Cal 1 Midterm #2 Questions

Question 1

Explain the Solution the value of c if ƒ is continuous in at the following interval:

        sin²7x/3x²     x ≠ 0    ƒ(x) =
         c                   x = 0

Answer 1

Start by recognizing that since ƒ(x) is continuous at the given interval, our
limit as x → 0 is equivalent to x = 0 so sin²7x/3x² = c.

Then consider the limit property that limit as x → 0 of sinax/x = a. This means that we should be able to manipulate sin²7x/3x² to match our limit property.

sin²7x/3x² → sin7x/3x ⋅ sin7x/x

Now re-apply the limit operator and solve each side individually:

limit as x → 0 of sin7x/3x
limit as x → 0 of (7/3)sin7x/7x = (7/3) ⋅ 1 = (7/3)

limit as x → 0 of sin7x/x = 7

Combine the new values of both sides which we’ll find equates to 49/3

Therefore c = 49/3

Question 2

Explain how to find h(4) and h’(4) when:

h(x) = ƒ(g(x))

Equation to g’ at (4,7) is y = 3x - 5
Equation to ƒ’ at (7,9) is y = -2x + 23

Answer 2

Start by identifying that g(4) = 7 and ƒ(7) = 9

Since we are given the equation of the tangent line of both we know that ƒ’(7) = -2 and g’(4) = 3

Now take the equation h(x) = ƒ(g(x)) and solve at h(4)

h(4) = ƒ(g(4)) → h(4) = ƒ(7) → h(4) = 9

Then take the derivative of h(x) and solve at h’(4)

h(x) = ƒ(g(x)) → h’(x) = ƒ’(g(x)) ⋅ g’(x)

h’(4) = ƒ’(g(4)) ⋅ g’(4) → h’(4) = ƒ’(7) ⋅ 3 →
h’(4) = -2 ⋅ 3 = -6

h(4) = 9
h’(4) = -6

Question 3

Explain how to find Critical Point of ƒ assuming a is constant:

ƒ(x) = x/√(x - a)

Answer 3

Note that Critical Points are when ƒ’(x) = 0 or undef, and at a point where ƒ(x) exists

Start by finding ƒ’(x):

ƒ’(x) = (√(x - a) - (x/2√(x - a))) / (x - a)

ƒ’(x) is undefined or 0 @:

√(x - a) - (x/2√(x - a)) = 0 → 2(x - a) = x → x = 2a

x - a = 0 → x = a

Since at x = a, ƒ(x) is undefined, our only Critical Point is at: x = 2a

Question 4

Explain how to find ƒ’(x) when:

ƒ(x) = (x² + 1)/eˣ

Answer 4

Use quotient rule:

ƒ’(x) = (2xeˣ - eˣ(x² + 1)) / (eˣ)²

Factor and simplify:

ƒ’(x) = (2xeˣ - eˣ(x² + 1)) / (eˣ)² →
ƒ’(x) = eˣ(2x - (x² + 1)) / (eˣ)² →
ƒ’(x) = (2x - x² - 1)) / (eˣ) → (- x² + 2x - 1) / (eˣ)

ƒ’(x) = - (x² - 2x + 1) / (eˣ) → - (x - 1)² / eˣ

Question 5

Explain the how to find the tangent line to ƒ⁻¹(x) at (8,32) when:

ƒ(x) = x³/⁵

Answer 5

Start by acknowledging that ƒ’⁻¹(8) = 1/ƒ’(32)

Now solve for ƒ’(x):

ƒ’(x) = ⅗x⁻²/⁵

Now solve at ƒ’(32) = 3/5(⁵√(32²)) = 3/5(4) = 3/20

Since ƒ’⁻¹(8) = 1/ƒ’(32), solve with new ƒ’(32) value:

ƒ’⁻¹(8) = 1/(3/20) = 20/3

Question 6

Determine ƒ’(x) when ƒ(x) = ln|cot x|

Answer 6

Start by taking the derivative of ƒ(x) using chain rule:

ƒ’(x) = 1/cot x ⋅ -csc²x → sinx/cosx ⋅ -1/sin²x

ƒ’(x) = -1/sinxcosx

Question 7

Explain how to find the derivative of ƒ(x) when:

ƒ(x) = log₁₀(2x)ᵉ

Answer 7

Since e is a constant, we may use the Chain Rule to solve the derivative:

ƒ’(x) = elog₁₀(2x)ᵉ⁻¹ ⋅ 2/(2x)ln(10)

Simplify the equation:

ƒ’(x) = elog₁₀(2x)ᵉ⁻¹/(xln(10))

Question 8

Explain the Critical Points of ƒ(x) = 1/x

Answer 8

Note that Critical Points exist when ƒ’(x) = 0

ƒ’(x) =0-1(x⁻²)

Since ƒ’(x) = 0 when x = 0 and ƒ(0) is undefined, there is No Critical Points

Question 9

Explain the tangent line to the graph of g at x = 0 when:

g(x) = ƒ(e³ˣ)

and the tangent line to f at x = 1 is y = 5 - 2x

Answer 9

Start by noting that since the eqxn of the tangent line of ƒ at x = 1 (ƒ’(1)) y = -2x + 5, we know ƒ’(1) = -2.

We can then use point slope formula to find the y₁ of ƒ:

Since y = -2x + 5 and y = -2(x - 1) + y₁ we set them equal to each other:

-2x + 5 = -2(x - 1) + y₁⁰3 = y₁

So ƒ(1) = 3

Now solve for g(0) using our new ƒ(x) values:

g(0) = ƒ(e⁰) = ƒ(1) = 3

Then take the derivative of g(x) and solve for g’(0):

g’(x) = ƒ’(e³ˣ) ⋅ 3e³ˣ → g’(0) = ƒ’(e⁰) ⋅ 3e⁰ →

g’(0) = ƒ’(1) ⋅ 3 = -2(3) = -6

Now solve for the eqxn to tangent line of g using our new g(0) and g’(0):

y = -6(x - 0) + 3 → y = -6x + 3

Question 10

How to find Absolute Min and Absolute Max over a particular interval ?

Answer 10

Start by taking the derivative of the function

Then solve with the derivative = 0. These are our critical points.

Then solve for the original function set at the endpoints of the interval and the critical points.

The highest resulting values will be our absolute maximums while the lowest resulting values will be our minimums.