x^2+3}{x^2-4}. Clearing denominators gives (y−2)x2=4y+3(y-2)x^2=4y+3, so we need 4y+3y−2≥0,y≠2.\frac{4y+3}{y-2}\ge 0,\qquad y\ne2. Sign analysis with numerator zero at −34-\frac34 and denominator zero at 22 shows the quotient is nonnegative exactly on (−∞,−34]∪(2,∞).(-\infty,-\frac34]\cup(2,\infty). The value −34-\frac34 is actually attained at x=0x=0. The value y=2y=2 is impossible because substituting it would require 3=−83=-8. Thus the range is exactly that interval, and the choice with that domain and range is correct. The range (−∞,2)∪(2,∞)(-\infty,2)\cup(2,\infty) is too large: it includes values between −34-\frac34 and2, such as 0, but for those values the quotient (4y+3)/(y−2)(4y+3)/(y-2) is negative, so no real xx exists. The range (−∞,−34]∪[2,∞)(-\infty,-\frac34]\cup[2,\infty) incorrectly brackets 2; since y=2y=2 can never be an output, that endpoint must be open. Also, with the incomplete domain ∣x∣>2|x|>2, all outputs are actually >2>2, so that first answer's range is not consistent either. For rational-function range questions, remember to rearrange into a condition on yy, use a sign chart, and test each endpoint by direct substitution.

Question 3

Let f(x)=x−1f(x)=\sqrt{x-1} and g(x)=x2−2xg(x)=x^2-2x. What are the domain and range of (f∘g)(x)(f\circ g)(x)?

  1. Domain: (−∞,1−2]∪[1+2,∞)(-\infty,1-\sqrt{2}]\cup[1+\sqrt{2},\infty); Range: (0,∞)(0,\infty)
  2. Domain: [1−2,1+2][1-\sqrt{2},1+\sqrt{2}]; Range: [0,∞)[0,\infty)
  3. Domain: (−∞,1−2]∪[1+2,∞)(-\infty,1-\sqrt{2}]\cup[1+\sqrt{2},\infty); Range: [0,∞)[0,\infty) (correct answer)
  4. Domain: [1−2,1+2][1-\sqrt{2},1+\sqrt{2}]; Range: (0,∞)(0,\infty)
Explanation: When you see a composition like (f∘g)(x)(f\circ g)(x), rewrite it first: f(g(x))=g(x)−1=x2−2x−1.f(g(x))=\sqrt{g(x)-1}=\sqrt{x^2-2x-1}. Now you are really asking for the domain and range of x2−2x−1\sqrt{x^2-2x-1}. The domain requires the radicand to be nonnegative: x2−2x−1≥0.x^2-2x-1\ge 0. Solve x2−2x−1=0x^2-2x-1=0 using the quadratic formula: x=1±2.x=1\pm \sqrt{2}. Since the parabola opens upward, the inequality x2−2x−1≥0x^2-2x-1\ge 0 is true outside the roots, so the domain is (−∞,1−2]∪[1+2,∞).(-\infty,1-\sqrt{2}]\cup[1+\sqrt{2},\infty). For the range, inside that domain the radicand can equal 00 at the two boundary points, so 0=0\sqrt{0}=0 is included. As xx moves away from the middle, x2−2x−1x^2-2x-1 grows without bound, so the square root also grows without bound. Thus the range is [0,∞)[0,\infty). The domain [1−2,1+2][1-\sqrt{2},1+\sqrt{2}] is tempting, but it represents where x2−2x−1≤0x^2-2x-1\le 0, i.e. where the square root would be negative inside — impossible. The choices with range (0,∞)(0,\infty) wrongly exclude 00, even though both endpoints of the domain make the radicand exactly 00. So only the combination with the outside-interval domain and closed range [0,∞)[0,\infty) is correct. Study tip: for quadratic\sqrt{\text{quadratic}}, sketch the parabola quickly. Domain is where the parabola is above the xx-axis; range starts at 00 if the parabola touches the axis inside the domain.

