Domain & Range - Algebra 3

1. Let $f(x)=5-3\sqrt{4-x^2}$. Which pair gives the domain and range of $f$?
   Answer: Domain $[-2,2]$, range $[-1,5]$ The radicand requires $4-x^2\ge0$, so $x\in[-2,2]$. As $x$ ranges over this interval, $\sqrt{4-x^2}$ ranges from 0 to 2. Therefore $5-3\sqrt{4-x^2}$ ranges from $5-6=-1$ to $5-0=5$. Starting the range at 2 uses the wrong coefficient, $[-4,4]$ is not the solution of $4-x^2\ge0$, and $[0,5]$ ignores the negative minimum.
2. What is the domain of $f(x)=\sqrt{x^2-5x+6}/(x-3)$?
   Answer: $(-\infty,2]\cup(3,\infty)$ The radicand factors as $(x-2)(x-3)$, so it is nonnegative when $x\le2$ or $x\ge3$. The denominator $x-3$ cannot be zero, so $x=3$ must be removed. Thus the domain is $(-\infty,2]\cup(3,\infty)$. Including $x=3$ makes the denominator zero; excluding $x=2$ is wrong because the radicand may be zero; using $x=3$ as the only exclusion ignores the radical restriction.
3. What are the domain and range of $f(x)=\frac{1}{\sqrt{x^2-4}}$?
   Answer: Domain: $(-\infty,-2)\cup(2,\infty)$; Range: $(0,\infty)$ When you see a function like $f(x)=\frac{1}{\sqrt{x^2-4}}$, ask two questions: What makes the denominator defined and nonzero? And can the output actually reach zero? The square root requires $x^2-4 \ge 0$, but because it's in a denominator, it must be strictly positive: $x^2-4>0$. Thus $|x|>2$, so the domain is $(-\infty,-2)\cup(2,\infty)$. For the range, the denominator is always a positive square root, so the fraction is always positive. As $x$ approaches $2$ or $-2$, the denominator approaches $0$, sending the function toward infinity; as $|x|$ grows, the fraction approaches $0$ but never equals it. So the range is $(0,\infty)$. The choice with domain $[-2,2]$ is reversed: inside that interval $x^2-4\le 0$, which makes the square root undefined (or zero). The choice with domain $(-\infty,-2]\cup[2,\infty)$ incorrectly includes $x=\pm2$, where the denominator becomes zero, and its range $[0,\infty)$ incorrectly includes $0$. The choice with domain $(-\infty,-2)\cup(2,\infty)$ gets the domain right but still lists range $[0,\infty)$, 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.
4. Let $f(x)=x^2-4x+5$ with domain $(-\infty,2]$. If $f^{-1}$ is the inverse function, what is the range of $f^{-1}$?
   Answer: $(-\infty,2]$ The range of $f^{-1}$ equals the domain of $f$, which is $(-\infty,2]$. The domain of $f^{-1}$ is the range of $f$: on $(-\infty,2]$, $f(x)\to\infty$ as $x\to-\infty$, and $f(2)=1$, so the range of $f$ is $[1,\infty)$. The interval $[2,\infty)$ is the domain of the increasing branch to the right of the vertex, not the range of $f^{-1}$, and $(-\infty,1]$ includes values below the minimum of $f$ while omitting values above 1.
5. Which option correctly states the domain and range of $f(x)=\sqrt{6+x-x^2}$?
   Answer: Domain: $[-2,3]$; Range: $[0,\frac{5}{2}]$ 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 $6+x-x^2 \ge 0$. Multiplying by $-1$ and reversing the inequality gives $x^2-x-6 \le 0$, which factors as $(x-3)(x+2)\le 0$. So $-2 \le x \le 3$, meaning the domain is $[-2,3]$. This immediately rules out the option with domain $(-\infty,-2]\cup[3,\infty)$, which has the inequality sign reversed, and the open-interval version $(-2,3)$, because the endpoints make the radicand exactly zero and must be included. For the range, complete the square on the radicand: $6+x-x^2 = \frac{25}{4} - \left(x-\frac12\right)^2$. The largest radicand value is $\frac{25}{4}$ at $x=\frac12$, and the smallest is $0$ at the endpoints. Therefore the square-root outputs run from $\sqrt0=0$ up to $\sqrt{\frac{25}{4}}=\frac52$. Since every value in between is achieved, the range is $[0,\frac52]$. The choice with range $[0,\frac52)$ is wrong because the maximum is actually reached at $x=\frac12$, and the choice with range $(0,\frac52]$ wrongly excludes $0$, even though $f(-2)=f(3)=0$. 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.
