Cal 1 Midterm #1 Questions Flashcards
Explain Solution to:
Limit as x → -∞ of -2/x³
Since the denominator is of a higher degree than the numerator and the numerator is constant, we know that -2 being divided by an increasingly large number can only = 0.
We can also place -2 behind the limit operator and solve lim x ➡ -∞ of 1/x³ which = 0. and -2(0) = 0.
Explain Solution to:
Limit as x → ∞ of (3x + 2)/(x² - 1)
We can start by dividing all the numbers by the highest degree in the denominator (x²). we find that x²/x² = 1, -1/X² = 0, 3x/x² = 0, and 2/x² = 0.
since 0/1 = 0 our answer is 0
What creates a Slanted Asymptote ?
When the degree in the numerator is exactly one more than the degree in the denominator
Ex: (2x² + 6x - 2)/(x+1)
Explain the Solution to:
Limit as x → ∞ of (2x² + 6x - 2)/(x+1)
Since the numerator is one degree more than the degree of the denominator, we can deduct that this limit has a slope. We can then divide each part by the value of x and solve their limits as they approach ∞:
x/x = 1, 1/x = 0, 2x²/x = 2x = ∞, 6x/x = 6, -2/x = 0. This gives us: (∞ + 6 - 0)/(1+0) = ∞
Explain how to find the slope of:
Limit as x → ∞ of (2x² + 6x - 2)/(x+1)
Nice that the numerator is one degree higher than the denominator so it must have a slope.
Perform polynomial long division of the numerator divided by the denominator. the slope and immediate Y intercept is the line which this function behaves like.
Explain solutionton to:
Limit as x → 2 of Cos(x² - 4 / x - 2)
- Factor the difference of squares in the numerator to leave us with (x + 2)
- ((2) + 2) = 4, and Cos(x) is continuous at Cos(4)
- Evaluate at Cos(4) (using radians)
Explain how to find k in the following Piecewise Function ?
𝘨(x) =
x² - 9 / x - 3; x ≠ 3
kx; x =3
Use the 3 conditions of continuity:
- (1) Find 𝘨(3): 𝘨(x) = kx → 𝘨(3) = 3k
- (2) Find Limit as x → 3 of 𝘨(x): 𝘨(x) = x² - 9 / x - 3 ➡ (x + 3) = 6
- (3) Set 𝘨(3) = Limit as x → 3 of 𝘨(x): 3k = 6 ➡ k = 2
k = 2
Explain Solution to:
Limit as x → -6 of
(x³ - 36x)/(x² + 6x)
- Factor the one x out of the numerator and denominator of the function
- Factor the numerator to x(x + 6)(x - 6) effectively eliminating the
(x + 6) in the denominator. - Substitute -6 into our new function: (-6 - 6), giving us f(x) = -12
Explain Solution to:
Limit as x → 2⁺ of (x² + 2x)/(x - 2)
- Since plugging in 2 (from the positive direction) gives us a # in the denominator that approaches an 0, and a positive numerator that’s larger than 0, we can assume:
Limit as x → 2⁺ of (x² + 2x)/(x - 2) = ∞
Explain Solution to:
Limit as x → ∞ of
(4x³ + 2x - 7)/(5x³ + 9x² - 7)
- Since x → ∞ we can divide each component by the highest degree x in the denominator.
