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Clemson University Calculus Questions

View attached explanation and answer. Let me know if you have any questions.Hey there, attached are the solutions to the requested questions, in pdf and doc formats.Let me know if you need any edits or clarifications.

HW 12
6.
The average cost function is the total cost divided by the number of units sold:
𝐢̅ (π‘₯) =

12,000 + 93π‘₯ + 0.02π‘₯ 2
π‘₯

For a rational functional, like the one above, when the degree of the numerator
polynomial is one more than the degree of the denominator polynomial, the function will have
an oblique asymptote and its equation is the quotient of the numerator and denominator
(excluding the remainder).
Divide all terms of the numerator by π‘₯:
𝐢̅ (π‘₯) =

12,000
+ 93 + 0.02π‘₯
π‘₯

The equation of the oblique asymptote is the quotient of the rational function:
𝑦 = 93 + 0.02π‘₯

7.
The average cost function is the total cost divided by the number of units sold:
𝐢̅ (π‘₯) =

11,000 + 89π‘₯ + 0.01π‘₯ 2
π‘₯

Divide all terms of the numerator by π‘₯:
𝐢̅ (π‘₯) =

11,000
+ 89 + 0.01π‘₯
π‘₯

The equation of the oblique asymptote is the quotient of the rational function:
𝑦 = 89 + 0.01π‘₯

8.
The average cost function is the total cost divided by the number of units sold:
𝐢̅ (𝑛) =

1,250 + 250𝑛 + 50𝑛2
𝑛

Divide all terms of the numerator by 𝑛:
𝐢̅ (𝑛) =

1,250
+ 250 + 50𝑛
π‘₯

The equation of the oblique asymptote is the quotient of the rational function:

𝑦 = 250 + 50𝑛

HW 13
1.

The absolute maximum or minimum of a continuous function on a closed interval can
only occur at the boundaries of the interval, at local extrema (when the tangent line is
horizontal), or at points where the function isn’t differentiable (sharp corners, like π‘₯ = 0 in 𝑦 =
|π‘₯|).
From this graph:
The absolute minimum occurs at π‘₯ = 1, the boundary of the closed interval [1,11], and
it is 2.
The absolute maximum occurs at π‘₯ = 11, the boundary of the closed interval [1,11],
and it is 9.
Note: there is a local maximum, at π‘₯ = 4, and a local minimum, at π‘₯ = 8, to this
function in the interval [1,11], but, as the graph shows, they’re not the absolute maximum or
minimum in this interval.

2.

From this graph:
The absolute minimum occurs at π‘₯ = 2, the boundary of the closed interval [2,6], and it
is 3.
The absolute maximum occurs at π‘₯ = 4, a local maximum, and it is 5.
Note: the local minimum, at π‘₯ = 8, isn’t the interval [2,6], so it couldn’t be the absolute
minimum of the function in this interval.

3.
Determine the first derivative of the function:
𝑓 β€² (π‘₯) = (5π‘₯ βˆ’ 6)β€²
Use the linearity property:
𝑓 β€² (π‘₯) = 5(π‘₯)β€² βˆ’ 6(1)β€²
Use the power rule:
𝑓 β€² (π‘₯) = 5(1) βˆ’ 6(0)
𝑓 β€² (π‘₯) = 5
Since the first derivative is a non-zero constant, the function won’t have local extrema.
The second derivative, from the constant rule, is 𝑓 β€²β€² (π‘₯) = 0.
The function 𝑓(π‘₯) is a linear function, so, it’s differentiable for any real values of π‘₯.
Evaluate the function at the boundaries of the closed intervals:
𝑓(0) = 5 βˆ™ 0 βˆ’ 6 = βˆ’6
𝑓(9) = 5 βˆ™ 9 βˆ’ 6 = 39

𝑓(βˆ’2) = 5 βˆ™ βˆ’2 βˆ’ 6 = βˆ’16
𝑓(3) = 5 βˆ™ 3 βˆ’ 6 = 9
On the interval, [0,9], the absolute minimum is βˆ’6 and it occurs at π‘₯ = 0; the absolute
maximum is 39 and it occurs at π‘₯ = 9.
On the interval, [βˆ’2,3], the absolute minimum is βˆ’16 and it occurs at π‘₯ = βˆ’2; the
absolute maximum is 9 and it occurs at π‘₯ = 3.

4.
Determine the first derivative of the function:
𝑓 β€² (π‘₯) = (π‘₯ 2 βˆ’ 2π‘₯ βˆ’ 9)β€²
Use the linearity property:
𝑓 β€² (π‘₯) = (π‘₯ 2 )β€² βˆ’ 2(π‘₯)β€² βˆ’ 9(1)β€²
Use the power rule:
𝑓 β€² (π‘₯) = (2π‘₯) βˆ’ 2(1) βˆ’ 9(0)
𝑓 β€² (π‘₯) = 2π‘₯ βˆ’ 2
Determine the x-coordinates of the local extrema:
𝑓 β€² (π‘₯) = 0
2π‘₯ βˆ’ 2 = 0
π‘₯=1
Determine the second derivative of the function:
𝑓 β€²β€² (π‘₯) = (2π‘₯ βˆ’ 2)β€²
Use the linearity property:
𝑓 β€²β€² (π‘₯) = 2(π‘₯)β€² βˆ’ 2(1)β€²
Use the power rule:
𝑓 β€²β€² (π‘₯) = 2(1) βˆ’ 2(0)
𝑓 β€²β€² (π‘₯) = 2
Since 𝑓 β€²β€² (1) > 0, the point at π‘₯ = 1 is a local minimum.
The function 𝑓(π‘₯) is a quadratic function, so, it’s differentiable for any real values of π‘₯.
Evaluate the function at the boundaries of the closed interval and at the local minimum:
𝑓(βˆ’1) = (βˆ’1)2 βˆ’ 2(βˆ’1) βˆ’ 9 = βˆ’6
𝑓(2) = (2)2 βˆ’ 2(2) βˆ’ 9 = βˆ’9
𝑓(1) = (1)2 βˆ’ 2(1) βˆ’ 9 = βˆ’10

On the interval, [βˆ’1,2], the absolute minimum is βˆ’10 and it occurs at π‘₯ = 1; the
absolute maximum is βˆ’6 and it occurs at π‘₯ = βˆ’1.

5.
On an open interval, a function can only have an absolute minimum if it doesn’t go to
negative infinity in this interval, similarly, it can only have an absolute maximum if it doesn’t
go to positive infinity in this interval.
For this function, on its…

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