Function Properties & Inverses - IB Mathematics Analysis and Approaches

1. Any function $f(x)$ can be written as the sum of an even function $f_e(x)$ and an odd function $f_o(x)$. For the function $f(x) = e^x + \sin(x)$, determine the even part, $f_e(x)$.
   Answer: $\cosh(x)$ The even part of a function $f(x)$ is given by the formula $f_e(x) = \frac{f(x) + f(-x)}{2}$. For $f(x) = e^x + \sin(x)$, we first find $f(-x)$: $$f(-x) = e^{-x} + \sin(-x) = e^{-x} - \sin(x)$$ Now, we substitute into the formula for $f_e(x)$: $$f_e(x) = \frac{(e^x + \sin(x)) + (e^{-x} - \sin(x))}{2}$$ $$f_e(x) = \frac{e^x + e^{-x}}{2}$$ This expression is the definition of the hyperbolic cosine function, $\cosh(x)$. Distractor A is part of the original function, not its even part. Distractor C is $\sinh(x)$, which is the even part of just $e^x$. Distractor D is the odd part of $f(x)$, which is $f_o(x) = \frac{f(x) - f(-x)}{2}$.
2. The function $f(x)$ is periodic with a fundamental period of 3. A new function is defined as $g(x) = 2f(\frac{x}{4} - 1) + 5$. What is the fundamental period of $g(x)$?
   Answer: 12 The period of a function is affected only by horizontal transformations (stretches, compressions, and reflections inside the function argument). Vertical shifts and stretches do not change the period. The transformation is of the form $Af(B(x-C))+D$. The new period $P'$ is related to the old period $P$ by $P' = P/|B|$. In this case, $f(x)$ has period $P=3$. The argument of $f$ in $g(x)$ is $\frac{x}{4} - 1 = \frac{1}{4}(x-4)$. So, $B = 1/4$. The new period is $P' = \frac{3}{|1/4|} = 3 \times 4 = 12$. Distractor A results from incorrectly multiplying the period by $B$. Distractor B ignores the horizontal stretch. Distractor D incorrectly involves the horizontal shift in the period calculation.
3. A function $f(x)$ is defined as $f(x) = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}$. Which property does this function have?
   Answer: It is an odd function. This question tests your understanding of even and odd functions, which is a key concept in function analysis. When you encounter piecewise functions like this, you need to check their symmetry properties by testing specific conditions. To determine if a function is odd, you must verify that $f(-x) = -f(x)$ for all values in the domain. Let's test this systematically. For any positive value $x > 0$, we have $f(x) = x^2$ and $f(-x) = -(-x)^2 = -x^2$. Also, $-f(x) = -x^2$. So $f(-x) = -f(x)$ holds for positive inputs. For negative values, if $x < 0$, then $-x > 0$, so $f(-x) = (-x)^2 = x^2$ and $-f(x) = -(-x^2) = x^2$. Again, $f(-x) = -f(x)$. Since this relationship holds for all values, the function is odd. Option A is incorrect because even functions satisfy $f(-x) = f(x)$, but here $f(1) = 1$ while $f(-1) = -1$, so $f(-x) \neq f(x)$. Option B is wrong because we've proven the function is odd. Option C is incorrect because this function has no repeating pattern over regular intervals—it's not periodic. When checking function symmetry, always test both the even condition ($f(-x) = f(x)$) and odd condition ($f(-x) = -f(x)$) systematically. Remember that odd functions have rotational symmetry about the origin, while even functions have reflectional symmetry about the y-axis.
