Function Notation & Inverses - IB Mathematics Analysis and Approaches

1. Let $f(x) = x+3$ and $g(x) = \frac{1}{x-1}$. The function $h$ is defined as $h(x) = (f \circ g)(x)$. Find the inverse function $h^{-1}(x)$.
   Answer: $h^{-1}(x) = 1 + \frac{1}{x-3}$ When you encounter composite functions and inverse functions together, you need to work systematically: first find the composite function, then find its inverse using the standard algebraic method. Start by finding $h(x) = (f \circ g)(x) = f(g(x))$. Since $g(x) = \frac{1}{x-1}$ and $f(x) = x + 3$, you substitute: $h(x) = f\left(\frac{1}{x-1}\right) = \frac{1}{x-1} + 3$. To find $h^{-1}(x)$, use the standard method. Let $y = \frac{1}{x-1} + 3$, then solve for $x$: - $y - 3 = \frac{1}{x-1}$ - $x - 1 = \frac{1}{y-3}$ - $x = 1 + \frac{1}{y-3}$ Therefore, $h^{-1}(x) = 1 + \frac{1}{x-3}$. Option A gives $\frac{1}{x-3}$, which misses the "+1" term entirely. Option B uses $\frac{1}{x+3}$, suggesting confusion about which operation to reverse when solving $y - 3 = \frac{1}{x-1}$. Option C gives $1 + \frac{1}{x+3}$, combining both errors: wrong sign in the denominator and missing the correct relationship. You can verify answer D by checking that $h(h^{-1}(x)) = x$: substituting $1 + \frac{1}{x-3}$ into $h$ should return $x$. Study tip: For composite function inverses, always find the composite first, then apply the inverse method. Don't try to find $(f^{-1} \circ g^{-1})(x)$ - that's not the same as $(f \circ g)^{-1}(x)$!
2. Find the domain of the function $f(x) = \sqrt{\ln(x+2)}$.
   Answer: $x \ge -1$ For the function to be defined, two conditions must be met. First, the argument of the natural logarithm must be positive: $x+2 > 0 \Rightarrow x > -2$. Second, the argument of the square root must be non-negative: $\ln(x+2) \ge 0$. To solve this inequality, we can exponentiate both sides: $e^{\ln(x+2)} \ge e^0$, which simplifies to $x+2 \ge 1$, or $x \ge -1$. We must satisfy both conditions simultaneously: $x > -2$ and $x \ge -1$. The intersection of these two conditions is $x \ge -1$. Thus, the domain is $[-1, \infty)$.
3. Let $f(x) = \sqrt{2x+3} - 1$. Find the range of $f^{-1}(x)$.
   Answer: $y \ge -3/2$ The range of the inverse function $f^{-1}(x)$ is the domain of the original function $f(x)$. To find the domain of $f(x) = \sqrt{2x+3} - 1$, we must ensure that the expression under the square root is non-negative. So, we require $2x+3 \ge 0$. Solving for $x$ gives $2x \ge -3$, which means $x \ge -3/2$. Therefore, the domain of $f(x)$ is $[-3/2, \infty)$. This is the range of $f^{-1}(x)$, so the range is $y \ge -3/2$.
4. The function $f$ is defined by $f(x) = \frac{3}{x+2} - 1$. What is the range of $f^{-1}(x)$?
   Answer: $y \ne -2$ The range of the inverse function $f^{-1}(x)$ is equal to the domain of the original function $f(x)$. The function $f(x) = \frac{3}{x+2} - 1$ is a rational function. Its domain is all real numbers except for values of $x$ that make the denominator zero. The denominator is $x+2$, which is zero when $x=-2$. Therefore, the domain of $f(x)$ is $\{x \in \mathbb{R} \mid x \ne -2\}$. This domain is the range of $f^{-1}(x)$. So, the range of $f^{-1}(x)$ is $\{y \in \mathbb{R} \mid y \ne -2\}$.
