Mathematics Exam Instructions - Parts 1 & 2

The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept.

The degree of a polynomial is determined by the highest power of the variable in the polynomial.

The leading coefficient of a polynomial is the coefficient of the term with the highest power of the variable.

The difference quotient is a way to calculate the average rate of change of a function over a small interval. It is defined as [f(x + h) - f(x)]/h, where h is a small change in x.

The secant of an angle in a right triangle is the ratio of the hypotenuse to the adjacent side.

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

The inequality x3 + x2 ≥ 2x can be rewritten as x3 + x2 - 2x ≥ 0. Factoring, we get x(x + 2)(x - 1) ≥ 0. The critical points are x = -2, x = 0, and x = 1. We can make a sign chart to see that the solution to the inequality is (−∞, −2] ∪ [0, 1].

To evaluate arccos(√3/2), we need to find the angle θ in the interval [0,π] such that cos(θ) = √3/2. This angle is θ = π/6.

The inverse of a function f(x) is a function that reverses the operation of f(x). In other words, if f(a) = b, then f⁻¹(b) = a. To find the inverse of a function, switch x and y in the equation and solve for y.

To find the inverse of a function, follow these steps:

1. Replace f(x) with y.
2. Swap x and y.
3. Solve the equation for y.
4. Replace y with f⁻¹(x).

The cotangent function (cot x) is the reciprocal of the tangent function (tan x). It is defined as cot x = cos x / sin x.

The cosecant function (csc x) is the reciprocal of the sine function (sin x). It is defined as csc x = 1/sin x.

The secant function (sec x) is the reciprocal of the cosine function (cos x). It is defined as sec x = 1/cos x.

To solve an exponential equation, use the following steps:

1. Isolate the exponential term.
2. Take the natural logarithm of both sides.
3. Solve for x.

The cosine function has a period of 2π. This means that the graph repeats every 2π units.

To find the solutions to a trigonometric equation, follow these steps:

1. Solve for the trigonometric function.
2. Find the general solution using the period of the function.
3. Find the solutions within the given interval.

The domain of a function is the set of all possible input values (x-values) for which the function is defined. It includes all the numbers that can be plugged into the function without causing any mathematical errors, such as division by zero or taking the square root of a negative number.

A function is even if its graph is symmetrical with respect to the y-axis. This means that for every point (x, y) on the graph, the point (-x, y) is also on the graph. Mathematically, f(x) is even if f(-x) = f(x) for all x in the domain of the function.

A function is odd if its graph is symmetrical with respect to the origin. This means that if (x, y) is on the graph, then (-x, -y) is also on the graph. Mathematically, f(x) is odd if f(-x) = -f(x) for all x in the domain of the function.

A vertical asymptote is a vertical line that the graph of a function approaches as the input approaches a certain value. This occurs when the function becomes unbounded (goes to infinity or negative infinity) as the input approaches the specific value.

A horizontal asymptote is a horizontal line that the graph of a function approaches as the input approaches positive or negative infinity. This indicates that the function has a limiting value as the input grows arbitrarily large or small.

The inverse function of a function f(x) is denoted as f⁻¹(x) and has the property that f⁻¹(f(x)) = x and f(f⁻¹(x)) = x for all x in the domain of the function. In essence, the inverse function 'undoes' the original function. This means that if we input a value x into f(x) and then input the output (f(x)) into its inverse f⁻¹(x), we get back the original input x.

The amplitude of a trigonometric function of the form A cos(Bx) or A sin(Bx) is the absolute value of A (i.e., |A|). It represents the maximum displacement of the function from its midline.

The period of a trigonometric function of the form A cos(Bx) or A sin(Bx) is 2π/B. It represents the length of one complete cycle of the function.

A linear equation that represents the relationship between the price of a cup of coffee (P) and the number of years since the cafe opened (t), where the slope (m) represents the rate of change in price per year and the y-intercept (b) represents the initial price.

The process of finding the composition of two functions, g and f, denoted as g ◦ f (x), where the output of the inner function f(x) is used as the input for the outer function g(x).

The amount of money needed to be invested initially to reach a specific future value after a certain time period with continuous compounding. This amount is calculated using the formula: A = Pe^(rt), where A is the future value, P is the initial investment, r is the annual interest rate, and t is the time period.

The rate at which a quantity changes over time. In the context of the problem, it represents the yearly increase in the price of a cup of coffee.

A type of interest calculation where interest is added to the principal continuously, meaning that the interest itself earns interest. It's modeled by the formula: A = Pe^(rt) where A is the final amount, P is the principal, r is the interest rate, and t is the time.

The initial price of a cup of coffee at the time the cafe opened.

The year when the cafe first opened.

The price of a cup of coffee after 10 years from the time the cafe opened.

The angle θ lies in Quadrant IV, where cosine is positive and sine is negative. To find tan θ, we can use the identity tan θ = sin θ / cos θ. First, we need to find sin θ. Using the Pythagorean identity, sin² θ + cos² θ = 1, we get sin² θ = 1 - cos² θ. Since θ is in Quadrant IV, sin θ is negative. Therefore, sin θ = -√(1 - cos² θ) = -√(1 - (2/3)²) = -√(5/9) = -√5/3. Finally, tan θ = sin θ / cos θ = (-√5/3) / (2/3) = -√5/2.

The function P(x) represents the perimeter of the rectangle. Since the rectangle lies in the first quadrant, the length of one side is x and the length of the other side is y. The perimeter is given by P(x) = 2(x + y). We are given that y = 6 - 2x. Substituting, we get P(x) = 2(x + (6 - 2x)) = 2(6 - x) = 12 - 2x.

To solve for x, we first need to use the property of logarithms that states logₐ(b) = c is equivalent to aᶜ = b. Applying this to our equation, we get 7^(2log₇(x) + 5) = 7⁹. Simplifying the left side, we get 7^(2log₇(x)) * 7⁵ = 7⁹. Using the property that logₐ(b) + logₐ(c) = logₐ(bc), we get 7^(log₇(x²)) * 7⁵ = 7⁹. Using the property that a^(logₐ(b)) = b, we get x² * 7⁵ = 7⁹. Dividing both sides by 7⁵, we get x² = 7⁴. Taking the square root of both sides, we get x = 7² = 49. Therefore, x = 49.

The zeros of a polynomial are the values of x for which the polynomial equals zero. Since the polynomial has zeros at x = 1 + √3 and x = 1 - √3, we can write the polynomial as f(x) = (x - (1 + √3))(x - (1 - √3)). Expanding this expression, we get f(x) = (x - 1 - √3)(x - 1 + √3) = x² - 2x + 1 - 3 = x² - 2x - 2.

The natural logarithm, ln(x), is the logarithm to the base e. ln(e) represents the logarithm of the base e to the base e. By definition, logₐ(a) = 1. Therefore, ln(e) = logₑ(e) = 1.

The period of the function tan(x) is π. Therefore, tan(5π/6) = tan(π/6). Using the unit circle, tan(π/6) = sin(π/6)/cos(π/6) = (1/2)/(√3/2) = 1/√3 = √3/3.

Secant and cosine are reciprocal functions. Therefore, sec(3π) = 1/cos(3π). Cosine has a period of 2π. Therefore, cos(3π) = cos(π) = -1. Therefore, sec(3π) = 1/cos(3π) = 1/(-1) = -1.

Sin(−225°) = sin(135°) because the sine function has a period of 360°. Using the unit circle, sin(135°) = √2/2.