4. Consider the periodic function $f(x) = \sin^2(x) - \cos^2(x)$. What is its fundamental period?
   Answer: $\pi$ When you encounter trigonometric functions with squared terms, look for opportunities to use fundamental identities to simplify and find the period more easily. Start by applying the double angle identity. Since $\cos(2x) = \cos^2(x) - \sin^2(x)$, we can rearrange this to get $\sin^2(x) - \cos^2(x) = -(\cos^2(x) - \sin^2(x)) = -\cos(2x)$. So $f(x) = \sin^2(x) - \cos^2(x) = -\cos(2x)$. Since the negative sign doesn't affect the period, we need to find the period of $\cos(2x)$. The general rule is that $\cos(bx)$ has period $\frac{2\pi}{|b|}$. Here, $b = 2$, so the period is $\frac{2\pi}{2} = \pi$. Let's examine why the other options are incorrect. Option A ($\frac{\pi}{2}$) would be the period if we had $\cos(4x)$, but our coefficient is only 2. Option B ($4\pi$) would correspond to $\cos(\frac{x}{2})$, suggesting a coefficient of $\frac{1}{2}$ rather than 2. Option C ($2\pi$) is the period of the basic $\cos(x)$ function, but fails to account for the coefficient 2 that doubles the frequency. You can verify: $f(x + \pi) = -\cos(2(x + \pi)) = -\cos(2x + 2\pi) = -\cos(2x) = f(x)$. Study tip: When you see squared trig functions, immediately think of double angle identities. They often simplify complex-looking expressions into basic trig functions whose periods are easier to determine.
5. The function $f(x) = \frac{ax+3}{2x-4}$, where $x \neq 2$, is self-inverse. Find the value of the constant $a$.
   Answer: 4 A function $f$ is self-inverse if $f^{-1}(x) = f(x)$. First, we find the inverse of $f(x)$. Let $y = \frac{ax+3}{2x-4}$. To find the inverse, we swap $x$ and $y$ and solve for $y$: $$x = \frac{ay+3}{2y-4}$$ $$x(2y-4) = ay+3$$ $$2xy - 4x = ay+3$$ $$2xy - ay = 4x+3$$ $$y(2x - a) = 4x+3$$ $$y = \frac{4x+3}{2x-a}$$ So, $f^{-1}(x) = \frac{4x+3}{2x-a}$. For $f$ to be self-inverse, we must have $f(x) = f^{-1}(x)$: $$\frac{ax+3}{2x-4} = \frac{4x+3}{2x-a}$$ By comparing coefficients, we see that we must have $a=4$. This also makes the denominators consistent, as $-4 = -a$ also gives $a=4$. Distractor A results from the common error of setting $a$ equal to the constant in the denominator, $-4$. Distractors B and C confuse coefficients.
6. Let $f(x)$ be a differentiable odd function. Which of the following statements about its derivative, $f'(x)$, must be true?
   Answer: $f'(x)$ is an even function. The definition of an odd function is $f(-x) = -f(x)$. To find the property of its derivative, we differentiate this identity with respect to $x$ using the chain rule on the left side: $$\frac{d}{dx}f(-x) = \frac{d}{dx}(-f(x))$$ $$f'(-x) \cdot \frac{d}{dx}(-x) = -f'(x)$$ $$f'(-x) \cdot (-1) = -f'(x)$$ $$-f'(-x) = -f'(x)$$ $$f'(-x) = f'(x)$$ This is the definition of an even function. Thus, the derivative of an odd function is an even function. Distractor A is incorrect; it applies to the derivative of an even function. Distractor C is not always true; consider $f(x) = \sin(x)$, which is odd. Its derivative is $f'(x) = \cos(x)$, and $f'(0) = \cos(0) = 1 \neq 0$. Distractor D is incorrect as we have proven $f'(x)$ must be even.
7. Let $f(x)$ be a non-zero odd function and $g(x)$ be a non-zero even function, both defined for all $x \in \mathbb{R}$. Let $h(x) = f(x) + g(x)$. Which expression is always equivalent to $h(-x)$?
