Calculus I Unit 1-3 Notations Flashcards
Change (y)
Δy = y₂ - y₁
or ƒ(x)₂ - ƒ(x)₁
Rate of Change (x)
Δx = x₂ - x₁
Average Rate of Change
Δy/Δx
Secant Line of a Function
Avg rate of change between 2 points/positions
Tangent Line of a Function
Slope at a curve of a particular point
(Instantaneous ROC at particular position)
Difference Equation
ƒ(x + h) - ƒ(x) / h
where h = Δx
Directional Limit Properties
If:
limit as x → a⁻ ƒ(x) = limit as x → a⁺ ƒ(x)
then:
limit as x → a ƒ(x) exists
Definition of
limit as x → a ƒ(x) = ∞
ƒ(x) grows arbitrarily large for all x sufficiently close to a
Definition of
limit as x → a ƒ(x) = -∞
ƒ(x) is negative and grows arbitrarily large for all x sufficiently close to a
Explain Solution for:
limit as x → 0⁺ (k + x) / x
where k is constant
Since k is constant and in the numerator, as the value of x decreases: the numerator approaches k while the denominator gets infinitely closer to 0.
limit as x → 0⁺ (k + x) / x = ∞
Explain Solution for:
limit as x → 0⁻ (k + x) / x
where k is constant
Since k is constant and in the numerator, as the value of x increases: the numerator approaches k while the denominator becomes infinitely smaller in magnitude through negative values
limit as x → 0⁻ (k + x) / x = -∞
Explain Solution for:
limit as x → 3⁺ (2 - 5x) / (x - 3)
Since as x approaches 3, the numerator becomes closer to -13, and the dominator becomes infinitely smaller from a positive directions
( -/+) = -
limit as x → 3⁺ (2 - 5x) / (x - 3) = -∞
Explain Solution for:
limit as x → 3⁻ (2 - 5x) / (x - 3)
Since as x approaches 3, the numerator becomes closer to -13, and the dominator becomes infinitely smaller from a negative directions ( -/-) = +
limit as x → 3⁺ (2 - 5x) / (x - 3) = ∞
How do you find the Vertical Asymptote of ƒ(x) ?
The Vertical Asymptote is the value of limit value of x where ƒ(x) = ∞
This is noted by x = a
Sum of Cubes Factorization:
(a³ + b³) = (a + b)(a² - ab + b²)
Difference of Cubes Factorization
(a³ - b³) = (a - b)(a² + ab + b²)
How do you find the horizontal asymptote of ƒ(x)
If a given Limit of ƒ(x) has an x → ± ∞ , the value of ƒ(x) = Horizontal Asymptote
Explain limit as x → ∞ of
(2 + 7x + 2x²)/(x²)
limit as x → ∞ = 2
Since there is a constant of 2 in the numerator while the highest degree (x²) is in both the numerator and denominator.
Both the numerator and denominator are divided by x² leaving us with 2
Explain solution to
limit as x → -∞ of:
√ (16x² + x) / (x)
Divide each part of the numerator and denominator by their respective highest degree
since the numerator is powered to an ‘odd’ degree and x → -∞ it will be divided by -x.
Since the denominator is powered by an even degree it will be divided by a x². Solve with new constant.
How do we know that a function doesn’t have an Horizontal Asymptote ?
The degree of X in the numerator is higher than the highest degree in the denominator
Explain the solution to:
limit as x → 0⁺ of (5eˣ) / (1 - eˣ)
limit as x → 0⁺ of (5eˣ) / (1 - eˣ): Since eⁿ when n > 0 is greater than 1, the denominator (1 - x) will always be negative. Since eˣ can never be negative, the numerator (5eˣ) must always be positive. As the value of x approaches 0 from the positive direction:
limit as x → 0⁺ = -∞
Explain the solution to:
limit as x → 0⁻ of (5eˣ) / (1 - eˣ)
limit as x → 0⁻ of (5eˣ) / (1 - eˣ): Since eⁿ when n < 0 is less than 1, the denominator will always be positive. Since eˣ can never be negative, the numerator (5eˣ) must always be positive. As the value of x approaches 0 from the positive direction:
limit as x → 0⁻ = ∞
Explain the Interval of:
limit as x → 0 of (5eˣ) / (1 - eˣ)
(-∞, 0) , (0,∞), Since the value of ƒ(x) exists at all numbers except when eˣ = 1 and
(e⁰ = 1)
What creates a Slanted Asymptote ?
A Slanted Asymptote is created when the degree of the independent variable in the numerator is EXACTLY ONE degree higher than the degree of the denominator.