6. Which interval notation gives the domain and range for the height during the flight?
   Answer: Domain: $[0,4]$; Range: $[0,144]$ 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 $h(t)=0$. $$-16t^2+32t+128=0 \Rightarrow t^2-2t-8=0 \Rightarrow (t-4)(t+2)=0$$ So $t=4$ or $t=-2$. Negative time doesn't make sense for this flight, so the domain is $[0,4]$. For the range, find the maximum height at the vertex: $$t=-\frac{32}{2(-16)}=1,\quad h(1)=-16+32+128=144$$ The lowest height during the flight is $0$ at touchdown, so the range is $[0,144]$. The choice with Domain $[0,4]$; Range $[0,144]$ is correct. The choice with Domain $[0,\infty)$ and Range $[0,144]$ forgets that the object stops being in flight after $t=4$. The choice with Domain $[0,4]$; Range $[128,144]$ incorrectly treats only the heights from launch to peak, ignoring the descent to the ground. The choice with Domain $[-2,4]$; Range $[0,144]$ includes the negative root $t=-2$, which represents time before launch. A reliable strategy: for any projectile motion question, solve $h(t)=0$ 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.
7. What is the range of $f(x)=(x^2+1)/(x^2+4)$?
   Answer: $[1/4,1)$ Set $y=(x^2+1)/(x^2+4)$. Rearranging gives $(y-1)x^2=1-4y$, so $x^2=(1-4y)/(y-1)$. This is nonnegative exactly when $1/4\le y<1$. The value $y=1$ would require $x^2+1=x^2+4$, impossible, and values below $1/4$ make the right side negative. Hence the range is $[1/4,1)$.
8. What are the domain and range of $g(x)=4-3\left(\frac{1}{2}\right)^{x+2}$?
   Answer: Domain: $(-\infty,\infty)$; Range: $(-\infty,4)$ 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, $g(x)=4-3\left(\frac12\right)^{x+2}$ is defined for every real $x$, so the domain is indeed $(-\infty,\infty)$. For the range, note that $\left(\frac12\right)^{x+2}$ is always positive but never zero. Multiplying by $-3$ makes that positive value negative, so $4-3(\text{positive})$ is always less than $4$. As $x\to\infty$, the exponential term approaches $0$, so $g(x)\to 4$ from below but never reaches $4$. As $x\to-\infty$, the exponential term grows without bound, so $g(x)\to-\infty$. Thus the range is $(-\infty,4)$. The choice with range $(-\infty,4]$ is a common trap: it treats the horizontal asymptote as an included value, but the exponential never equals zero, so $g(x)$ never equals $4$. The choices with range $(4,\infty)$ and $[4,\infty)$ get the direction backwards — subtracting a positive quantity from $4$ always gives values below $4$, 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 $y=4$), 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.
9. What is the range of $f(x)=x^2+4x+7$ when the domain is restricted to $[-1,3]$?
   Answer: $[4,28]$ The vertex is at $x=-2$, outside the given domain, so the function is increasing on $[-1,3]$. Evaluate the endpoints: $f(-1)=4$ and $f(3)=28$. Hence the range is $[4,28]$. The value 3 is the function value at the vertex, which is not in the domain; 7 is the $y$-intercept; 31 is not produced by any $x$ in the interval.
10. A ball is thrown upward from a 64-foot building with an initial velocity of 48 ft/s. Its height after $t$ seconds is $h(t)=-16t^2+48t+64$, until it hits the ground. What is the range of $h$?
    Answer: $[0,100]$ The ball hits the ground when $-16t^2+48t+64=0$, or $(t-4)(t+1)=0$, so the relevant domain is $0\le t\le4$. The maximum height occurs at $t=48/(32)=1.5$, giving $h(1.5)=100$, and the minimum height on this domain is 0. Thus the range is $[0,100]$. Starting at 64 ignores the flight up and down; $[0,\infty)$ ignores the ground and maximum.
11. What is the range of $f(x)=(2x+1)/(x-3)$?
    Answer: $(-\infty,2)\cup(2,\infty)$ Solving $y=(2x+1)/(x-3)$ for $x$ gives $x(y-2)=3y+1$. If $y=2$, this equation becomes $0=7$, so 2 is not in the range. Every other $y$ produces a valid $x$. The excluded value is the horizontal asymptote $y=2$, not the vertical asymptote $x=3$, not a sign error, and not the $y$-intercept.
12. What is the domain of $f(x)=\sqrt{4-x^2}/(x^2-1)$?
    Answer: $[-2,-1)\cup(-1,1)\cup(1,2]$ The square root requires $4-x^2\ge0$, so $x\in[-2,2]$. The denominator cannot be zero, so $x\ne\pm1$. Removing $-1$ and $1$ from $[-2,2]$ gives $[-2,-1)\cup(-1,1)\cup(1,2]$. Including $\pm1$ makes the denominator zero; excluding $\pm2$ drops valid points where the radicand is zero; and an interval missing $(-1,1)$ would incorrectly require the denominator to be positive.
