One-to-One Functions - Math 3

The function $f(x) = |x - a| + |x - b|$ where $a < b$ is defined for all real $x$ . On which interval is $f$ guaranteed to be one-to-one?
Answer: $[b, \infty)$ , where the derivative exists and is constant, ensuring strict monotonicity We need to analyze $f(x) = |x - a| + |x - b|$ where $a < b$ . For $x < a$ : both $(x-a) < 0$ and $(x-b) < 0$ , so $f(x) = -(x-a) - (x-b) = -2x + (a+b)$ , which decreases with slope $-2$ . For $a \leq x \leq b$ : $(x-a) \geq 0$ and $(x-b) \leq 0$ , so $f(x) = (x-a) - (x-b) = b-a$ , which is constant. For $x > b$ : both $(x-a) > 0$ and $(x-b) > 0$ , so $f(x) = (x-a) + (x-b) = 2x - (a+b)$ , which increases with slope $2$ . Therefore: on $(-\infty, a]$ , $f$ decreases (one-to-one); on $[a,b]$ , $f$ is constant (not one-to-one unless $a=b$ ); on $[b, \infty)$ , $f$ increases (one-to-one). Choice B is correct. Choice A works but choice B represents a cleaner interval. Choice C is wrong because constant functions aren't one-to-one. Choice D is incorrect because it spans both decreasing and constant regions.
A function $f: \mathbb{R} \to \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y) + 2xy$ for all $x, y \in \mathbb{R}$ . If $f(1) = 3$ , determine whether $f$ can be one-to-one.
Answer: No, because the functional equation forces $f(x) = x^2 + 2x$ , and quadratics are not globally one-to-one Let's find $f$ explicitly. Setting $y = 0$ : $f(x + 0) = f(x) + f(0) + 2x \cdot 0$ , so $f(x) = f(x) + f(0)$ , which gives $f(0) = 0$ . Setting $x = y = 1$ : $f(2) = f(1) + f(1) + 2 = 6 + 2 = 8$ . Setting $x = 2, y = -1$ : $f(1) = f(2) + f(-1) + 2(2)(-1) = 8 + f(-1) - 4$ , so $3 = 4 + f(-1)$ , giving $f(-1) = -1$ . By induction and the functional equation's structure, we can show $f(x) = x^2 + 2x$ . Verification: $f(x+y) = (x+y)^2 + 2(x+y) = x^2 + 2xy + y^2 + 2x + 2y = (x^2 + 2x) + (y^2 + 2y) + 2xy = f(x) + f(y) + 2xy$ ✓. Since $f(x) = x^2 + 2x = x(x+2)$ , we have $f'(x) = 2x + 2 = 0$ when $x = -1$ . The function decreases on $(-\infty, -1)$ and increases on $(-1, \infty)$ , so it's not globally one-to-one. For example, $f(-2) = 4 - 4 = 0 = f(0)$ .
Consider $f(x) = e^{x^2 - 2x}$ . To find the maximal intervals on which $f$ is one-to-one, which approach yields the correct answer?
Answer: Find where $f'(x) = 0$ : this gives $x = 1$ , so $f$ is one-to-one on $(-\infty, 1]$ and $[1, \infty)$ separately When determining where a function is one-to-one, you need to find intervals where the function is strictly monotonic (either always increasing or always decreasing). The key insight is that a function can only be one-to-one on an interval if it never "turns around" – that is, if its derivative doesn't change sign. The correct approach is to find where $f'(x) = 0$ to locate critical points. For $f(x) = e^{x^2 - 2x}$ , using the chain rule: $f'(x) = e^{x^2 - 2x} \cdot (2x - 2) = e^{x^2 - 2x} \cdot 2(x - 1)$ . Since $e^{x^2 - 2x} > 0$ always, $f'(x) = 0$ only when $x = 1$ . For $x < 1$ , $f'(x) < 0$ (decreasing), and for $x > 1$ , $f'(x) > 0$ (increasing). Therefore, $f$ is one-to-one on $(-\infty, 1]$ and $[1, \infty)$ separately, making A correct. Option B is wrong because while $e^u$ is one-to-one as a function of $u$ , when $u = x^2 - 2x$ is not one-to-one in $x$ , the composition $f(x) = e^{x^2 - 2x}$ won't be one-to-one either. Option C incorrectly assumes that having a minimum value makes a function one-to-one – this confuses range restrictions with injectivity. Option D correctly identifies the monotonic intervals but wrongly suggests this doesn't determine the answer. Remember: To find maximal one-to-one intervals, always locate critical points using the derivative, then determine where the function is strictly monotonic between these points.
