One-to-One & Invertibility - Algebra 3

Let $f(x)=x^4-2x^2+1$ . On which of the following domains is $f$ one-to-one?
Answer: $(-\infty,-1]$ Whenever a question asks where a function is one-to-one, imagine drawing horizontal lines: each must intersect the graph at most once on that domain. Here $f(x)=x^4-2x^2+1=(x^2-1)^2$ , an even function symmetric about the $y$ -axis, so intervals that contain both $x$ and $-x$ will generally fail. On $(-\infty,-1]$ , $f$ is strictly decreasing. Its derivative $f'(x)=4x^3-4x=4x(x-1)(x+1)$ is negative for every $x<-1$ , so as $x$ moves from left to right, $f$ falls from $\infty$ to $0$ without repeating a value. Thus this domain is one-to-one. $[-1,1]$ fails because it is symmetric about $0$ ; for instance, $f(-1)=f(1)=0$ . $[0,\infty)$ fails because the graph decreases from $(0,1)$ to $(1,0)$ , then increases, so the horizontal line $y=\frac14$ hits both $x=\sqrt{\frac12}$ and $x=\sqrt{\frac32}$ . $(-\infty,0]$ fails for the mirror reason: $y=\frac14$ hits $-\sqrt{\frac12}$ and $-\sqrt{\frac32}$ . Each wrong domain includes a turning point, causing the function to repeat output values; the trap is assuming a function is one-to-one just because it passes the vertical line test. Study tip: factor the function, then sketch or use the derivative to identify intervals where it is strictly monotonic. Only those intervals can be one-to-one.
How many of the following statements are true? I. If a function is strictly decreasing on its entire domain, then it is one-to-one. II. If a function is one-to-one on its entire domain, then it must be either strictly increasing or strictly decreasing on that domain. III. Restricting a quadratic function to one side of its axis of symmetry makes it one-to-one. IV. If $f$ is one-to-one and $f(3)=7$ , then $f^{-1}(7)=3$ .
Answer: 3 Statements I, III, and IV are true. Strictly decreasing functions give distinct outputs for distinct inputs. A quadratic restricted to one side of its axis is strictly monotonic. The inverse statement is the definition of inverse value. Statement II is false: for example, $f(x)=1/x$ on $x\ne0$ is one-to-one but is not increasing or decreasing over its entire disconnected domain.
Let $f$ be defined by $f(x)=\frac{x^2-4}{x-2}$ on the domain $x\neq 2$ . Which statement about $f$ is true?
Answer: $f$ is one-to-one, and $f^{-1}(x)=x-2$ with domain $x\neq 4$ . Whenever you see a rational function and are asked about inverses, simplify the function first, but keep the original domain restriction in mind. Here $f(x)=\frac{x^2-4}{x-2}=x+2$ , valid only for $x\neq 2$ . The range is therefore all real numbers except $4$ : if $y=x+2$ and $x\neq 2$ , then $y\neq 4$ . Since for every $y\neq 4$ exactly one $x=y-2$ maps to it, $f$ is one-to-one, and its inverse is $f^{-1}(x)=x-2$ with domain $x\neq 4$ . The distractor offering $f^{-1}(x)=x+2$ simply confuses $f$ with $f^{-1}$ ; the inverse must undo adding $2$ , so it must subtract $2$ . The distractor that gives $f^{-1}(x)=x-2$ but domain $x\neq 2$ confuses the domain of $f$ with the domain of the inverse; because $4$ is not in the range of $f$ , the inverse must exclude $4$ , not $2$ . The choice that says $f$ is not one-to-one because it is undefined at $x=2$ misses the point: a single missing point in the domain does not destroy injectivity; it simply removes the corresponding output $4$ from the range, so an inverse function exists on the domain $x\neq 4$ . When attacking inverse-function questions, always simplify and compute the range. The domain of the inverse is exactly the range of the original function;it has nothing to do with where the original function is undefined. Make a habit of asking "What outputs can $f$ actually produce?" before choosing the inverse's domain.
The function $f(x)=x^5+x^3+x$ is one-to-one on $\mathbb{R}$ . If $g=f^{-1}$ , what is $g(3)$ ?