   Answer: $g(x) - f(x)$ By definition, an odd function satisfies $f(-x) = -f(x)$ and an even function satisfies $g(-x) = g(x)$. We can substitute these into the expression for $h(-x)$: $$h(-x) = f(-x) + g(-x)$$ $$h(-x) = -f(x) + g(x)$$ Rearranging the terms gives: $$h(-x) = g(x) - f(x)$$ Therefore, option C is the correct answer. Distractor A is incorrect because it implies $h(x)$ is even, which is not generally true. Distractor B is incorrect because it implies $h(x)$ is odd, which is also not generally true. Distractor D is incorrect as it results from a sign error in the substitution.
8. What is the fundamental period of the function $f(x) = |\sin(2x)|$?
   Answer: $\frac{\pi}{2}$ The function $g(x) = \sin(x)$ has a fundamental period of $2\pi$. The function $h(x) = \sin(2x)$ has a period of $\frac{2\pi}{2} = \pi$. Taking the absolute value of the sine function reflects the negative portions of the graph across the x-axis. This causes the function to repeat more frequently. The period of $|\sin(u)|$ is $\pi$, which is half the period of $\sin(u)$. Therefore, the period of $f(x) = |\sin(2x)|$ is half the period of $\sin(2x)$. Half of $\pi$ is $\frac{\pi}{2}$. We can verify: $f(x+\frac{\pi}{2}) = |\sin(2(x+\frac{\pi}{2}))| = |\sin(2x+\pi)| = |-\sin(2x)| = |\sin(2x)| = f(x)$. Distractor C is the period of $\sin(2x)$ without the absolute value. Distractor A halves the period twice. Distractor D is the period of $\sin(x)$.
9. Let $f(x) = \cos(4x) + \tan(6x)$. Find the fundamental period of $f(x)$.
   Answer: $\frac{\pi}{2}$ First, find the period of each term. The period of $\cos(kx)$ is $\frac{2\pi}{|k|}$ and the period of $\tan(kx)$ is $\frac{\pi}{|k|}$. Period of $\cos(4x)$ is $P_1 = \frac{2\pi}{4} = \frac{\pi}{2}$. Period of $\tan(6x)$ is $P_2 = \frac{\pi}{6}$. The period of the sum of two periodic functions is the least common multiple (LCM) of their individual periods. We need to find LCM$(\frac{\pi}{2}, \frac{\pi}{6})$. Let the period be $P$. We need $P = n_1 P_1 = n_2 P_2$ for the smallest positive integers $n_1, n_2$. $n_1 \frac{\pi}{2} = n_2 \frac{\pi}{6} \implies 3n_1 = n_2$. The smallest positive integers satisfying this are $n_1=1$ and $n_2=3$. Thus, the period is $P = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}$. Alternatively, $P = 3 \cdot \frac{\pi}{6} = \frac{\pi}{2}$. Distractor A is the period of the second term only. Distractor D is a common multiple of the periods but not the least common multiple. Distractor B is an incorrect calculation.
10. Let $f(x) = x^3 + x - 2$. The inverse function $f^{-1}(x)$ exists. Find the value of $(f^{-1})'(0)$.
    Answer: $\frac{1}{4}$ We use the formula for the derivative of an inverse function: $(f^{-1})'(y_0) = \frac{1}{f'(x_0)}$, where $y_0 = f(x_0)$. Here, we want to find $(f^{-1})'(0)$, so $y_0 = 0$. First, we need to find the value of $x_0$ such that $f(x_0) = 0$. $$x_0^3 + x_0 - 2 = 0$$ By inspection, we can see that $x_0 = 1$ is the solution, since $1^3 + 1 - 2 = 0$. Next, we need to find the derivative of $f(x)$: $$f'(x) = 3x^2 + 1$$ Now, we evaluate the derivative at $x_0 = 1$: $$f'(1) = 3(1)^2 + 1 = 4$$ Finally, we apply the formula: $$(f^{-1})'(0) = \frac{1}{f'(1)} = \frac{1}{4}$$ Distractor A is $f'(0)$. Distractor B is $f'(1)$, but forgets to take the reciprocal. Distractor D is an incorrect calculation of the derivative.