5. Find the range of the function $f(x) = \frac{2x-1}{x+3}$.
   Answer: ${y \in \mathbb{R} \mid y \ne 2}$ One way to find the range of a function is to find the domain of its inverse. Let $y = \frac{2x-1}{x+3}$. To find the inverse, swap $x$ and $y$: $x = \frac{2y-1}{y+3}$. Now solve for $y$: $x(y+3) = 2y-1 \Rightarrow xy + 3x = 2y-1 \Rightarrow 3x+1 = 2y-xy \Rightarrow 3x+1 = y(2-x) \Rightarrow y = \frac{3x+1}{2-x}$. The domain of this inverse function, $f^{-1}(x)$, is all real numbers except where the denominator is zero, i.e., $x \ne 2$. Since the domain of $f^{-1}$ is the range of $f$, the range of $f(x)$ is ${y \in \mathbb{R} \mid y \ne 2}$. Alternatively, for large $|x|$, $f(x)$ approaches the ratio of the leading coefficients, $y=2/1=2$, which is the horizontal asymptote.
6. Let $f(x) = 2x - 5$. If $(f \circ g)(x) = 4x^2 + 6x - 3$, find the expression for $g(x)$.
   Answer: $2x^2 + 3x + 1$ We are given $f(x) = 2x - 5$ and $f(g(x)) = 4x^2 + 6x - 3$. By the definition of $f$, we have $f(g(x)) = 2g(x) - 5$. So we can set the two expressions for $f(g(x))$ equal to each other: $2g(x) - 5 = 4x^2 + 6x - 3$. Now, we solve for $g(x)$. Add 5 to both sides: $2g(x) = 4x^2 + 6x + 2$. Divide by 2: $g(x) = \frac{4x^2 + 6x + 2}{2} = 2x^2 + 3x + 1$.
7. The function $f$ is defined by $f(x) = \ln(3x-5)$. What is the domain of the inverse function, $f^{-1}$?
   Answer: $x \in \mathbb{R}$ The domain of the inverse function $f^{-1}$ is equal to the range of the original function $f$. The function is $f(x) = \ln(3x-5)$. First, consider the domain of $f$: we need the argument of the logarithm to be positive, so $3x-5 > 0$, which means $x > 5/3$. Now, consider the range of $f$. As $x$ approaches $5/3$ from the right, $3x-5$ approaches $0^+$, and $\ln(3x-5)$ approaches $-\infty$. As $x$ approaches $\infty$, $3x-5$ approaches $\infty$, and $\ln(3x-5)$ approaches $\infty$. Therefore, the range of $f$ is $(-\infty, \infty)$, which is $\mathbb{R}$. This range is the domain of $f^{-1}$.
8. The function $f(x) = e^{2x} - 3$. The domain of $f$ is restricted to $x \ge 0$. What is the domain of $f^{-1}(x)$?
   Answer: $x \ge -2$ The domain of the inverse function $f^{-1}$ is the range of the original function $f$ with its restricted domain. The function is $f(x) = e^{2x} - 3$ with domain $x \ge 0$. The function $g(x) = e^{2x}$ is an increasing function. Therefore, the minimum value of $f(x)$ on its domain will occur at the minimum value of $x$, which is $x=0$. At $x=0$, $f(0) = e^{2(0)} - 3 = e^0 - 3 = 1 - 3 = -2$. As $x$ increases towards infinity, $e^{2x}$ increases towards infinity, and so does $f(x)$. Thus, the range of $f$ on the domain $x \ge 0$ is $[-2, \infty)$. This range is the domain of $f^{-1}(x)$.