Answer: 1 By definition, $g(3)$ is the input whose output is 3. Since $f(1)=1^5+1^3+1=3$ , the inverse value is $g(3)=1$ . The value 273 is $f(3)$ , not $g(3)$ , and 3 is not generally the inverse of 3.
Let $f(x)=2x+3$ if $x<0$ , and $f(x)=x^2+1$ if $x\ge0$ . Which statement best describes $f$ ?
Answer: $f$ is not one-to-one because $f(-1)=f(0)=1$ . Whenever you see a piecewise function and are asked whether it is one-to-one, don't just check each formula separately. A function is one-to-one only if no output value is shared by two different input values across the entire domain. Here, look at the boundary between the pieces: for $x<0$ , $f(-1)=2(-1)+3=1$ , and for $x\ge0$ , $f(0)=0^2+1=1$ . Since $-1\ne0$ but $f(-1)=f(0)$ , the function repeats an output, so it is not one-to-one. The choice saying "not one-to-one because $f(-1)=f(0)=1$ " gives exactly that evidence and is correct. The choice saying "one-to-one because each piece is one-to-one" misses the point: each piece can be one-to-one on its own, but the pieces can still overlap in output values across the boundary. The choice saying "one-to-one because its range is all real numbers" confuses being onto with being one-to-one; having range all real numbers means the function hits every real number, not that every output comes from only one input. Finally, the choice saying "not one-to-one because $x^2+1$ is not one-to-one on $[0,\infty)$ " has the right conclusion but the wrong reason: on $[0,\infty)$ , $x^2+1$ is actually increasing and therefore one-to-one. The real failure is the overlap between the two pieces. For piecewise one-to-one questions, evaluate each formula at boundary points and compare outputs. A single duplicated $y$ -value is enough to disprove one-to-one.
Let $f(x)=2^{x-3}-5$ . What are the domain and range of $f^{-1}$ ?
Answer: domain $x>-5$ , range all real numbers Whenever you're asked about the domain and range of an inverse function, remember that the inverse swaps the roles of inputs and outputs. So the domain of $f^{-1}$ equals the range of $f$ , and the range of $f^{-1}$ equals the domain of $f$ . Here, $f(x)=2^{x-3}-5$ is an exponential function. Its domain is all real numbers because you can plug in any $x$ . Its range is $y>-5$ , since $2^{x-3}$ is always positive, so subtracting 5 gives values strictly greater than $-5$ , approaching $-5$ as a horizontal asymptote but never reaching it. Therefore, for $f^{-1}$ , the domain is $x>-5$ (the range of $f$ ), and the range is all real numbers (the domain of $f$ ). The choice "domain $x\ge -5$ , range all real numbers" is tempting but incorrect because $-5$ is not actually achieved by $f$ , so it cannot be in the domain of $f^{-1}$ . The choice "domain all real numbers, range $y>-5$ " describes the original function $f$ , not its inverse — it reverses the two. "Domain $x>5$ , range all real numbers" uses the wrong sign; the horizontal asymptote is at $-5$ , not $5$ , and there is no shift by $+5$ in this function. Study tip: for inverse functions, just swap the domain and range of the original. Also, watch for open versus closed endpoints—exponential functions never actually touch their horizontal asymptote.
Let $f(x)=4^{x-1}+2$ . Which statement is true?
Answer: $f$ is one-to-one; its range is $(2,\infty)$ , and $f^{-1}(x)=\log_4(x-2)+1$ has domain $x>2$ . The base-4 exponential $4^{x-1}$ is strictly increasing and always positive, so $f$ is one-to-one with range $(2,\infty)$ . Solving $y=4^{x-1}+2$ gives $x-1=\log_4(y-2)$ , so $f^{-1}(x)=\log_4(x-2)+1$ with domain $x>2$ . The range cannot include 2, and the logarithmic expression with $x+2$ has the wrong signs.
The function $f(x)=|x+3|+|x-2|+kx$ , where $k$ is a real constant, is one-to-one for which set of $k$ -values?