A function $h(x)$ satisfies the property that $h(a) = h(b)$ implies $a = b$ for all $a, b$ in its domain $D$ . If $D = [-3, 0) \cup (0, 3]$ , which statement about $h(x)$ is necessarily true?
Answer: $h(x)$ is one-to-one on $D$ , and this property is unaffected by the discontinuity at $x = 0$ The given property ' $h(a) = h(b)$ implies $a = b$ for all $a, b$ in domain $D$ ' is precisely the definition of a one-to-one function. This property holds regardless of whether the domain is connected or has discontinuities. The fact that $x = 0$ is not in the domain doesn't affect the one-to-one property, as this property only concerns pairs of points that are both in the domain. Choice B incorrectly assumes disconnected domains prevent one-to-one functions. Choice C contradicts the given condition. Choice D incorrectly focuses on boundary values rather than the given logical condition.
A student claims that the function $g(x) = \frac{x}{x^2 + 1}$ is one-to-one on the interval $[-1, 1]$ because it passes through the origin and appears to be increasing near $x = 0$ . Which analysis most accurately evaluates this claim?
Answer: The claim is incorrect because $g'(x) = \frac{1-x^2}{(x^2+1)^2}$ changes sign within $[-1, 1]$ , but the function is still one-to-one there To determine if $g(x) = \frac{x}{x^2 + 1}$ is one-to-one on $[-1, 1]$ , we find $g'(x) = \frac{1-x^2}{(x^2+1)^2}$ . This derivative is positive for $|x| < 1$ and equals zero at $x = \pm 1$ . Since $g'(x) > 0$ throughout $(-1, 1)$ , the function is strictly increasing on $[-1, 1]$ , making it one-to-one despite the student's flawed reasoning. Choice A gives correct conclusion but insufficient reasoning. Choice C is irrelevant to the given interval. Choice D addresses continuity, not one-to-one property. The student's reasoning about behavior 'near zero' is insufficient for the entire interval.
Consider the function $f(x) = x^3 + ax^2 + bx + c$ where $a$ , $b$ , and $c$ are constants. For this function to be one-to-one on its entire domain $(-\infty, \infty)$ , which condition on the discriminant of $f'(x)$ must be satisfied?
Answer: The discriminant $4a^2 - 12b$ must be negative to ensure $f'(x)$ has no real roots When you encounter questions about one-to-one functions, you need to think about when a function passes the horizontal line test—meaning no horizontal line intersects the graph more than once. For polynomial functions, this connects directly to the behavior of the derivative. A cubic function $f(x) = x^3 + ax^2 + bx + c$ is one-to-one when it's always increasing (since the leading coefficient is positive). This happens when $f'(x) \geq 0$ for all $x$ , and more specifically, when $f'(x) > 0$ everywhere except possibly at isolated points. The derivative is $f'(x) = 3x^2 + 2ax + b$ , which is a quadratic with positive leading coefficient. For $f'(x)$ to never become negative, it cannot have two distinct real roots (which would create a valley where $f'(x) < 0$ ). The discriminant of this quadratic is $(2a)^2 - 4(3)(b) = 4a^2 - 12b$ . When this discriminant is negative, $f'(x)$ has no real roots and stays positive everywhere, ensuring $f(x)$ is strictly increasing and therefore one-to-one. Option A is wrong because two distinct roots would create intervals where $f'(x) < 0$ , making the function decrease. Option B incorrectly assumes all cubics with positive leading coefficients are one-to-one—this ignores the critical role of the middle terms. Option C is wrong because a repeated root would create a point where $f'(x) = 0$ , potentially allowing the function to have a horizontal tangent and fail the one-to-one test. Remember: for polynomials to be one-to-one, focus on when the derivative maintains its sign. The discriminant tells you about the roots of that derivative.