13. What are the domain and range of this revenue function in interval notation?
    Answer: Domain: $[0,100]$; Range: $\left[0,\frac{800}{7}\right]$ Whenever you see a domain and range question, first check whether the domain is explicitly restricted. Here the problem states $0\le p\le 100$, so the domain is immediately $[0,100]$. For the range, determine how $R(p)$ behaves on that interval. Since $R'(p)=\frac{600}{(p+5)^2}>0$, the function is increasing, so the minimum occurs at $p=0$: $R(0)=0$. The maximum occurs at $p=100$: $R(100)=\frac{120(100)}{105}=\frac{800}{7}$. Thus the range is $\left[0,\frac{800}{7}\right]$, matching the choice with Domain $[0,100]$; Range $\left[0,\frac{800}{7}\right]$. The distractor with range $\left(0,\frac{800}{7}\right]$ incorrectly excludes $0$, forgetting that selling at price ($0) gives revenue ($0). The choice with Domain $[0,\infty)$; Range $[0,120)$ confuses the given restricted domain with the function's horizontal asymptote behavior; $R(p)$ approaches but never reaches $120$ as $p\to\infty$, and $p$ is capped at $100$ anyway. The choice with Domain $[0,100]$; Range $[0,120)$ gets the domain right but uses the asymptotic upper bound instead of computing the actual maximum at $p=100$. 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.
14. What are the domain and range of $f(x)=\ln(x^2-1)$?
    Answer: Domain: $(-\infty,-1)\cup(1,\infty)$; Range: $(-\infty,\infty)$ 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 $x^2-1>0$. Solving $x^2>1$ gives $x<-1$ or $x>1$, so the domain is $(-\infty,-1)\cup(1,\infty)$. Next, ask what values the inside can produce: as $x$ moves through that domain, $x^2-1$ can take every positive number, from just above $0$ all the way to $\infty$. Since $\ln(t)$ can output any real number when $t>0$, the range is all real numbers, $(-\infty,\infty)$. That matches the choice with domain $(-\infty,-1)\cup(1,\infty)$ and range $(-\infty,\infty)$. The choice using closed brackets $[-1,1]$ at the endpoints is tempting, but $x=\pm1$ makes $x^2-1=0$, and $\ln(0)$ is undefined. The choice with range $(0,\infty)$ 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 $(-1,1)$ is exactly backwards: that interval is where $x^2-1<0$, which makes the logarithm undefined, not defined. A solid habit: for any logarithm, first solve "inside $>0$" 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.
15. What is the range of $f(x)=2^x+2^{-x}$?
    Answer: $[2,\infty)$ For every real $x$, $2^x>0$ and $2^{-x}>0$. Since $(2^{x/2}-2^{-x/2})^2\ge0$, $2^x+2^{-x}\ge2$, with equality at $x=0$. The expression is unbounded above as $x\to\infty$ or $x\to-\infty$, so the range is $[2,\infty)$. A lower bound of 0 or 1 ignores the minimum of the sum, and $(0,2]$ reverses the inequality.
16. Let $f(x)=\sqrt{x-1}$ and $g(x)=x^2-2x$. What are the domain and range of $(f\circ g)(x)$?
    Answer: Domain: $(-\infty,1-\sqrt{2}]\cup[1+\sqrt{2},\infty)$; Range: $[0,\infty)$ When you see a composition like $(f\circ g)(x)$, rewrite it first: $$f(g(x))=\sqrt{g(x)-1}=\sqrt{x^2-2x-1}.$$ Now you are really asking for the domain and range of $\sqrt{x^2-2x-1}$. The domain requires the radicand to be nonnegative: $$x^2-2x-1\ge 0.$$ Solve $x^2-2x-1=0$ using the quadratic formula: $$x=1\pm \sqrt{2}.$$ Since the parabola opens upward, the inequality $x^2-2x-1\ge 0$ is true outside the roots, so the domain is $$(-\infty,1-\sqrt{2}]\cup[1+\sqrt{2},\infty).$$ For the range, inside that domain the radicand can equal $0$ at the two boundary points, so $\sqrt{0}=0$ is included. As $x$ moves away from the middle, $x^2-2x-1$ grows without bound, so the square root also grows without bound. Thus the range is $[0,\infty)$. The domain $[1-\sqrt{2},1+\sqrt{2}]$ is tempting, but it represents where $x^2-2x-1\le 0$, i.e. where the square root would be negative inside — impossible. The choices with range $(0,\infty)$ wrongly exclude $0$, even though both endpoints of the domain make the radicand exactly $0$. So only the combination with the outside-interval domain and closed range $[0,\infty)$ is correct. Study tip: for $\sqrt{\text{quadratic}}$, sketch the parabola quickly. Domain is where the parabola is above the $x$-axis; range starts at $0$ if the parabola touches the axis inside the domain.