9. Given $f(x) = 2x+1$ and $g(x) = x^3$, find the value of $(g \circ f^{-1})(9)$.
   Answer: 64 The expression $(g \circ f^{-1})(9)$ means $g(f^{-1}(9))$. First, we need to find the value of $f^{-1}(9)$. Let $f^{-1}(9) = k$. By the definition of an inverse function, $f(k) = 9$. So, $2k+1 = 9$, which gives $2k=8$ and $k=4$. Thus, $f^{-1}(9) = 4$. Now, we substitute this result into $g$: $g(f^{-1}(9)) = g(4)$. Since $g(x) = x^3$, $g(4) = 4^3 = 64$.
10. Let $f(x) = \ln(x)$ and $g(x) = 9-x^2$. Find the domain of the function $h(x) = f(g(x))$.
    Answer: $(-3, 3)$ When you encounter a composite function like $h(x) = f(g(x))$, finding its domain requires understanding how the functions work together. The domain of a composite function is restricted by both the inner function's domain and the requirement that the inner function's output must be valid input for the outer function. Here, $f(x) = \ln(x)$ requires $x > 0$ (you can't take the natural logarithm of zero or negative numbers), and $g(x) = 9 - x^2$ is defined for all real numbers. Since $h(x) = f(g(x)) = \ln(9 - x^2)$, you need $9 - x^2 > 0$ for the logarithm to be defined. Solving $9 - x^2 > 0$: rearrange to get $9 > x^2$, which means $x^2 < 9$. Taking square roots gives $|x| < 3$, or equivalently, $-3 < x < 3$. This is the interval $(-3, 3)$. Answer A) $x > 0$ would be correct if you only considered $f(x)$'s domain in isolation, ignoring that we're using $g(x)$ as the input. Answer B) $[-3, 3]$ includes the endpoints where $g(-3) = g(3) = 0$, but $\ln(0)$ is undefined. Answer C) $(-\infty, -3) \cup (3, \infty)$ represents where $g(x) < 0$, which would make the logarithm undefined. Remember: for composite functions, work from the inside out. The inner function's output becomes the outer function's input, so check that this relationship produces valid results throughout the domain.
11. The function $f(x) = -2(x+3)^2 + 5$. Which of the following is a possible domain for $f(x)$ on which it has an inverse function?
    Answer: $[-3, \infty)$ A function has an inverse if and only if it is one-to-one. The graph of $f(x) = -2(x+3)^2 + 5$ is a parabola with its vertex at $(-3, 5)$ and opening downwards. A parabola is not one-to-one over its entire domain. To make it one-to-one, we must restrict its domain to one side of its axis of symmetry, $x=-3$. The function is increasing on $(-\infty, -3]$ and decreasing on $[-3, \infty)$. Any interval that is a subset of one of these two will work. Choice C, $[-3, \infty)$, is the portion of the domain where the function is strictly decreasing, so it is one-to-one on this domain and thus has an inverse.
12. The range of the function $f(x) = a + \ln(x-b)$ is $(-\infty, \infty)$ and its domain is $(2, \infty)$. The graph of $f(x)$ passes through the point $(3, 5)$. Find the value of $a$.
    Answer: 5 When working with logarithmic functions, you need to understand how transformations affect the domain and range. The function $f(x) = a + \ln(x-b)$ is a vertical shift of $\ln(x-b)$ by $a$ units. Since the domain is $(2, \infty)$, the expression $(x-b)$ must be positive when $x > 2$. This means $x - b > 0$ when $x > 2$, so $b = 2$. The natural logarithm $\ln(x-2)$ has range $(-\infty, \infty)$ when $x > 2$, and adding the constant $a$ doesn't change this range, confirming our function has the correct range. Now you can use the given point $(3, 5)$ to find $a$. Substituting into the function: $f(3) = a + \ln(3-2) = a + \ln(1) = a + 0 = a = 5$ Since $\ln(1) = 0$, we get $a = 5$. Looking at the wrong answers: Choice A (2) likely comes from confusing $a$ with $b$, since $b = 2$ based on the domain. Choice B (3) might result from mixing up the x-coordinate of the given point with the parameter $a$. Choice C ($e^5$) represents a common error of exponentiating both sides incorrectly or confusing the inverse relationship between logarithms and exponentials. Study tip: With logarithmic transformations, always identify the horizontal shift first using the domain, then use given points to find the vertical shift. The domain tells you what's inside the logarithm, while specific points reveal the vertical translation.
