Functions

• Case 1: $\lfloor x \rfloor < -2 \implies \lfloor x \rfloor \le -3 \implies x < -2$
• Case 2: $\lfloor x \rfloor > 3 \implies \lfloor x \rfloor \ge 4 \implies x \ge 4$

• Maximum value occurs at the left endpoint $t = -1$: \[ g(-1) = (-1 - 2)^2 + 3 = (-3)^2 + 3 = 9 + 3 = 12 \]
• Minimum value occurs at the right endpoint $t = 1$: \[ g(1) = (1 - 2)^2 + 3 = (-1)^2 + 3 = 1 + 3 = 4 \] End{itemize} Therefore, the range of the function over its valid domain is the closed interval $[4, 12]$. Final Answer: Range $\in [4, 12]$.

1. Constraints on the base $x$: The base must be strictly positive and cannot equal $1$: \[ x > 0 \quad \text{and} \quad x \neq 1 \]
2. Constraint on the argument: The argument must be strictly positive: \[ \cos x > 0 \]

• For $k = 0$: The interval is $(-\frac{\pi}{2}, \frac{\pi}{2})$. Restricting this to positive numbers ($x > 0$) gives $(0, \frac{\pi}{2})$. We must also remove $x = 1$, which splits this section into $(0, 1) \cup (1, \frac{\pi}{2})$.
• For $k \ge 1$: The entire interval is positive, so it is included unchanged.

1. Injectivity Test: Let us analyze the derivative of the function to check its monotonicity: \[ f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x-1)^2 \] For all real numbers $x$, $(x-1)^2 \ge 0$, which means $f'(x) \ge 0$ everywhere. Since the derivative is non-negative and only vanishes at a single isolated point ($x=1$), the function is strictly increasing across its entire domain. A strictly monotonic function never repeats an output value, which proves it is One-One (Injective).
2. Surjectivity Test: The function is an odd-degree polynomial expression ($n=3$). Let us look at its behavior at the limits of the domain: \[ \lim_{x \to \infty} f(x) = \infty \quad \text{and} \quad \lim_{x \to -\infty} f(x) = -\infty \] Since the polynomial is continuous, the Intermediate Value Theorem guarantees that it takes every real value between negative and positive infinity. Thus, $\text{Range}(f) = (-\infty, \infty) = \mathbb{R}$. Since the calculated range matches the given codomain ($\mathbb{R}$), the function is Onto (Surjective).

• Test for Injectivity: Let us choose two symmetric points around zero, $x_1 = 2$ and $x_2 = -2$. Evaluate their images: $f(2) = 2^2 = 4$ and $f(-2) = (-2)^2 = 4$. Since distinct domain inputs produce identical outputs ($f(2) = f(-2)$), the function is Many-One, not injective.
• Test for Surjectivity: The function squares real inputs, so its outputs are always non-negative: $\text{Range}(f) = [0, \infty)$. The problem states that the codomain is explicitly restricted to $[0, \infty)$. Since the range equals the codomain, the function is Onto (Surjective).

• The first element has exactly $n$ available choices in the target set.
• The second element has exactly $n-1$ choices remaining (since one was taken and the function must be one-one).
• The third element has $n-2$ choices remaining, and so on.

• Part (a): Evaluate $(f \circ g)(x) = f(g(x))$. Substitute the full expression for $g(x)$ into the input slot of function $f$: \[ f(g(x)) = f(x^2 - 2) \] Apply the rule for $f$, which multiplies its input by 2 and adds 1: \[ f(g(x)) = 2(x^2 - 2) + 1 = 2x^2 - 4 + 1 = 2x^2 - 3 \]
• Part (b): Evaluate $(g \circ f)(x) = g(f(x))$. Substitute the expression for $f(x)$ into the input slot of function $g$: \[ g(f(x)) = g(2x + 1) \] Apply the rule for $g$, which squares its input and subtracts 2: \[ g(f(x)) = (2x + 1)^2 - 2 \] Expand the perfect square expression: \[ g(f(x)) = (4x^2 + 4x + 1) - 2 = 4x^2 + 4x - 1 \]