Question 4

What are the domain and range of g(x)=4−3(12)x+2g(x)=4-3\left(\frac{1}{2}\right)^{x+2}?

  1. Domain: (−∞,∞)(-\infty,\infty); Range: (−∞,4)(-\infty,4) (correct answer)
  2. Domain: (−∞,∞)(-\infty,\infty); Range: (−∞,4](-\infty,4]
  3. Domain: (−∞,∞)(-\infty,\infty); Range: (4,∞)(4,\infty)
  4. Domain: (−∞,∞)(-\infty,\infty); Range: [4,∞)[4,\infty)
Explanation: Whenever you see an exponential function like this one, start by separating the two questions: the domain is usually all real numbers, and the range depends on the vertical shift and the sign of the coefficient. Here, g(x)=4−3(12)x+2g(x)=4-3\left(\frac12\right)^{x+2} is defined for every real xx, so the domain is indeed (−∞,∞)(-\infty,\infty). For the range, note that (12)x+2\left(\frac12\right)^{x+2} is always positive but never zero. Multiplying by −3-3 makes that positive value negative, so 4−3(positive)4-3(\text{positive}) is always less than 44. As x→∞x\to\infty, the exponential term approaches 00, so g(x)→4g(x)\to 4 from below but never reaches 44. As x→−∞x\to-\infty, the exponential term grows without bound, so g(x)→−∞g(x)\to-\infty. Thus the range is (−∞,4)(-\infty,4). The choice with range (−∞,4](-\infty,4] is a common trap: it treats the horizontal asymptote as an included value, but the exponential never equals zero, so g(x)g(x) never equals 44. The choices with range (4,∞)(4,\infty) and [4,∞)[4,\infty) get the direction backwards — subtracting a positive quantity from 44 always gives values below 44, never above. They likely come from misreading the minus sign as a shift upward rather than a reflection. For similar questions, identify the horizontal asymptote first (here y=4y=4), then decide whether the graph approaches it from above or below based on the coefficient's sign. That instantly gives you the correct open or closed endpoint.

Question 5

An object is launched from a height of 128 feet. Its height after tt seconds is h(t)=−16t2+32t+128h(t)=-16t^2+32t+128, and it stays in the air until it hits the ground.

Which interval notation gives the domain and range for the height during the flight?

  1. Domain: [0,∞)[0,\infty); Range: [0,144][0,144]
  2. Domain: [0,4][0,4]; Range: [0,144][0,144] (correct answer)
  3. Domain: [0,4][0,4]; Range: [128,144][128,144]
  4. Domain: [−2,4][-2,4]; Range: [0,144][0,144]
Explanation: When you see a projectile height function, the domain is the time the object is actually in the air, and the range is every height it reaches from launch to landing. Start by finding when it hits the ground: set h(t)=0h(t)=0. −16t2+32t+128=0⇒t2−2t−8=0⇒(t−4)(t+2)=0-16t^2+32t+128=0 \Rightarrow t^2-2t-8=0 \Rightarrow (t-4)(t+2)=0 So t=4t=4 or t=−2t=-2. Negative time doesn't make sense for this flight, so the domain is [0,4][0,4]. For the range, find the maximum height at the vertex: t=−322(−16)=1,h(1)=−16+32+128=144t=-\frac{32}{2(-16)}=1,\quad h(1)=-16+32+128=144 The lowest height during the flight is 00 at touchdown, so the range is [0,144][0,144]. The choice with Domain [0,4][0,4]; Range [0,144][0,144] is correct. The choice with Domain [0,∞)[0,\infty) and Range [0,144][0,144] forgets that the object stops being in flight after t=4t=4. The choice with Domain [0,4][0,4]; Range [128,144][128,144] incorrectly treats only the heights from launch to peak, ignoring the descent to the ground. The choice with Domain [−2,4][-2,4]; Range [0,144][0,144] includes the negative root t=−2t=-2, which represents time before launch. A reliable strategy: for any projectile motion question, solve h(t)=0h(t)=0 for the domain, find the vertex for the maximum height, and use the ground height as the minimum. Watch for negative time roots — they are extraneous in real-world flight problems.