Answer: $k > 2$ or $k < -2$ Write the function by intervals: for $x<-3$ , $f(x)=(k-2)x-1$ ; for $-3\le x<2$ , $f(x)=kx+5$ ; for $x\ge2$ , $f(x)=(k+2)x+1$ . Since these linear pieces meet continuously, the function is one-to-one exactly when all three slopes are strictly positive or all three are strictly negative. All positive requires $k-2>0$ , $k>0$ , and $k+2>0$ , so $k>2$ . All negative requires $k-2<0$ , $k<0$ , and $k+2<0$ , so $k<-2$ . Thus the correct set is $k>2$ or $k<-2$ . The option with $k\ge2$ or $k\le-2$ incorrectly includes $k=2$ , where the left piece is constant, and $k=-2$ , where the right piece is constant. The option allowing all real $k$ except $-2$ and $2$ includes values such as $k=0$ , where the middle interval is constant. The interval $-2<k<2$ makes the outer slopes opposite in sign, so the graph turns around and repeats values.
The function $f$ is defined by $f(x)=x^2-6x+10$ with domain $x\ge3$ . If $f^{-1}$ is the inverse of $f$ , which expression and domain are correct?
Answer: $f^{-1}(x)=3+\sqrt{x-1}$ , domain $x\ge1$ When you see a question about inverse functions, your first move is to remember what an inverse does: it undoes the original function, swapping inputs and outputs. Since the domain of $f$ is $x\ge3$ , its range becomes the domain of $f^{-1}$ . Complete the square: $f(x)=(x-3)^2+1$ . With $x\ge3$ , the output is at least $1$ , so the range of $f$ is $y\ge1$ —therefore the domain of $f^{-1}$ must be $x\ge1$ . To find the rule, set $y=(x-3)^2+1$ , solve: $x=3\pm\sqrt{y-1}$ . Because the original domain is $x\ge3$ , choose the plus branch: $f^{-1}(x)=3+\sqrt{x-1}$ . That matches the correct choice. The wrong answer $3-\sqrt{x-1}$ is the negative branch; it would be the inverse if the domain of $f$ were $x\le3$ , not $x\ge3$ . The choice with $3+\sqrt{x+1}$ and domain $x\ge -1$ comes from misreading the completing-the-square step: if you thought you needed $x+1$ , you likely shifted the constant term in the wrong direction. The choice with $3+\sqrt{x-1}$ but domain $x\ge3$ confuses the range of $f$ with the domain of $f$ —the inverse's domain should be the original function's range, not its domain. No fifth option appears here, but the same logic handles any similar distractor.estrength A quick habit: whenever you find an inverse of a quadratic, always check two things—which branch the original domain forces, and whether you used the original function's range as the inverse's domain. That one check eliminates most common traps on inverses.
Which function is one-to-one on all real numbers?
Answer: $f(x)=\frac{x^3}{x^2+1}$ The quadratic factors as $(x-1)^2$ , so it has symmetric outputs. The absolute-value expression is constant on $[-1,1]$ , and cosine is periodic. For $f(x)=\frac{x^3}{x^2+1}$ , if $f(a)=f(b)$ , then $(a-b)(a^2b^2+a^2+ab+b^2)=0$ . The second factor is positive when $a\ne b$ , so $a=b$ . Thus it is one-to-one.
Let $f(x)=\frac{x+2}{x^2-4}$ , with domain $x\ne \pm2$ . Which statement is true?
Answer: $f$ is one-to-one; its range is all real numbers except $0$ and $-\frac14$ , and $f^{-1}(x)=2+\frac1x$ for $x\ne0,-\frac14$ . For $x\ne\pm2$ , the function simplifies to $\frac{1}{x-2}$ , which is one-to-one. The value $0$ is missing because the numerator is never zero. Also, $x=-2$ is excluded from the domain, so the output $-\frac14$ is missing from the range. Thus the inverse domain must exclude both 0 and $-\frac14$ . The other choices omit one of the range restrictions or use the wrong inverse formula.
Let $f(x)=3+\sqrt{2x-4}$ , with domain $x\ge2$ . Which statement about $f^{-1}$ is true?