To find the inverse function, set the equation equal to $y$ and isolate $x$ in terms of $y$: \[ y = x^2 - 4x + 9 \] Complete the square on the right-hand side to make isolating $x$ simpler: \[ y = (x^2 - 4x + 4) + 5 \implies y = (x - 2)^2 + 5 \] Subtract 5 from both sides of the equation: \[ (x - 2)^2 = y - 5 \] Take the square root of both sides. This introduces a choice of signs: \[ x - 2 = \pm\sqrt{y - 5} \implies x = 2 \pm\sqrt{y - 5} \] To choose the correct sign, check the specified domain of the original function ($x \in (2, \infty)$). Since $x$ must be strictly greater than 2, we must choose the positive sign branch: \[ x = 2 + \sqrt{y - 5} \] Now swap the variable labels to write the final inverse function in terms of $x$: \[ f^{-1}(x) = 2 + \sqrt{x - 5} \] The domain of this inverse function is $(5, \infty)$, which matches the range of the original function. Final Answer: $f^{-1}(x) = 2 + \sqrt{x - 5}$.

Substitute the full expression for $f(x)$ back into itself: \[ f(f(x)) = \frac{f(x) + 1}{f(x) - 1} = \frac{\left(\frac{x+1}{x-1}\right) + 1}{\left(\frac{x+1}{x-1}\right) - 1} \] Multiply both the numerator and the denominator by the common factor $(x-1)$ to simplify the complex fraction: \[ f(f(x)) = \frac{(x + 1) + 1(x - 1)}{(x + 1) - 1(x - 1)} \] Expand and combine the terms in the numerator and denominator: \[ \text{Numerator} = x + 1 + x - 1 = 2x \] \[ \text{Denominator} = x + 1 - x + 1 = 2 \] Divide the simplified terms: \[ f(f(x)) = \frac{2x}{2} = x \] Since $f(f(x)) = x$, the function is its own inverse ($f = f^{-1}$). This type of function is called an involution. Final Answer: $x$.

To test the symmetry of the function, replace $x$ with $-x$ throughout the expression: \[ f(-x) = \log_e \left( -x + \sqrt{(-x)^2 + 1} \right) = \log_e \left( \sqrt{x^2 + 1} - x \right) \] To find the relationship between $f(-x)$ and $f(x)$, let us rationalize the expression inside the logarithm by multiplying the numerator and denominator by its conjugate $(\sqrt{x^2 + 1} + x)$: \[ \sqrt{x^2 + 1} - x = \frac{(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x} = \frac{(x^2 + 1) - x^2}{x + \sqrt{x^2 + 1}} = \frac{1}{x + \sqrt{x^2 + 1}} \] Substitute this reciprocal back into the logarithm expression: \[ f(-x) = \log_e \left( \frac{1}{x + \sqrt{x^2 + 1}} \right) = \log_e \left( x + \sqrt{x^2 + 1} \right)^{-1} \] Apply the logarithmic power rule to bring the exponent $-1$ out to the front: \[ f(-x) = -1 \cdot \log_e \left( x + \sqrt{x^2 + 1} \right) = -f(x) \] Since $f(-x) = -f(x)$, the function satisfies the algebraic definition of an odd function. Final Answer: $f(x)$ is an Odd function.

According to the functional decomposition theorem, any function can be split into even and odd components. The formula for the even component is: \[ E(x) = \frac{f(x) + f(-x)}{2} \] Substitute our given function $f(x) = e^x$ into this formula: \[ E(x) = \frac{e^x + e^{-x}}{2} \] This specific expression forms the definition of the hyperbolic cosine function ($\cosh x$). We can verify it is even by checking $E(-x) = \frac{e^{-x} + e^{-(-x)}}{2} = \frac{e^{-x} + e^x}{2} = E(x)$. Final Answer: $E(x) = \frac{e^x + e^{-x}}{2}$.

To determine the parity of the product function $h(x)$, replace $x$ with $-x$: \[ h(-x) = f(-x) \cdot g(-x) \] Apply the given parity properties for each individual function ($f(-x) = -f(x)$ and $g(-x) = g(x)$): \[ h(-x) = (-f(x)) \cdot (g(x)) \] Factor out the negative sign: \[ h(-x) = -(f(x) \cdot g(x)) = -h(x) \] Since $h(-x) = -h(x)$, the product function satisfies the definition of an odd function. Final Answer: The product function $h(x)$ is Odd.