11. Let $f(x) = e^{2x}$. The graph of $f$ is restricted to the domain $x \ge 0$. Its inverse is $g(x)$. What is $g'(e^2)$?
    Answer: $\frac{1}{2e^2}$ We can solve this in two ways. Method 1: Using the formula $g'(y_0) = \frac{1}{f'(x_0)}$ where $g=f^{-1}$ and $y_0 = f(x_0)$. Here $y_0 = e^2$. We need to find $x_0$ such that $f(x_0) = e^2$. $$e^{2x_0} = e^2 \implies 2x_0 = 2 \implies x_0 = 1$$ Now find the derivative of $f(x)$: $f'(x) = 2e^{2x}$. Evaluate at $x_0=1$: $f'(1) = 2e^{2(1)} = 2e^2$. Therefore, $g'(e^2) = \frac{1}{f'(1)} = \frac{1}{2e^2}$. Method 2: Find the inverse function explicitly. Let $y = e^{2x}$. Swap variables: $x = e^{2y}$. Solve for $y$: $\ln x = 2y \implies y = \frac{1}{2}\ln x$. So $g(x) = \frac{1}{2}\ln x$. Now find the derivative of $g(x)$: $g'(x) = \frac{1}{2} \cdot \frac{1}{x} = \frac{1}{2x}$. Evaluate at $x = e^2$: $g'(e^2) = \frac{1}{2e^2}$. Both methods yield the same result. Distractor D is $f'(1)$, forgetting the reciprocal. Distractor B misses the factor of 2 from the chain rule. Distractor C is an algebraic error.
12. The function $f(x) = 3\cos(2x) - 1$ is restricted to the domain $D = [0, \frac{\pi}{2}]$ to ensure it is one-to-one. What is the domain of the inverse function $f^{-1}(x)$?
    Answer: $[-4, 2]$ The domain of the inverse function $f^{-1}(x)$ is the range of the original function $f(x)$ on its restricted domain. The restricted domain of $f(x)$ is $x \in [0, \frac{\pi}{2}]$. For this interval of $x$, the argument of the cosine function, $2x$, is in the interval $[0, \pi]$. Over the interval $[0, \pi]$, the function $\cos(u)$ takes all values from $\cos(0)=1$ down to $\cos(\pi)=-1$. So the range of $\cos(2x)$ on the given domain is $[-1, 1]$. Now we find the range of $f(x) = 3\cos(2x) - 1$. The maximum value is $3(1) - 1 = 2$. The minimum value is $3(-1) - 1 = -4$. So, the range of $f(x)$ on the domain $D$ is $[-4, 2]$. This is the domain of $f^{-1}(x)$. Distractor B comes from a sign error $1 \pm 3$. Distractor C is the range of $\cos(2x)$, not $f(x)$. Distractor D is the range of $f^{-1}(x)$, not its domain.
13. The function $f(x) = x^2 - 4x + 5$ is defined for $x \ge 2$. Find the inverse function $f^{-1}(x)$.
    Answer: $f^{-1}(x) = 2 + \sqrt{x-1}$ First, complete the square for $f(x)$: $f(x) = (x^2 - 4x + 4) + 1 = (x-2)^2 + 1$. The domain of $f$ is $[2, \infty)$, so its range is $[1, \infty)$. The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. So, for $f^{-1}$, the domain is $[1, \infty)$ and the range is $[2, \infty)$. To find the inverse, let $y = (x-2)^2 + 1$ and swap $x$ and $y$: $$x = (y-2)^2 + 1$$ Now, solve for $y$: $$x-1 = (y-2)^2$$ $$\pm\sqrt{x-1} = y-2$$ $$y = 2 \pm\sqrt{x-1}$$ Since the range of $f^{-1}(x)$ must be $[2, \infty)$, we must choose the positive root. Therefore, $f^{-1}(x) = 2 + \sqrt{x-1}$. Distractor A uses the negative root, which would give the incorrect range. Distractor B has an algebra error under the square root. Distractor C has a sign error on the constant term.