Two students are debating whether $g(x) = \ln(x^2 + 1)$ is one-to-one on $\mathbb{R}$ . Student A argues it's not one-to-one because $x^2 + 1$ is even. Student B argues it is one-to-one because the natural logarithm is always one-to-one. Which analysis is most mathematically sound?
Answer: Student A is correct; since $g(-x) = g(x)$ for all $x$ , the function is even and cannot be one-to-one Student A correctly identifies that $g(x) = \ln(x^2 + 1)$ is an even function because $g(-x) = \ln((-x)^2 + 1) = \ln(x^2 + 1) = g(x)$ . Any even function that is not constant cannot be one-to-one on $\mathbb{R}$ because it produces the same output for $x$ and $-x$ when $x \neq 0$ . Student B's reasoning is flawed because while $\ln(u)$ is one-to-one, $u = x^2 + 1$ is not one-to-one as a function of $x$ . Choice B incorrectly applies composition properties. Choice C is partially correct about intervals but doesn't address the students' reasoning. Choice D suggests an unnecessary approach when symmetry arguments suffice.
Let $f(x) = \sqrt{x^2 - 6x + 8}$ and consider its natural domain. If we restrict $f$ to make it one-to-one, which of the following represents the largest possible domain for such a restriction?
Answer: $[4, \infty)$ , using only the right branch where the function increases monotonically First, find the natural domain: $x^2 - 6x + 8 \geq 0$ , which factors as $(x-2)(x-4) \geq 0$ . This gives domain $(-\infty, 2] \cup [4, \infty)$ . The expression $x^2 - 6x + 8 = (x-3)^2 - 1$ has vertex at $x = 3$ . On $(-\infty, 2]$ , we have $x - 3 < 0$ , so $(x-3)^2$ decreases as $x$ increases toward 2, making $f(x)$ decrease. On $[4, \infty)$ , we have $x - 3 > 0$ , so $(x-3)^2$ increases as $x$ increases, making $f(x)$ increase. Both intervals alone give one-to-one functions, but $[4, \infty)$ is larger than $(-\infty, 2]$ . Choice A is wrong because the complete domain isn't one-to-one ( $f(1) = f(5) = \sqrt{3}$ ). Choice B uses an invalid domain since $f(3)$ is undefined. Choice C gives a smaller domain than choice D.
Consider $h(x) = \ln(x^2 - 4x + 5)$ . To determine if $h$ is one-to-one on its natural domain, which analysis is correct?
Answer: The function $h$ is one-to-one on $[2, \infty)$ where $x^2 - 4x + 5$ increases, but not on $(-\infty, \infty)$ First, note that $x^2 - 4x + 5 = (x-2)^2 + 1 \geq 1 > 0$ for all real $x$ , so $h$ is defined on all of $\mathbb{R}$ . While $\ln$ is one-to-one, the composition $h(x) = \ln(g(x))$ is one-to-one only if $g(x) = x^2 - 4x + 5$ is one-to-one. Since $g(x)$ is a parabola opening upward with vertex at $x = 2$ , it decreases on $(-\infty, 2]$ and increases on $[2, \infty)$ . Therefore $g$ is not one-to-one on $\mathbb{R}$ (e.g., $g(1) = g(3) = 2$ ), so $h$ is not one-to-one on $\mathbb{R}$ . However, $h$ is one-to-one on $[2, \infty)$ where $g$ is strictly increasing. Choice A incorrectly assumes composition of one-to-one functions is always one-to-one. Choice C misapplies the discriminant. Choice D incorrectly focuses on the sign of $\ln$ output rather than injectivity.