• Step 1: Find the period $T_1$ of $\sin(3x)$. The standard period of sine is $2\pi$. Divide by the input scaling factor $k = 3$: \[ T_1 = \frac{2\pi}{3} \]
• Step 2: Find the period $T_2$ of $\cos(2x)$. The standard period of cosine is $2\pi$. Divide by the input scaling factor $k = 2$: \[ T_2 = \frac{2\pi}{2} = \pi \]

Let us analyze the base function. The standard fractional part function, defined as $\{x\} = x - \lfloor x \rfloor$, repeats its values over a constant interval of length 1. Thus, its standard fundamental period is: \[ T_{\text{base}} = 1 \] The problem introduces an input scaling multiplier of $k = 5$. Apply the scaling theorem, which states that multiplying the input variable divides the fundamental period by that factor: \[ T_{\text{new}} = \frac{T_{\text{base}}}{|k|} = \frac{1}{5} = 0.2 \] Final Answer: $\frac{1}{5}$.

Let us analyze the individual components first. Taking the absolute value of a trigonometric function folds its negative waves upward, which cuts its period in half: the period of $|\sin x|$ is $\pi$, and the period of $|\cos x|$ is $\pi$. The LCM of these periods is $\pi$.
However, notice that $\sin x$ and $\cos x$ are complementary functions ($\sin(\frac{\pi}{2} + x) = \cos x$). For symmetric expressions involving complementary functions, the true fundamental period can be smaller than the LCM. Let us test if half of the LCM angle ($\frac{\pi}{2}$) satisfies the periodicity definition: \[ f\left(x + \frac{\pi}{2}\right) = \left|\sin\left(x + \frac{\pi}{2}\right)\right| + \left|\cos\left(x + \frac{\pi}{2}\right)\right| \] Apply the trigonometric reduction identities: \[ \sin\left(x + \frac{\pi}{2}\right) = \cos x \quad \text{and} \quad \cos\left(x + \frac{\pi}{2}\right) = -\sin x \] Substitute these expansions back into the expression for the shifted function: \[ f\left(x + \frac{\pi}{2}\right) = |\cos x| + |-\sin x| = |\cos x| + |\sin x| = f(x) \] The function values match exactly over intervals of length $\frac{\pi}{2}$. Therefore, the true fundamental period is $\frac{\pi}{2}$, not $\pi$. Final Answer: $\frac{\pi}{2}$.

• Branch 1: $\lfloor x \rfloor = 2 \implies 2 \le x < 3$
• Branch 2: $\lfloor x \rfloor = 3 \implies 3 \le x < 4$

Let us first analyze the range of the inner quadratic expression: \[ y = x^2 - 2x + 2 \] Complete the square to find the vertex coordinates: \[ y = (x - 1)^2 + 1 \] For all real numbers $x$, $(x-1)^2 \ge 0$. Adding 1 means the expression is always strictly positive: \[ x^2 - 2x + 2 \ge 1 \] Now evaluate this result using the definition of the signum function. The signum function maps inputs based on their sign: \[ \text{sgn}(t) = \begin{cases} -1 & \text{if } t < 0 \\ 0 & \text{if } t = 0 \\ 1 & \text{if } t > 0 \end{cases} \] Since the inner quadratic expression is strictly positive ($y \ge 1 > 0$) for all real inputs, the signum function outputs a constant value of $1$ everywhere. Thus, the range contains only a single element. Final Answer: Range $\in \{1\}$.

• Case 1: $x$ is an integer ($x \in \mathbb{Z}$). By definition, the greatest integer of any integer is just the number itself. Thus, $\lfloor x \rfloor = x$ and $\lfloor -x \rfloor = -x$: \[ E = x + (-x) = 0 \]
• Case 2: $x$ is not an integer ($x \notin \mathbb{Z}$). We can write $x$ as the sum of an integer part and a fractional part: $x = n + f$, where $n \in \mathbb{Z}$ and $f \in (0, 1)$. Thus, $\lfloor x \rfloor = n$.
  Now evaluate the negative input expression $-x$: \[ -x = -n - f = (-n - 1) + (1 - f) \] Since $f \in (0, 1)$, the new fractional part $(1-f)$ also lies between 0 and 1. Therefore, the greatest integer of this negative expression is: \[ \lfloor -x \rfloor = -n - 1 \] Sum the values from both parts: \[ E = n + (-n - 1) = -1 \]

• Step 1: Start with the base parent graph: $y_0 = |x|$. This is a V-shaped graph with its vertex at the origin $(0,0)$.
• Step 2: Analyze the horizontal change inside the absolute value braces: $x \to x - 3$. This horizontal modification shifts the entire curve 3 units to the right along the x-axis. The vertex moves from $(0,0)$ to $(3,0)$.
• Step 3: Analyze the vertical change outside the braces: $+ 2$. This vertical modification shifts the entire curve 2 units upward along the y-axis. The vertex moves from $(3,0)$ to its final position at $(3,2)$.

