One-to-One Functions - College Algebra

1. The inverse function $f^{-1}(x)$ exists for $f(x) = \sqrt{x+3} - 2$ with domain $[-3, \infty)$. If $g(x) = [f^{-1}(x)]^2 + 1$, what is the domain of $g(x)$?
   Answer: $[-2, \infty)$ because this represents the range of the original function $f(x)$ 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.
2. Function $f(x) = x^3 + ax^2 + bx + c$ passes through points $(-1, 2)$, $(0, 3)$, and $(1, 6)$. If this cubic function is one-to-one, which constraint must be satisfied?
   Answer: The values $a$ and $b$ must satisfy $4a^2 < 12b$ to prevent local extrema 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.
3. A function $m(x)$ satisfies $m(x+4) = m(x)$ for all $x$ in its domain, and $m(x)$ is continuous. Which statement about the one-to-one property of $m(x)$ is correct?
   Answer: $m(x)$ cannot be one-to-one unless it is constant, because periodicity creates repeated output values 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.
4. Consider functions $p(x) = 2^x$ and $q(x) = \log_2(x)$. The composition $r(x) = p(q(x-1)) + 3$ has what relationship to the one-to-one property?
   Answer: $r(x)$ is one-to-one because it equals $x-1+3 = x+2$ for $x > 1$, which is strictly increasing 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.
5. Consider the transformation $T(x) = \frac{1}{2}f(x-3) + 1$ where $f(x)$ is a known one-to-one function. Which statement correctly describes the one-to-one property of $T(x)$?
   Answer: $T(x)$ is one-to-one because all transformations applied preserve the horizontal line test 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.

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