Answer: $f^{-1}(x)=\frac{(x-3)^2+4}{2}$ , with domain $x\ge3$ . The range of $f$ is $[3,\infty)$ , so the inverse domain is $x\ge3$ . Solving $y=3+\sqrt{2x-4}$ gives $y-3=\sqrt{2x-4}$ , so $(y-3)^2=2x-4$ and $x=\frac{(y-3)^2+4}{2}$ . The choices with $x+3$ or $-4$ are algebraic sign errors, and defining the inverse for all real $x$ ignores the range of the original function.
The function $f(x)=2x^2-8x+5$ is restricted to one side of its vertex so that it is one-to-one. Which restriction makes the range of $f^{-1}$ equal to $[2,\infty)$ ?
Answer: $[2,\infty)$ The range of an inverse function is the domain of the original function. The vertex is at $x=2$ , and the right-side restriction $[2,\infty)$ makes $f$ strictly increasing. The left-side restriction $(-\infty,2]$ would make the inverse range $(-\infty,2]$ . The interval $[0,\infty)$ is not one-sided about the vertex and is not one-to-one.
For what real value of $a$ is $f(x)=\frac{ax+1}{x-2}$ its own inverse? Assume the natural domain $x\ne2$ .
Answer: $a=2$ Solving $y=\frac{ax+1}{x-2}$ for $x$ gives $x=\frac{2y+1}{y-a}$ , so $f^{-1}(x)=\frac{2x+1}{x-a}$ . Setting $f^{-1}(x)=f(x)$ gives $\frac{ax+1}{x-2}=\frac{2x+1}{x-a}$ . Cross-multiplying and equating coefficients yields $a=2$ . The other values do not make the two formulas identical.
Let $f(x)=\frac{x}{|x|+1}$ , with domain all real numbers. Which of the following is the inverse of $f$ ?
Answer: $f^{-1}(x)=\frac{x}{1+x}$ for $-1<x<0$ , and $f^{-1}(x)=\frac{x}{1-x}$ for $0\le x<1$ . For $x<0$ , $f(x)=\frac{x}{1-x}$ , whose range is $(-1,0)$ , and solving gives $f^{-1}(x)=\frac{x}{1+x}$ . For $x\ge0$ , $f(x)=\frac{x}{1+x}$ , whose range is $[0,1)$ , and solving gives $f^{-1}(x)=\frac{x}{1-x}$ . The swapped formulas ignore which piece produces negative outputs, and the single formulas do not correctly invert both branches.
For which real value(s) of $k$ does $f(x)=x^2+2x+k$ , restricted to the domain $x\le -1$ , have an inverse function?
Answer: All real $k$ , because changing $k$ shifts the graph vertically and does not change one-to-one behavior. The vertex of $x^2+2x+k$ is at $x=-1$ . On $x\le -1$ , the function is strictly decreasing, so every horizontal line meets the graph at most once. The constant $k$ only shifts the graph vertically; it changes the range but does not affect whether the function is one-to-one. Therefore every real $k$ works. The other choices confuse the location or range of the vertex with injectivity.
The one-to-one function $f$ satisfies $f(2)=5$ , $f(5)=7$ , and $f(7)=2$ . If $g$ is the inverse of $f$ , what is $g(f(7))+g(2)$ ?
Answer: 14 Whenever you see a function and its inverse, focus on reversing the input-output pairs. Since $f$ gives $2\to 5$ , $5\to 7$ , and $7\to Car2$ , its inverse $g$ gives $5\to Car2$ , $7\to Car5$ , and $2\to Car7$ . Now evaluate carefully. The inner value is $f(7)=2$ . Therefore $g(f(7))=g(2)$ , and from the inverse pairs $g(2)=7$ . The second term is also $g(2)=7$ . So \[ g(f(7))+g(2)=7+7=14. The choice $5$ comes from confusing $g$ with $f$—for example, using $f(2)=5$ to claim $g(2)=5$. But $f(2)=5$ actually means $g(5)=2$, not the other way around. The choice $7$ ignores the second term: it gives the value of $g(f(7))$ alone, or it treats $+g(2)$ as adding zero. The choice "Cannot be determined" overlooks that although not every value of $f$ is given, the values needed here are fully determined by the inverted pairs. Study tip: when inverse-function values appear, first write down the reversed pairs from the given facts. That simple step stops the most common trap on this exam: using the original function's table where you needthe inverse's.