• The graph of $f(x) = |x|$ consists of two straight lines forming a V-shape that starts at $(0,0)$ and opens upward.
• Let us rewrite the quadratic function $g(x)$ by completing the square: \[ g(x) = -(x^2 - 4x + 4) - 3 + 4 = -(x - 2)^2 + 1 \] This is a parabola with its vertex at $(2, 1)$ that opens downward because of the negative leading coefficient. Its x-intercepts occur where $-(x-2)^2 + 1 = 0 \implies (x-2)^2 = 1 \implies x = 1 \text{ or } x = 3$.

• At $x = 1$: $f(1) = |1| = 1$, and $g(1) = 1$. Both functions equal 1 at $x=1$, so $(1,1)$ is a point of intersection.
• For the region $x > 1$, the downward parabola curves back down toward the x-axis, crossing it at $x=3$. Meanwhile, the line $y=x$ rises continuously. The curves must cross exactly once more between $x=1$ and $x=2$.
• For negative values of $x$ ($x < 0$), $f(x)$ is positive while $g(x)$ lies entirely below the x-axis, so they never intersect.

• Step 1: For the positive domain ($x \ge 0$), the graph of $f(|x|)$ is identical to the original graph of $f(x)$. Therefore, keep the entire right side of the graph unchanged.
• Step 2: For the negative domain ($x < 0$), the graph is defined as $f(-x)$, which is the horizontal reflection of the positive side across the y-axis. Therefore, erase the left side of the original graph completely.
• Step 3: Take the right side of the graph (from Step 1) and reflect it across the vertical y-axis onto the left side.

• Part (a): Test for Commutativity. Evaluate $b * a$ by reversing the elements in the formula: \[ b * a = ba + 2 \] Since standard multiplication of rational numbers is commutative ($ab = ba$), we can see that: \[ a * b = ab + 2 = ba + 2 = b * a \] Since $a * b = b * a$ for all elements, the operation is Commutative.
• Part (b): Test for Associativity. Let us evaluate and compare the two grouping configurations: \[ \text{Left Grouping: } (a * b) * c = (ab + 2) * c \] Apply the operation rule treating $(ab+2)$ as the first element: \[ (ab + 2) * c = (ab + 2)c + 2 = abc + 2c + 2 \] \[ \text{Right Grouping: } a * (b * c) = a * (bc + 2) \] Apply the operation rule treating $(bc+2)$ as the second element: \[ a * (bc + 2) = a(bc + 2) + 2 = abc + 2a + 2 \] Comparing the two results, $abc + 2c + 2 \neq abc + 2a + 2$ unless $a = c$. Since the results are not equal for all elements, the operation is not Associative.

By definition, an element $e$ is the identity element if it satisfies the condition $a * e = a$ for all elements $a$ in the set. Apply the operation rule to the expression $a * e$: \[ a + e + ae = a \] Subtract $a$ from both sides of the equation: \[ e + ae = 0 \] Factor out the identity variable $e$: \[ e(1 + a) = 0 \] Since the domain explicitly excludes $-1$ ($a \neq -1$), the term $(1+a)$ is never zero. Therefore, we can divide both sides by $(1+a)$ to solve for $e$: \[ e = 0 \] Let us verify this result by checking the other side: $0 * a = 0 + a + 0 \cdot a = a$. The identity element is confirmed. Final Answer: The identity element is $e = 0$.

By definition, an element $b$ is the inverse of $a$ if combining them returns the identity element ($a * b = e$). From the previous problem, we know the identity element is $e = 0$. Set up the inverse equation: \[ a * b = 0 \implies a + b + ab = 0 \] Group the terms containing $b$ to isolate it: \[ b + ab = -a \implies b(1 + a) = -a \] Divide both sides by the factor $(1+a)$ (which is valid since $a \neq -1$): \[ b = -\frac{a}{1 + a} \] Thus, the inverse of any element $a$ is given by the formula $-\frac{a}{1+a}$. Final Answer: $a^{-1} = -\frac{a}{1 + a}$.