Question 6

What are the domain and range of f(x)=1x2−4f(x)=\frac{1}{\sqrt{x^2-4}}?

  1. Domain: [−2,2][-2,2]; Range: (0,∞)(0,\infty)
  2. Domain: (−∞,−2]∪[2,∞)(-\infty,-2]\cup[2,\infty); Range: [0,∞)[0,\infty)
  3. Domain: (−∞,−2)∪(2,∞)(-\infty,-2)\cup(2,\infty); Range: [0,∞)[0,\infty)
  4. Domain: (−∞,−2)∪(2,∞)(-\infty,-2)\cup(2,\infty); Range: (0,∞)(0,\infty) (correct answer)
Explanation: When you see a function like f(x)=1x2−4f(x)=\frac{1}{\sqrt{x^2-4}}, ask two questions: What makes the denominator defined and nonzero? And can the output actually reach zero? The square root requires x2−4≥0x^2-4 \ge 0, but because it's in a denominator, it must be strictly positive: x2−4>0x^2-4>0. Thus ∣x∣>2|x|>2, so the domain is (−∞,−2)∪(2,∞)(-\infty,-2)\cup(2,\infty). For the range, the denominator is always a positive square root, so the fraction is always positive. As xx approaches 22 or −2-2, the denominator approaches 00, sending the function toward infinity; as ∣x∣|x| grows, the fraction approaches 00 but never equals it. So the range is (0,∞)(0,\infty). The choice with domain [−2,2][-2,2] is reversed: inside that interval x2−4≤0x^2-4\le 0, which makes the square root undefined (or zero). The choice with domain (−∞,−2]∪[2,∞)(-\infty,-2]\cup[2,\infty) incorrectly includes x=±2x=\pm2, where the denominator becomes zero, and its range [0,∞)[0,\infty) incorrectly includes 00. The choice with domain (−∞,−2)∪(2,∞)(-\infty,-2)\cup(2,\infty) gets the domain right but still lists range [0,∞)[0,\infty), forgetting that a reciprocal of a positive square root can never be zero. Only the pairing of the strict domain and strictly positive range matches the function. On exam day, remember: for rational functions with even roots, set the radicand strictly greater than zero if it's in the denominator. Then check whether the output can hit zero — reciprocals of positive quantities never do.

Question 7

Which option correctly states the domain and range of f(x)=6+x−x2f(x)=\sqrt{6+x-x^2}?

  1. Domain: (−∞,−2]∪[3,∞)(-\infty,-2]\cup[3,\infty); Range: [0,∞)[0,\infty)
  2. Domain: [−2,3][-2,3]; Range: [0,52][0,\frac{5}{2}] (correct answer)
  3. Domain: [−2,3][-2,3]; Range: [0,52)[0,\frac{5}{2})
  4. Domain: (−2,3)(-2,3); Range: (0,52](0,\frac{5}{2}]
Explanation: When you see a square-root function, the first move is always the same: the expression inside the radical must be nonnegative. That inequality gives you the domain, and once you know the values the radicand can take, the square root tells you the range. Here, set 6+x−x2≥06+x-x^2 \ge 0. Multiplying by −1-1 and reversing the inequality gives x2−x−6≤0x^2-x-6 \le 0, which factors as (x−3)(x+2)≤0(x-3)(x+2)\le 0. So −2≤x≤3-2 \le x \le 3, meaning the domain is [−2,3][-2,3]. This immediately rules out the option with domain (−∞,−2]∪[3,∞)(-\infty,-2]\cup[3,\infty), which has the inequality sign reversed, and the open-interval version (−2,3)(-2,3), because the endpoints make the radicand exactly zero and must be included. For the range, complete the square on the radicand: 6+x−x2=254−(x−12)26+x-x^2 = \frac{25}{4} - \left(x-\frac12\right)^2. The largest radicand value is 254\frac{25}{4} at x=12x=\frac12, and the smallest is 00 at the endpoints. Therefore the square-root outputs run from 0=0\sqrt0=0 up to 254=52\sqrt{\frac{25}{4}}=\frac52. Since every value in between is achieved, the range is [0,52][0,\frac52]. The choice with range [0,52)[0,\frac52) is wrong because the maximum is actually reached at x=12x=\frac12, and the choice with range (0,52](0,\frac52] wrongly excludes 00, even though f(−2)=f(3)=0f(-2)=f(3)=0. On exam day, remember: solve the radicand inequality for domain, then complete the square to find the radicand's maximum and minimum before taking the square root.