Let $f(x)=x^2+1$ with domain $x\ge0$ , and let $g(x)=\sqrt{x-1}$ with domain $x\ge1$ . Which statement correctly verifies that $g$ is the inverse of $f$ ?
Answer: $f(g(x))=x$ for $x\ge1$ and $g(f(x))=x$ for $x\ge0$ . When you see an inverse-function question, remember: two functions are inverses only if their compositions return $x$ , and only on the correct domains. Start by testing both compositions. Since $g(x)=\sqrt{x-1}$ , we have $f(g(x))=(\sqrt{x-1})^2+1=x-1+1=x.$ But $g(x)$ is only defined for $x\ge1$ , so this holds for $x\ge1$ , not for all real $x$ . Next, $g(f(x))=\sqrt{(x^2+1)-1}=\sqrt{x^2}=|x|.$ Because $f$ 's domain is $x\ge0$ , $|x|=x$ here, so $g(f(x))=x$ for $x\ge0$ . Therefore, the statement " $f(g(x))=x$ for $x\ge1$ and $g(f(x))=x$ for $x\ge0$ " correctly verifies that $g$ is the inverse of $f$ . Why are the others wrong? The statement using only $f(0)=1$ and $g(1)=0$ checks just one point, not every input. The claim that $f(g(x))=x$ for all real $x$ ignores that $g$ is undefined for $x<1$ . The claim that $g(f(x))=x$ for all real $x$ fails for negative $x$ , because $\sqrt{x^2}=|x|$ , not $x$ . Study tip: to verify inverses, always check both compositions and respect the given domains. One matching pair of values is never enough.
Which condition guarantees that a function $f$ has an inverse?
Answer: For all $a,b$ in the domain of $f$ , if $f(a)=f(b)$ , then $a=b$ . Whenever you see a question about inverse functions, your first thought should be: an inverse exists only if each output comes from exactly one input. That is, the function must be one-to-one. The condition "for all $a,b$ in the domain of $f$ , if $f(a)=f(b)$ , then $a=b$ " is the precise definition of one-to-one. It guarantees that when you reverse the function, there is no ambiguity about which input produced a given output. The vertical line test only tells you that a graph represents a function, not that the function has an inverse. An even function with domain all real numbers, such as $f(x)=x^2$ , actually fails the one-to-one condition because $f(1)=f(-1)$ , so it cannot have an inverse. And just having the range equal the domain says nothing about whether different inputs could collide; for example, a constant function on a one-element domain technically works, but on a larger domain it fails because many inputs share the same output. So when you see inverse-function questions, ask yourself: Could two different inputs produce the same output? If yes, no inverse. Graphically, check the horizontal line test — any horizontal line should hit the curve at most once. That visual habit will help you quickly eliminate tempting but incomplete conditions.
Which function has an inverse whose domain is $x>2$ ?
Answer: $f(x)=2^x+2$ Whenever you see a question about the domain of an inverse function, translate it immediately: the domain of the inverse is exactly the range of the original function. So you need to find which function has range $x>2$ (more precisely, $y>2$ ). The function $f(x)=2^x+2$ is correct because $2^x$ is always positive, so every output is greater than $2$ . Its range is $(2,\infty)$ , meaning the inverse has domain $x>2$ . Now look at the wrong choices. The function $f(x)=\log_2(x-2)$ has domain $x>2$ , but its range is all real numbers, so its inverse has domain all real numbers — this is a classic trap where you confuse the original domain with the inverse's domain. The function $f(x)=2^x-2$ has range $(-2,\infty)$ , so its inverse's domain is $x>-2$ , not $x>2$ . Finally, $f(x)=\log_2(x+2)$ has domain $x>-2$ and range all real numbers, so its inverse's domain is also all real numbers. Your study tip: when a question asks about the inverse's domain, always ask "what is the range of the original function?" Remember that exponential functions have ranges like $(0,\infty)$ , and vertical shifts move that range up or down. Log functions, on the other hand, always have range all real numbers.

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