13. The function $f(x) = \frac{ax+3}{x-2}$, where $x \ne 2$, is a self-inverse function. Find the value of $a$.
    Answer: 2 A function is self-inverse if $f(x) = f^{-1}(x)$. First, find the inverse of $f(x)$. Let $y = \frac{ax+3}{x-2}$. Swap variables: $x = \frac{ay+3}{y-2}$. Now, solve for $y$: $x(y-2) = ay+3 \Rightarrow xy - 2x = ay + 3 \Rightarrow xy - ay = 2x+3 \Rightarrow y(x-a) = 2x+3 \Rightarrow y = \frac{2x+3}{x-a}$. So, $f^{-1}(x) = \frac{2x+3}{x-a}$. For $f(x) = f^{-1}(x)$, we need $\frac{ax+3}{x-2} = \frac{2x+3}{x-a}$. Comparing rational functions, the coefficients of $x$ in the numerators must be equal: $a = 2$. Also, the constant terms in the numerators must be equal: $3 = 3$ ✓. Finally, the denominators must have the same form, so $-2 = -a$, giving $a = 2$. All conditions confirm $a = 2$.
14. A function $f$ is such that $f(x+1) = 2f(x) - f(x-1)$ for all $x \in \mathbb{R}$. Given that $f(1)=5$ and $f(2)=8$, find $f(4)$.
    Answer: 14 The recurrence relation can be rearranged as $f(x+1) - f(x) = f(x) - f(x-1)$. This shows that the difference between the function values at consecutive integers is constant. This means the sequence of values $f(1), f(2), f(3), ...$ forms an arithmetic progression. The common difference is $d = f(2) - f(1) = 8 - 5 = 3$. We can find the next terms: $f(3) = f(2) + d = 8 + 3 = 11$. $f(4) = f(3) + d = 11 + 3 = 14$. Alternatively, use the given formula directly. For $x=2$, $f(3) = 2f(2) - f(1) = 2(8) - 5 = 16 - 5 = 11$. For $x=3$, $f(4) = 2f(3) - f(2) = 2(11) - 8 = 22 - 8 = 14$.
15. Let $g(x) = \frac{x}{x-1}$. Which expression is equivalent to $(g \circ g)(x)$?
    Answer: x The expression $(g \circ g)(x)$ means $g(g(x))$. To find this, we substitute $g(x)$ into $g(x)$ itself. $g(g(x)) = \frac{g(x)}{g(x)-1}$. Now substitute the expression for $g(x)$: $g(g(x)) = \frac{\frac{x}{x-1}}{\frac{x}{x-1} - 1}$. To simplify, first work on the denominator: $\frac{x}{x-1} - 1 = \frac{x}{x-1} - \frac{x-1}{x-1} = \frac{x-(x-1)}{x-1} = \frac{1}{x-1}$. Now the full expression becomes $\frac{\frac{x}{x-1}}{\frac{1}{x-1}}$. This simplifies to $\frac{x}{x-1} \cdot \frac{x-1}{1} = x$. This shows that $g(x)$ is a self-inverse function.
16. The point $(a, b)$ lies on the graph of $y=f(x)$. Which point must lie on the graph of $y = 3f^{-1}(x/2)$?
    Answer: $(2b, 3a)$ If the point $(a, b)$ lies on the graph of $y=f(x)$, then by definition, $f(a) = b$. From the definition of an inverse function, this implies that $f^{-1}(b) = a$. Let the new function be $g(x) = 3f^{-1}(x/2)$. We want to find a point on its graph. We know the value of $f^{-1}$ when its input is $b$. So, we can set the input of $f^{-1}$ in the expression for $g(x)$ equal to $b$. Let $x/2 = b$, which implies $x = 2b$. Now, we evaluate $g(x)$ at $x=2b$: $y = g(2b) = 3f^{-1}((2b)/2) = 3f^{-1}(b)$. Since we know $f^{-1}(b) = a$, we have $y = 3a$. Therefore, the point $(2b, 3a)$ must lie on the graph of $y = 3f^{-1}(x/2)$.
