Cal 1 Midterm #1 Questions

Question 1

Explain Solution to:

Limit as x → -∞ of -2/x³

Answer 1

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.

Question 2

Explain Solution to:

Limit as x → ∞ of (3x + 2)/(x² - 1)

Answer 2

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

Question 3

What creates a Slanted Asymptote ?

Answer 3

When the degree in the numerator is exactly one more than the degree in the denominator

Ex: (2x² + 6x - 2)/(x+1)

Question 4

Explain the Solution to:

Limit as x → ∞ of (2x² + 6x - 2)/(x+1)

Answer 4

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) = ∞

Question 4

Explain how to find the slope of:

Limit as x → ∞ of (2x² + 6x - 2)/(x+1)

Answer 4

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.

Question 5

Explain solutionton to:

Limit as x → 2 of Cos(x² - 4 / x - 2)

Answer 5

• 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)

Question 6

Explain how to find k in the following Piecewise Function ?

𝘨(x) =
x² - 9 / x - 3; x ≠ 3
kx; x =3

Answer 6

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

Question 7

Explain Solution to:

Limit as x → -6 of
(x³ - 36x)/(x² + 6x)

Answer 7

• 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

Question 8

Explain Solution to:

Limit as x → 2⁺ of (x² + 2x)/(x - 2)

Answer 8

• 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) = ∞

Question 9

Explain Solution to:

Limit as x → ∞ of
(4x³ + 2x - 7)/(5x³ + 9x² - 7)

Answer 9

• 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

Question 10

Explain Solution to:

Position of Rocket: 𝘚(t) = 2t - t² where s = position, t = seconds

Find the average velocity over interval [3, 5]

Answer 10

• 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

Question 11

Find the difference quotient for the function ƒ(x) = x² + 2x

Answer 11

• 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

Question 12

Explain how to find ƒ’(a) and the Eqxn of the Tangent Line:

ƒ(x) = -x³ and a = -2

Answer 12

• 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

Question 13

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

Answer 13

• 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

Question 14

Explain Solution to:
limit x → a √(9 + ƒ(x)) / 𝘨(x)

when limit x → a ƒ(x) = 7 & limit x → a 𝘨(x) = 4

Answer 14

Substitute all the f(x) and g(x) for their limits and solve:

√(9 + 7) / 4 → √(16) / 4 → 4/4 → 1

Question 15

Determine and Explain the Limit to:

limit x → -2⁻ of (x³ - 6x² + 8x)/(x⁴ - 4x²)

Answer 15

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

-∞

Question 16

Assume ƒ is cont. on the interval [5,9], with ƒ(5) = -2, and ƒ(9) = 4. What must be true about this function ?

Answer 16

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

Question 17

Under what conditions is a function not differentiable ?

Answer 17

Sharp turns at a point ( |x - 1| )
Lack of continuity (Break at a point)
Cusp at a point
Vertical Tangents

Question 18

Find the derivative of the function ƒ(x) = eˣ at (1,e)

Answer 18

This is a trick question as it only asks for the derivative.

d/dx [eˣ] = eˣ

Question 19

Explain the Solution to:

limit x → 5 of (x - 3) / (x - 4)²(x - 5)

Answer 19

Since x → 5 produces a negative infinity from the left and a positive infinity from the right, this limit will never exist.

Question 20

Explain the Solution to:

limit x → ∞ of e⁻⁵ˣ

Answer 20

Any number to the an arbitrarily large negative power approaches 0

Question 21

Explain the continuity of ƒ(x) if:

ƒ(x) = (x² + x - 2) / (x² + 5x + 6)

Answer 21

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.

Question 22

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) ?

Answer 22

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.