- Since the degree is equal on both sides we can easily conclude that that f(x) = 4/5
Explain Solution to:
Position of Rocket: 𝘚(t) = 2t - t² where s = position, t = seconds
Find the average velocity over interval [3, 5]
- Solve for 𝘚(3) and 𝘚(5) to obtain our y₁ & y₂
- Set up Vₐᵥ₉ = (𝘚(5) - 𝘚(3))/(x₂ - x₁) →
(-15 - (-3))/2 → -12/2 = -6m/s
Find the difference quotient for the function ƒ(x) = x² + 2x
- Identify the difference quotient: ƒ(x+𝘩) - ƒ(x) / 𝘩
- Plug the function into the difference quotient:
(x+𝘩)² + 2(x+𝘩) - (x² + 2x) / 𝘩 - The solve to obtain:
(x² + 2x𝘩 + 𝘩² + 2x + 2𝘩 - x² - 2x) / 𝘩 - Then cancel out the repeat variables (x² and 2x):
(2x𝘩 + 𝘩² + 2𝘩) / 𝘩 - Divide the numerator by 𝘩:
2x + 𝘩 + 2
Explain how to find ƒ’(a) and the Eqxn of the Tangent Line:
ƒ(x) = -x³ and a = -2
- To find ƒ’(a), first solve for ƒ’(x) using the definition of the derivative:
limit as h → 0 of ƒ(x+𝘩) - ƒ(x) / 𝘩 plugging in our ƒ(x) = -x³
ƒ’(x) = -3x²
- Then solve for ƒ’(a) by plugging in a = -2 into our new function:
ƒ’(a) = -3(-2)² = -12 - The Eqxn of the Tangent Line is then solved using ƒ’(a) = -12 as our slope/mₜₐₙ | a = -2 as our x₁ | and ƒ(a) = 8 as our y₁
y = -12(x + 2) + 8 → y = -12x - 24 + 8
Explain how to find a & b in the following Piecewise Function ? assuming ƒ is continuous at x = 1
ƒ(x) =
-4x; x < 1
x² - b; x = 1
4a; x > 1
- First solve for ƒ(1) since it’s our given:
ƒ(1) = 1² - b = 1 - b - Then solve for limit x → 1⁻ since this allows substitution
limit x → 1⁻ of -4x = -4 - Now set limit x → 1⁻ ƒ(x) = limit x → 1⁺ ƒ(x) and solve for a
-4 = 4a → a = -1 - Now set limit x → 1 ƒ(x) = ƒ(1) and solve for b
-4 = 1 - b → b = 5
Explain Solution to:
limit x → a √(9 + ƒ(x)) / 𝘨(x)
when limit x → a ƒ(x) = 7 & limit x → a 𝘨(x) = 4
Substitute all the f(x) and g(x) for their limits and solve:
√(9 + 7) / 4 → √(16) / 4 → 4/4 → 1
Determine and Explain the Limit to:
limit x → -2⁻ of (x³ - 6x² + 8x)/(x⁴ - 4x²)
Start by factoring out an x in the numerator and x² in the denominator.
Factor out the numerator into x(x-4)(x-2)
Factor the denominator into x²(x+2)(x-2)
Cancel out the x/x² and the (x-2)/(x-2) leaving us with
(x-4) / x(x+2)
Plug -2 into the numerator (-2 - 4) = -6
Plug -2 into the outer x, and a number close to but less than -2 into the inner x → -2(-2.00001 + 2)
Since the numerator is negative and the denominator is a positive that gets significantly close to 0, our answer will be
-∞
Assume ƒ is cont. on the interval [5,9], with ƒ(5) = -2, and ƒ(9) = 4. What must be true about this function ?
Limit x → 7 of ƒ(x) = ƒ(7)
The equation ƒ(x) = 0 has at least one solution
There exists some c in the interval such that ƒ(c) = a number between -2 and 4
Under what conditions is a function not differentiable ?
Sharp turns at a point ( |x - 1| )
Lack of continuity (Break at a point)
Cusp at a point
Vertical Tangents
Find the derivative of the function ƒ(x) = eˣ at (1,e)
This is a trick question as it only asks for the derivative.
d/dx [eˣ] = eˣ
Explain the Solution to:
limit x → 5 of (x - 3) / (x - 4)²(x - 5)
Since x → 5 produces a negative infinity from the left and a positive infinity from the right, this limit will never exist.
Explain the Solution to:
limit x → ∞ of e⁻⁵ˣ
Any number to the an arbitrarily large negative power approaches 0
Explain the continuity of ƒ(x) if:
ƒ(x) = (x² + x - 2) / (x² + 5x + 6)
Factor out the numerator and denominator to obtain:
ƒ(x) = (x + 2)(x - 1) / (x + 2)(x + 3), we notice that ƒ(x) is not continuous at neither -2 or -3 so:
(-∞ , -3)(-3 , -2)(-2 , ∞) is our interval of continuity.
Explain how to solve: A function 𝑓 is even if 𝑓(−𝑥) = 𝑓(𝑥) for all 𝑥 in the domain of 𝑓. Suppose 𝑓 is even, with:
limit x → 2⁺ 𝑓(𝑥) = 5
limit x → 2⁻ 𝑓(𝑥) = 8
What is the limit limit x → -2⁺ 𝑓(𝑥) ?
View this function as a graph 𝑓(−𝑥) describe mirror of 𝑓(𝑥). This would mean that as long as the |x| is less than or equal to 2 our 𝑓(x) = 8. if our |x| value is higher than 2 𝑓(x) = 5.