17. The function $f$ is defined by $f(x) = x^2 - 4x + 7$ for $x \ge 2$. Find the expression for $f^{-1}(x)$.
    Answer: $f^{-1}(x) = 2 + \sqrt{x-3}$ To find the inverse function, let $y = f(x)$, swap $x$ and $y$, and solve for $y$. Start with $y = x^2 - 4x + 7$. Swap variables: $x = y^2 - 4y + 7$. To solve for $y$, complete the square for the terms involving $y$: $x = (y^2 - 4y + 4) - 4 + 7$, which simplifies to $x = (y-2)^2 + 3$. Rearrange to isolate $y$: $x-3 = (y-2)^2$, so $y-2 = \pm\sqrt{x-3}$, and $y = 2 \pm \sqrt{x-3}$. The original function has a domain $x \ge 2$. This becomes the range of the inverse function. So, we must have $f^{-1}(x) \ge 2$. We must choose the positive sign to satisfy this condition. Therefore, $f^{-1}(x) = 2 + \sqrt{x-3}$.
18. Let $f(x) = \sqrt{x-1}$ and $g(x) = \frac{1}{x^2-9}$. Find the domain of the composite function $(f \circ g)(x)$.
    Answer: $[-\sqrt{10}, -3) \cup (3, \sqrt{10}]$ The composite function is $f(g(x)) = \sqrt{\frac{1}{x^2-9}-1}$. For $f(g(x))$ to be defined, two conditions must be met. First, $x$ must be in the domain of $g$, so $x^2-9 \ne 0$, meaning $x \ne \pm 3$. Second, the input to $f$, which is $g(x)$, must be in the domain of $f$. The domain of $f(u) = \sqrt{u-1}$ is $u \ge 1$. Thus, we require $g(x) \ge 1$, which means $\frac{1}{x^2-9} \ge 1$. We can solve this by considering $\frac{1}{x^2-9} - 1 \ge 0$, which is $\frac{1-(x^2-9)}{x^2-9} \ge 0$, or $\frac{10-x^2}{x^2-9} \ge 0$. The critical values are $x = \pm\sqrt{10}$ and $x = \pm 3$. Testing intervals reveals the inequality holds for $-\sqrt{10} \le x < -3$ and $3 < x \le \sqrt{10}$. So the domain is $[-\sqrt{10}, -3) \cup (3, \sqrt{10}]$.
19. Let $f(x) = \sqrt{x+4}$ for $x \ge -4$. Find the value of $x$ for which $f(x) = f^{-1}(x)$.
    Answer: $\frac{1+\sqrt{17}}{2}$ The intersection of the graph of a function and its inverse often lies on the line $y=x$. So we can solve $f(x)=x$ to find the point(s) of intersection. $\sqrt{x+4} = x$. Squaring both sides gives $x+4 = x^2$. Rearranging gives the quadratic equation $x^2 - x - 4 = 0$. Using the quadratic formula, $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)} = \frac{1 \pm \sqrt{1+16}}{2} = \frac{1 \pm \sqrt{17}}{2}$. Since we squared the equation, we must check for extraneous solutions. The original equation was $\sqrt{x+4}=x$, which implies $x$ must be non-negative. Therefore, we must take the positive root: $x = \frac{1+\sqrt{17}}{2}$. We can check that $\sqrt{17}$ is slightly more than 4, so $x$ is approx $5.1/2 = 2.55$, which is positive.

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