Question 8

An online store estimates that its weekly revenue, in thousands of dollars, from selling an item at price pp dollars is R(p)=120pp+5R(p)=\frac{120p}{p+5}, for 0≤p≤1000\le p\le 100.

What are the domain and range of this revenue function in interval notation?

  1. Domain: [0,100][0,100]; Range: (0,8007]\left(0,\frac{800}{7}\right]
  2. Domain: [0,∞)[0,\infty); Range: [0,120)[0,120)
  3. Domain: [0,100][0,100]; Range: [0,8007]\left[0,\frac{800}{7}\right] (correct answer)
  4. Domain: [0,100][0,100]; Range: [0,120)[0,120)
Explanation: Whenever you see a domain and range question, first check whether the domain is explicitly restricted. Here the problem states 0≤p≤1000\le p\le 100, so the domain is immediately [0,100][0,100]. For the range, determine how R(p)R(p) behaves on that interval. Since R′(p)=600(p+5)2>0R'(p)=\frac{600}{(p+5)^2}>0, the function is increasing, so the minimum occurs at p=0p=0: R(0)=0R(0)=0. The maximum occurs at p=100p=100: R(100)=120(100)105=8007R(100)=\frac{120(100)}{105}=\frac{800}{7}. Thus the range is [0,8007]\left[0,\frac{800}{7}\right], matching the choice with Domain [0,100][0,100]; Range [0,8007]\left[0,\frac{800}{7}\right]. The distractor with range (0,8007]\left(0,\frac{800}{7}\right] incorrectly excludes 00, forgetting that selling at price ($0) gives revenue ($0). The choice with Domain [0,∞)[0,\infty); Range [0,120)[0,120) confuses the given restricted domain with the function's horizontal asymptote behavior; R(p)R(p) approaches but never reaches 120120 as p→∞p\to\infty, and pp is capped at 100100 anyway. The choice with Domain [0,100][0,100]; Range [0,120)[0,120) gets the domain right but uses the asymptotic upper bound instead of computing the actual maximum at p=100p=100. A quick study tip: when a function is continuous and monotonic on a closed interval, evaluate the endpoints to find the range; don't rely on asymptotes or intuition.

Question 9

What are the domain and range of f(x)=x−2x+3f(x)=\sqrt{\frac{x-2}{x+3}}?