14. Let $f$ be a non-zero even function and $g$ be a non-zero odd function. Which of the following statements must be true?
    Answer: $(g \circ f)$ is an even function. Let's test each statement using the definitions $f(-x) = f(x)$ (even) and $g(-x) = -g(x)$ (odd). A: $(f \circ g)(-x) = f(g(-x)) = f(-g(x))$. Since $f$ is even, $f(-y) = f(y)$. So, $f(-g(x)) = f(g(x)) = (f \circ g)(x)$. This shows $f \circ g$ is an even function, so A is false. B: $(g \circ f)(-x) = g(f(-x))$. Since $f$ is even, $f(-x) = f(x)$. So, $g(f(-x)) = g(f(x)) = (g \circ f)(x)$. This shows $g \circ f$ is an even function, so B is true. C: For an odd function $g$, $g(0) = g(-0) = -g(0)$, which implies $2g(0)=0$ and so $g(0)=0$. For an even function $f$, there is no restriction on the value of $f(0)$. For example, $f(x) = x^2+1$ is even but $f(0)=1$. So $f(0)=g(0)$ is not always true. C is false. D: $(f+g)(-x) = f(-x)+g(-x) = f(x) - g(x)$. This is not equal to $(f+g)(x)$ or $-(f+g)(x)$ in general, so $f+g$ is neither odd nor even (unless one function is zero, which is ruled out). D is false.
15. Let $f(x) = \ln(x)$ for $x>0$ and $g(x) = x^2+1$. The function $h(x) = (f \circ g)(x)$ is defined on a restricted domain $D$ to be one-to-one. Which of the following could be a valid expression for $h^{-1}(x)$?
    Answer: $\sqrt{e^x - 1}$ First, find the composite function $h(x)$: $h(x) = f(g(x)) = f(x^2+1) = \ln(x^2+1)$. The natural domain is $x \in \mathbb{R}$ since $x^2+1 > 0$. The function $h(x)$ is even and therefore not one-to-one. To make it one-to-one, we must restrict its domain, for example, to $D=[0, \infty)$. On this domain, the range of $h(x)$ is $[\ln(1), \infty) = [0, \infty)$. To find the inverse, let $y = \ln(x^2+1)$ and swap variables: $x = \ln(y^2+1)$. Now, solve for $y$: $$e^x = y^2+1$$ $$y^2 = e^x - 1$$ $$y = \pm\sqrt{e^x - 1}$$ The range of $h^{-1}(x)$ must be the restricted domain of $h(x)$, which we chose as $D=[0, \infty)$. Therefore, we must choose the positive root: $h^{-1}(x) = \sqrt{e^x - 1}$. Distractor B confuses the order of operations. Distractor C has algebraic errors. Distractor D is not a function, and an inverse must be a function.
16. Given that $f(x)$ is an odd function and $\int_0^3 f(x) \,dx = 5$, what is the value of $\int_{-3}^3 (2f(x) + 3) \,dx$?
    Answer: 18 We can split the integral using the property of linearity: $$\int_{-3}^3 (2f(x) + 3) \,dx = \int_{-3}^3 2f(x) \,dx + \int_{-3}^3 3 \,dx = 2\int_{-3}^3 f(x) \,dx + \int_{-3}^3 3 \,dx$$ A key property of odd functions is that their definite integral over a symmetric interval $[-a, a]$ is zero. Since $f(x)$ is odd, $\int_{-3}^3 f(x) \,dx = 0$. The given information $\int_0^3 f(x) \,dx = 5$ is consistent with this, as $\int_{-3}^0 f(x) \,dx = -5$. So, the first term is $2 \times 0 = 0$. The second term is the integral of a constant: $$\int_{-3}^3 3 \,dx = [3x]_{-3}^3 = 3(3) - 3(-3) = 9 - (-9) = 18$$ The total value is $0 + 18 = 18$. Distractor A calculates $2 \times 5$ and ignores the rest. Distractor C incorrectly computes $2\int_0^3 f(x)dx + 18 = 2(5)+18=28$. Distractor D incorrectly treats $f(x)$ as an even function, where $\int_{-3}^3 f(x) dx = 2\int_0^3 f(x) dx = 10$, leading to $2(10) + 18 = 38$.