17. The function $f(x)=\begin{cases} x^2+1, & x<0,\\ 3-x, & 0\le x\le 4,\\ 2x-5, & x>4 \end{cases}$ is defined for all real numbers. Which option gives the domain and range of the function?
    Answer: Domain: $(-\infty,\infty)$; Range: $[-1,\infty)$ Whenever you see a piecewise-defined function, start by asking: Do the pieces cover every real $x$? Here the pieces cover $x<0$, $0\le x\le 4$, and $x>4$, whose union is all real numbers. So the domain is $(-\infty,\infty)$. Now collect the possible outputs. For $x<0$, $x^2+1>1$, so those outputs are $(1,\infty)$. For $0\le x\le 4$, $3-x$ decreases from $3$ to $-1$, giving $[-1,3]$. For $x>4$, $2x-5>3$, giving $(3,\infty)$. The union of $(1,\infty)$, $[-1,3]$, and $(3,\infty)$ is $[-1,\infty)$: the value $-1$ is included because $f(4)=-1$, and every value above $-1$ appears in at least one piece. The domain $[0,\infty)$ is incorrect because it ignores the $x<0$ piece entirely. The range $[1,\infty)$ is incorrect because it misses all the middle-piece outputs from $-1$ up to $1$, like $f(4)=-1$ or $f(2)=1$. The range $(-1,\infty)$ is very close but wrong because it excludes $-1$, even though the function actually reaches $-1$ at $x=4$. Only Domain $(-\infty,\infty)$; Range $[-1,\infty)$ 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.
18. Consider $f(x)=\sqrt{1/(x^2-9)}$. Which statement correctly identifies its domain and range?
    Answer: Domain $(-\infty,-3)\cup(3,\infty)$, range $(0,\infty)$ The denominator must be positive because it is under a square root in a reciprocal, so $x^2-9>0$, giving $|x|>3$. For every such $x$, the output is positive; it approaches 0 at infinity and becomes arbitrarily large near $\pm3$. Thus the range is $(0,\infty)$, not including 0. Including $\pm3$ makes the original denominator zero, and $(-3,3)$ is where $x^2-9<0$.
19. Let $f(x)=x$ for $x<0$, $f(x)=x^2$ for $0\le x\le2$, and $f(x)=x+3$ for $x>2$. What is the range of $f$?
    Answer: $(-\infty,4]\cup(5,\infty)$ For $x<0$, outputs cover $(-\infty,0)$. For $0\le x\le2$, outputs cover $[0,4]$. For $x>2$, outputs cover $(5,\infty)$. Combining gives $(-\infty,4]\cup(5,\infty)$. The value 5 is never produced because the second branch ends at 4 and the third branch starts just above 5; $(-\infty,5)$ incorrectly omits all outputs greater than 5.
20. Let $r(x)=\frac{2x^2+3}{x^2-4}$. Which option gives the domain and range of this function in interval notation?
    Answer: Domain: $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$; Range: $(-\infty,-\frac{3}{4}]\cup(2,\infty)$ Whenever you see a rational function and need its range, solve for $x$ in terms of $y$ rather than guessing. The domain is found first: $x^2-4=0$ at $x=\pm2$, so the domain is $$(-\infty,-2)\cup(-2,2)\cup(2,\infty).$$ A domain like $(-\infty,-2)\cup(2,\infty)$ drops the entire valid interval $(-2,2)$; for example, $x=0$ gives $r(0)=-\frac34$, so it cannot be excluded. For range, set $y=\frac{2x^2+3}{x^2-4}$. Clearing denominators gives $(y-2)x^2=4y+3$, so we need $$\frac{4y+3}{y-2}\ge 0,\qquad y\ne2.$$ Sign analysis with numerator zero at $-\frac34$ and denominator zero at $2$ shows the quotient is nonnegative exactly on $$(-\infty,-\frac34]\cup(2,\infty).$$ The value $-\frac34$ is actually attained at $x=0$. The value $y=2$ is impossible because substituting it would require $3=-8$. Thus the range is exactly that interval, and the choice with that domain and range is correct. The range $(-\infty,2)\cup(2,\infty)$ is too large: it includes values between $-\frac34$ and2, such as 0, but for those values the quotient $(4y+3)/(y-2)$ is negative, so no real $x$ exists. The range $(-\infty,-\frac34]\cup[2,\infty)$ incorrectly brackets 2; since $y=2$ can never be an output, that endpoint must be open. Also, with the incomplete domain $|x|>2$, all outputs are actually $>2$, so that first answer's range is not consistent either. For rational-function range questions, remember to rearrange into a condition on $y$, use a sign chart, and test each endpoint by direct substitution.

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