Consider the function $f(x) = x^3 - 3x^2 + 2x$ on the interval $[0, 3]$ . To determine if $f$ is one-to-one on this domain, which of the following approaches would provide the most definitive conclusion?
Answer: Verify that $f$ has no repeated values by solving $f'(x) = 0$ and checking function behavior at critical points To determine if a function is one-to-one, we need to check if it ever has the same output for different inputs. For $f(x) = x^3 - 3x^2 + 2x$ , we find $f'(x) = 3x^2 - 6x + 2$ . Setting $f'(x) = 0$ gives critical points, and we must check if the function has local maxima and minima that could create repeated values. Choice A is incorrect because $f'(x)$ is not always positive on $[0,3]$ . Choice C only checks three points, which is insufficient. Choice D is incorrect because the leading coefficient alone doesn't determine one-to-one behavior on a restricted domain.
A function $h(x)$ is defined piecewise as $h(x) = \begin{cases} x^2 - 4 & \text{if } x \leq 0 \\ -x + 4 & \text{if } x > 0 \end{cases}$ . For $h(x)$ to be one-to-one on its entire domain, which modification would be necessary?
Answer: Restrict the domain to $x \geq 0$ to use only the linear portion of the function For the current piecewise function, the first piece $x^2 - 4$ on $x \leq 0$ gives values from $-4$ to $+\infty$ , while the second piece $-x + 4$ on $x > 0$ gives values from $(-\infty, 4)$ . These ranges overlap significantly, meaning the same y-value can come from both pieces, violating the one-to-one property. Choice C restricts to only the linear piece, which is strictly decreasing and therefore one-to-one. Choice A doesn't solve the overlap problem. Choice B creates two increasing pieces that still overlap. Choice D is incorrect because the function has repeated y-values.
Consider the function $f(x) = |x - 2| + |x + 1|$ on different domains. Which statement about the one-to-one property of this function is correct?
Answer: $f(x)$ is one-to-one on $(-\infty, -1]$ and $[2, \infty)$ , but not on any interval containing $[-1, 2]$ To analyze $f(x) = |x - 2| + |x + 1|$ , we consider the critical points $x = -1$ and $x = 2$ . For $x \leq -1$ : $f(x) = (2-x) + (-x-1) = 1-2x$ (decreasing). For $-1 < x < 2$ : $f(x) = (2-x) + (x+1) = 3$ (constant). For $x \geq 2$ : $f(x) = (x-2) + (x+1) = 2x-1$ (increasing). The function is one-to-one on $(-\infty, -1]$ and $[2, \infty)$ where it's strictly monotonic, but not on any interval containing part of $[-1, 2]$ where it's constant. Choice A incorrectly states the monotonicity. Choice B misses the decreasing portion. Choice D is false about absolute value functions in general.
A function $f$ is defined implicitly by $x^3 + y^3 = 6xy$ near the point $(3, 3)$ . In a neighborhood of this point, is $y$ a one-to-one function of $x$ ?
Answer: Yes, because implicit differentiation gives $\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x} \neq 0$ near $(3,3)$ Using implicit differentiation on $x^3 + y^3 = 6xy$ : $3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$ . Solving for $\frac{dy}{dx}$ : $3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$ , so $\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$ . At $(3,3)$ : $\frac{dy}{dx} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1 \neq 0$ . By the implicit function theorem, since $\frac{\partial}{\partial y}(x^3 + y^3 - 6xy) = 3y^2 - 6x = 27 - 18 = 9 \neq 0$ at $(3,3)$ , there exists a neighborhood where $y$ is a unique function of $x$ . Since $\frac{dy}{dx} \neq 0$ in this neighborhood (by continuity), the function is strictly monotonic and hence one-to-one locally. Choice C mentions the implicit function theorem but doesn't verify the conditions. Choice B is incorrect about intersecting branches locally. Choice D is unnecessarily restrictive.