  1. Domain: (−3,2](-3,2]; Range: [0,∞)[0,\infty)
  2. Domain: (−∞,−3)∪[2,∞)(-\infty,-3)\cup[2,\infty); Range: [0,∞)[0,\infty)
  3. Domain: (−∞,−3)∪[2,∞)(-\infty,-3)\cup[2,\infty); Range: [0,1)∪(1,∞)[0,1)\cup(1,\infty) (correct answer)
  4. Domain: (−∞,−3)∪(2,∞)(-\infty,-3)\cup(2,\infty); Range: [0,1)∪(1,∞)[0,1)\cup(1,\infty)
Explanation: Whenever you see a square root of a rational expression, start with the domain: the inside must be nonnegative and the denominator cannot be zero. Set x−2x+3≥0.\frac{x-2}{x+3}\ge 0. The sign of the quotient changes at x=−3x=-3 and x=2x=2. It is positive on (−∞,−3)(-\infty,-3) and on [2,∞)[2,\infty), zero at x=2x=2, and undefined at x=−3x=-3. So the domain is (−∞,−3)∪[2,∞)(-\infty,-3)\cup[2,\infty). the range, rewrite the radicand:x−2x+3=1−5x+3.\frac{x-2}{x+3}=1-\frac{5}{x+3}.For x<−3x<-3, this quantity takes every value greater than 11, giving (1,∞)(1,\infty). For x≥2x\ge2, it starts at 00 at x=2x=2 and approaches 11 from below, never reaching it, giving [0,1)[0,1). Thus the radicand takes values [0,1)∪(1,∞)[0,1)\cup(1,\infty), and taking square roots preserves these intervals, so the range is [0,1)∪(1,∞)[0,1)\cup(1,\infty). The choice saying domain (−3,2](-3,2] and range [0,∞)[0,\infty) misses the valid outer interval and wrongly includes the entire range. The choice with the correct domain but range [0,∞)[0,\infty) fails to notice that y=1y=1 would require x−2x+3=1\frac{x-2}{x+3}=1, which has no solution. The choice with domain (−∞,−3)∪(2,∞)(-\infty,-3)\cup(2,\infty) has the right range but incorrectly excludes x=2x=2, where the radicand is 00 and the function is defined. Always test boundary points: include endpoints where the radicand equals zero, and solve equations like "can this equal 1?" to avoid range traps.

Question 10

Let f(x)=∣x−2∣−3f(x)=|x-2|-3, with the domain restricted to −1<x≤5-1<x\le 5. What is the range of the function?

  1. [−3,0][-3,0] (correct answer)
  2. (−3,0](-3,0]
  3. [−3,0)[-3,0)
  4. (−3,0)(-3,0)
Explanation: When you see an absolute-value function with a restricted domain, first find the vertex and then check the domain endpoints. Here f(x)=∣x−2∣−3f(x)=|x-2|-3 has its vertex at x=2x=2, which is inside the interval −1<x≤5-1<x\le5, so the minimum is f(2)=−3f(2)=-3. Moving toward the included endpoint x=5x=5, the output grows to f(5)=0f(5)=0, so 0 is also in the range. Moving toward x=−1x=-1, the output approaches 0 again, but x=−1x=-1 is not included; however, 0 is already attained at x=5x=5, so that doesn't matter. Because the function is continuous over the domain, every value between −3-3 and 00 appears. Therefore the range is [−3,0][-3,0]. Each other choice is missing something important: (−3,0](-3,0] excludes −3-3, but the included point x=2x=2 produces exactly −3-3. [−3,0)[-3,0) excludes 00, but the included point x=5x=5 produces exactly 00. (−3,0)(-3,0) excludes both endpoints, yet both −3-3 and 00 are actually attained. Study tip: whenever the domain is restricted, check whether the vertex and each endpoint are included. Whether those key x-values are in the domain decides whether the range brackets are open or closed.

Question 11

The function $$f(x)=\begin{cases} x^2+1, & x<0,\ 3-x, & 0\le x\le 4,\ 2x-5, & x>4 \end{cases}

