College Algebra Quiz: One To One Functions

What this quiz covers

This quiz focuses on One To One Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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.

Question 1

1. $T(x)$ is one-to-one because horizontal shifts and vertical shifts preserve the one-to-one property
2. $T(x)$ is one-to-one because all transformations applied preserve the horizontal line test (correct answer)
3. $T(x)$ is not one-to-one because the vertical compression by factor $\frac{1}{2}$ violates the horizontal line test
4. $T(x)$ is not one-to-one because the combination of compression and shifts creates repeated output values

Explanation: If $f(x)$ is one-to-one, then $T(x) = \frac{1}{2}f(x-3) + 1$ is also one-to-one. Let's verify: suppose $T(x_1) = T(x_2)$ for some $x_1 \neq x_2$. Then $\frac{1}{2}f(x_1-3) + 1 = \frac{1}{2}f(x_2-3) + 1$, which simplifies to $\frac{1}{2}f(x_1-3) = \frac{1}{2}f(x_2-3)$, so $f(x_1-3) = f(x_2-3)$. Since $f$ is one-to-one, this implies $x_1-3 = x_2-3$, so $x_1 = x_2$, contradicting our assumption. The horizontal shift, vertical compression by positive factor, and vertical shift all preserve injectivity. Choice A is incomplete (missing compression), C is incorrect (compression by positive factor preserves one-to-one), and D is false.

Question 2

1. $m(x)$ cannot be one-to-one unless it is constant, because periodicity creates repeated output values (correct answer)
2. $m(x)$ can be one-to-one if its domain is restricted to an interval of length less than 4
3. $m(x)$ is one-to-one on its natural domain because the horizontal line test applies to each period separately
4. $m(x)$ cannot be one-to-one because continuous periodic functions must have at least one local maximum and minimum

Explanation: A periodic function with period 4 means $m(x) = m(x+4)$ for all $x$. If $m$ is not constant, then there exist values $a$ and $b$ with $a \neq b$ such that $m(a) \neq m(b)$. But then $m(a) = m(a+4) = m(a+8) = \ldots$ and $m(b) = m(b+4) = m(b+8) = \ldots$. For the function to be one-to-one on its entire domain, each output value can be achieved at most once. However, if the domain includes multiple periods, then $m(a) = m(a+4k)$ for integer $k$, violating the one-to-one property unless $m$ is constant. Choice B discusses restriction (different question), C misunderstands the horizontal line test globally, and D makes unnecessary assumptions about extrema.

Question 3

1. $(-\infty, \infty)$ because squaring the inverse function eliminates all domain restrictions
2. $[-3, \infty)$ because $g(x)$ inherits the same domain restrictions as $f(x)$
3. $[1, \infty)$ because $g(x)$ squares the inverse function and adds 1, shifting the minimum value
4. $[-2, \infty)$ because this represents the range of the original function $f(x)$ (correct answer)

Explanation: When dealing with composite functions involving inverse functions, the key insight is understanding how domains and ranges transform. The domain of $g(x) = [f^{-1}(x)]^2 + 1$ depends entirely on where $f^{-1}(x)$ is defined. To find this, you need the range of the original function $f(x) = \sqrt{x+3} - 2$. Since $f(x)$ has domain $[-3, \infty)$, the smallest input is $x = -3$, giving $f(-3) = \sqrt{-3+3} - 2 = 0 - 2 = -2$. As $x$ increases without bound, $\sqrt{x+3}$ grows without bound, so $f(x)$ approaches infinity. Therefore, the range of $f(x)$ is $[-2, \infty)$. The domain of $f^{-1}(x)$ equals the range of $f(x)$, which is $[-2, \infty)$. Since $g(x) = [f^{-1}(x)]^2 + 1$, and squaring and adding constants don't create additional restrictions, $g(x)$ has the same domain as $f^{-1}(x)$: $[-2, \infty)$. Option A incorrectly assumes squaring eliminates domain restrictions—it doesn't expand where the function is originally undefined. Option B confuses the domain of $f(x)$ with the domain of $g(x)$; these are different because we're working with the inverse. Option C misinterprets how the operations affect the domain rather than the range. Remember: when finding the domain of a composition involving an inverse function, always identify the range of the original function first—this becomes your inverse function's domain and determines everything that follows.

Question 4

1. $r(x)$ is one-to-one because it equals $x-1+3 = x+2$ for $x > 1$, which is strictly increasing (correct answer)
2. $r(x)$ is one-to-one because it equals $x-1+3 = x+2$ for $x > 0$, which is a linear function
3. $r(x)$ is not one-to-one because the domain restriction to $x > 1$ creates a limited range of outputs
4. $r(x)$ is not one-to-one because compositions of inverse functions cancel in a way that creates repeated values

Explanation: First, let's find $r(x) = p(q(x-1)) + 3 = 2^{\log_2(x-1)} + 3$. Since $2^{\log_2(u)} = u$ for $u > 0$, we have $r(x) = (x-1) + 3 = x + 2$. However, the domain is restricted by $q(x-1) = \log_2(x-1)$, which requires $x-1 > 0$, so $x > 1$. On the domain $(1, \infty)$, $r(x) = x + 2$ is strictly increasing, hence one-to-one. Choice B has the wrong domain ($x > 0$ instead of $x > 1$), choice C incorrectly suggests domain restrictions prevent one-to-one property, and choice D misunderstands how inverse function compositions work.

Question 5

1. The discriminant of $f'(x) = 0$ must be negative, ensuring no critical points exist
2. The discriminant of $f'(x) = 0$ must be negative, ensuring $f'(x) > 0$ for all $x$
3. The values $a$ and $b$ must satisfy $4a^2 < 12b$ to prevent local extrema (correct answer)
4. The coefficient $a$ must equal zero to ensure the function is strictly increasing

Explanation: For a cubic function to be one-to-one, it must be strictly monotonic, which means $f'(x)$ should never equal zero (or have at most one zero that's not a local extremum). Given the three points, we can solve: $f(0) = c = 3$, $f(-1) = -1 + a - b + 3 = 2$ gives $a - b = 0$, and $f(1) = 1 + a + b + 3 = 6$ gives $a + b = 2$. Solving: $a = 1, b = 1$. So $f(x) = x^3 + x^2 + x + 3$ and $f'(x) = 3x^2 + 2x + 1$. For no real critical points, the discriminant $\Delta = 4 - 12 = -8 < 0$. In general, for $f'(x) = 3x^2 + 2ax + b$, we need $\Delta = 4a^2 - 12b < 0$, or $4a^2 < 12b$. Choices A and B are redundant, and D is too restrictive.