17. Given that $f(x) = 2x + \cos(x)$ is a one-to-one function, find the value of $(f^{-1})'(1)$.
    Answer: $\frac{1}{2}$ We use the formula for the derivative of an inverse function: $(f^{-1})'(y_0) = \frac{1}{f'(x_0)}$, where $y_0 = f(x_0)$. In this problem, we want to find $(f^{-1})'(1)$, so $y_0 = 1$. First, we need to find the value of $x_0$ such that $f(x_0) = 1$. $$2x_0 + \cos(x_0) = 1$$ By inspection, we can see that $x_0 = 0$ is the solution, since $2(0) + \cos(0) = 0 + 1 = 1$. Next, we find the derivative of $f(x)$: $$f'(x) = 2 - \sin(x)$$ Now, we evaluate the derivative at $x_0 = 0$: $$f'(0) = 2 - \sin(0) = 2 - 0 = 2$$ Finally, we apply the formula: $$(f^{-1})'(1) = \frac{1}{f'(0)} = \frac{1}{2}$$ Distractor D is the value of $f'(0)$, but forgets to take the reciprocal. Distractor C could be the value of $f(0)$. Distractor A is a plausible calculation error.
18. The domain of the function $f(x) = \arcsin(\ln x)$ is restricted to $[a, b]$ so that its inverse $f^{-1}(x)$ exists and is defined for all values in its domain. What is the value of $ab$?
    Answer: 1 First, we must find the natural domain of $f(x) = \arcsin(\ln x)$. The domain of $\arcsin(u)$ is $u \in [-1, 1]$. Therefore, we must have $-1 \le \ln x \le 1$. Exponentiating with base $e$ across the inequality gives: $$e^{-1} \le x \le e^1$$ So, the domain of $f(x)$ is $[1/e, e]$. This means $a = 1/e$ and $b = e$. The function $f(x)$ is strictly increasing on this domain, so it is one-to-one and its inverse exists without further restriction. The question asks for the value of $ab$: $$ab = \left(\frac{1}{e}\right) \cdot (e) = 1$$ The distractors are based on misunderstanding the domain constraints or the properties of logarithms and exponentials.
19. The function $f(x) = (x-c)^2 + d$ is restricted to the domain $x \ge c$. Its inverse is $f^{-1}(x) = 3 + \sqrt{x-5}$. What are the values of $c$ and $d$?
    Answer: $c=3, d=5$ Let's find the inverse of the general function $f(x)=(x-c)^2+d$ with domain $x \ge c$. The range is $y \ge d$. Let $y=(x-c)^2+d$. Swap variables: $x=(y-c)^2+d$. Solve for $y$: $x-d=(y-c)^2 \implies y-c = \pm\sqrt{x-d} \implies y = c \pm\sqrt{x-d}$. The range of the inverse must be the domain of the original, which is $y \ge c$. So we must choose the positive root: $f^{-1}(x) = c + \sqrt{x-d}$. We are given that $f^{-1}(x) = 3 + \sqrt{x-5}$. By comparing the two forms, we can equate the corresponding parts: $c = 3$ $-d = -5 \implies d = 5$. Therefore, $c=3$ and $d=5$. Distractor B confuses the values of c and d. Distractors C and D involve sign errors.

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