The function $h(x) = \frac{ax + b}{cx + d}$ where $ad - bc \neq 0$ is defined on $\mathbb{R} \setminus \{-\frac{d}{c}\}$ . For which condition on the parameters is $h$ guaranteed to be one-to-one on its domain?
Answer: $ad - bc \neq 0$ , which is the condition already given and sufficient for injectivity A rational function of the form $h(x) = \frac{ax + b}{cx + d}$ (a Möbius transformation) is one-to-one on its domain if and only if $ad - bc \neq 0$ . This condition ensures the function is invertible. To verify: if $h(x_1) = h(x_2)$ , then $\frac{ax_1 + b}{cx_1 + d} = \frac{ax_2 + b}{cx_2 + d}$ . Cross-multiplying and simplifying leads to $(ad - bc)(x_1 - x_2) = 0$ . If $ad - bc \neq 0$ , then $x_1 = x_2$ . Choice A is incorrect because the sign of $ad - bc$ doesn't matter for injectivity. Choice B is wrong because $c \neq 0$ alone isn't sufficient. Choice D is incorrect because the function can be one-to-one even when $c = 0$ (making it linear) as long as $a \neq 0$ .
Consider the inverse trigonometric function $g(x) = \arctan(x^2 - 1)$ . On which of the following intervals is $g$ one-to-one?
Answer: $[1, \infty)$ , where $x^2 - 1$ is strictly increasing and $\arctan$ preserves this property When analyzing whether a composite function like $g(x) = \arctan(x^2 - 1)$ is one-to-one, you need to examine both the inner function $x^2 - 1$ and how the outer function $\arctan$ behaves with it. Since $\arctan$ is strictly increasing on all real numbers, $g(x)$ will be one-to-one precisely where the inner function $x^2 - 1$ is one-to-one. The parabola $x^2 - 1$ decreases on $(-\infty, 0]$ and increases on $[0, \infty)$ , so it's only one-to-one on intervals that don't cross $x = 0$ . Answer C is correct because on $[1, \infty)$ , the function $x^2 - 1$ is strictly increasing. Since $\arctan$ is also strictly increasing, their composition $g(x)$ remains strictly increasing and therefore one-to-one on this interval. Answer A is wrong because $x^2 - 1$ is not one-to-one over all real numbers—it takes the same values on both sides of $x = 0$ . For example, $g(-2) = g(2) = \arctan(3)$ . Answer B fails because while $g(x) \leq 0$ on $[-1, 1]$ , the function still isn't one-to-one since $x^2 - 1$ decreases then increases on this interval, creating the same duplicate value problem. Answer D is incorrect because although $x^2 - 1 \geq 0$ on $(-\infty, -1] \cup [1, \infty)$ , this union includes both decreasing and increasing portions of the parabola, so $g$ won't be one-to-one across the entire set. Study tip: For composite functions to be one-to-one, focus on finding intervals where the inner function is monotonic (strictly increasing or decreasing).
A function $g$ is defined such that $g(2x - 1) = x^3 + 1$ for all real numbers $x$ . To determine if $g$ is one-to-one on its natural domain, which approach is most mathematically sound?
Answer: Assume $g(a) = g(b)$ and show $a = b$ using the given functional equation directly The most direct approach is to use the definition of one-to-one. If $g(a) = g(b)$ , we need to show $a = b$ . From the given equation, if $a$ and $b$ are in the range of $2x-1$ (which is all real numbers), then $a = 2x_1 - 1$ and $b = 2x_2 - 1$ for some $x_1, x_2$ . Then $g(a) = x_1^3 + 1$ and $g(b) = x_2^3 + 1$ . If these are equal, then $x_1^3 = x_2^3$ , so $x_1 = x_2$ , which means $a = b$ . Choice A requires finding $g(x)$ explicitly, which is unnecessary. Choice B incorrectly assumes composition preserves injectivity in this direction. Choice C misunderstands the relationship between the functions.

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