  1. Domain: [0,∞)[0,\infty); Range: [−1,∞)[-1,\infty)
  2. Domain: (−∞,∞)(-\infty,\infty); Range: [1,∞)[1,\infty)
  3. Domain: (−∞,∞)(-\infty,\infty); Range: (−1,∞)(-1,\infty)
  4. Domain: (−∞,∞)(-\infty,\infty); Range: [−1,∞)[-1,\infty) (correct answer)
Explanation: Whenever you see a piecewise-defined function, start by asking: Do the pieces cover every real xx? Here the pieces cover x<0x<0, 0≤x≤40\le x\le 4, and x>4x>4, whose union is all real numbers. So the domain is (−∞,∞)(-\infty,\infty). Now collect the possible outputs. For x<0x<0, x2+1>1x^2+1>1, so those outputs are (1,∞)(1,\infty). For 0≤x≤40\le x\le 4, 3−x3-x decreases from 33 to −1-1, giving [−1,3][-1,3]. For x>4x>4, 2x−5>32x-5>3, giving (3,∞)(3,\infty). The union of (1,∞)(1,\infty), [−1,3][-1,3], and (3,∞)(3,\infty) is [−1,∞)[-1,\infty): the value −1-1 is included because f(4)=−1f(4)=-1, and every value above −1-1 appears in at least one piece. The domain [0,∞)[0,\infty) is incorrect because it ignores the x<0x<0 piece entirely. The range [1,∞)[1,\infty) is incorrect because it misses all the middle-piece outputs from −1-1 up to 11, like f(4)=−1f(4)=-1 or f(2)=1f(2)=1. The range (−1,∞)(-1,\infty) is very close but wrong because it excludes −1-1, even though the function actually reaches −1-1 at x=4x=4. Only Domain (−∞,∞)(-\infty,\infty); Range [−1,∞)[-1,\infty) matches. Study tip: for a piecewise function, find the range of each piece separately, then union them. Pay special attention to endpoints — include a value only if one of the pieces actually reaches it.

Question 12

Consider the function f(x)=x2−9x2−4x+3f(x)=\frac{x^2-9}{x^2-4x+3}. Which option correctly describes both the domain and the range of this function?

  1. Domain: (−∞,1)∪(1,∞)(-\infty,1)\cup(1,\infty); Range: (−∞,1)∪(1,∞)(-\infty,1)\cup(1,\infty)
  2. Domain: (−∞,1)∪(1,3)∪(3,∞)(-\infty,1)\cup(1,3)\cup(3,\infty); Range: (−∞,1)∪(1,3)∪(3,∞)(-\infty,1)\cup(1,3)\cup(3,\infty) (correct answer)
  3. Domain: (−∞,1)∪(1,3)∪(3,∞)(-\infty,1)\cup(1,3)\cup(3,\infty); Range: (−∞,1)∪(1,∞)(-\infty,1)\cup(1,\infty)
  4. Domain: (−∞,1)∪(1,3)∪(3,∞)(-\infty,1)\cup(1,3)\cup(3,\infty); Range: (−∞,3)∪(3,∞)(-\infty,3)\cup(3,\infty)
Explanation: Whenever you see a rational function, factor first:
f(x)=x2−9x2−4x+3=(x−3)(x+3)(x−1)(x−3).f(x)=\frac{x^2-9}{x^2-4x+3}=\frac{(x-3)(x+3)}{(x-1)(x-3)}.
The factor x−3x-3 cancels, but the original function is still undefined at x=3x=3, and x=1x=1 makes the denominator zero too. So the domain is all real numbers except 11 and 33:
(−∞,1)∪(1,3)∪(3,∞).(-\infty,1)\cup(1,3)\cup(3,\infty). After canceling, f(x)=x+3x−1f(x)=\frac{x+3}{x-1} for x≠1,3x\neq1,3. To find the range, solve y=x+3x−1y=\frac{x+3}{x-1} for xx:
y(x−1)=x+3⇒x=y+3y−1.y(x-1)=x+3 \Rightarrow x=\frac{y+3}{y-1}.
This shows y≠1y\neq1, since that would make the denominator zero. But the original also excludes x=3x=3, and at x=3x=3 the simplified expression equals 33. Since x=3x=3 is not allowed, y=3y=3 is never produced. Thus the range also excludes 11 and 33, giving the same set as the domain. The domain-only-correct option that lists range as (−∞,1)∪(1,∞)(-\infty,1)\cup(1,\infty) misses that 33 is not in the range. The option with domain (−∞,1)∪(1,∞)(-\infty,1)\cup(1,\infty) is wrong because it incorrectly includes x=3x=3 in the domain. The range (−∞,3)∪(3,∞)(-\infty,3)\cup(3,\infty) is wrong because it incorrectly includes 11 in the range. Study tip: after simplifying a rational function, always re-check excluded xx-values in the original, and solve for xx in terms of yy to find excluded yy-values.