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HW 12

6.

The average cost function is the total cost divided by the number of units sold:

πΆΜ
(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₯:

πΆΜ
(π₯) =

12,000

+ 93 + 0.02π₯

π₯

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

π¦ = 93 + 0.02π₯

7.

The average cost function is the total cost divided by the number of units sold:

πΆΜ
(π₯) =

11,000 + 89π₯ + 0.01π₯ 2

π₯

Divide all terms of the numerator by π₯:

πΆΜ
(π₯) =

11,000

+ 89 + 0.01π₯

π₯

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

π¦ = 89 + 0.01π₯

8.

The average cost function is the total cost divided by the number of units sold:

πΆΜ
(π) =

1,250 + 250π + 50π2

π

Divide all terms of the numerator by π:

πΆΜ
(π) =

1,250

+ 250 + 50π

π₯

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

π¦ = 250 + 50π

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 isnβt differentiable (sharp corners, like π₯ = 0 in π¦ =

|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, theyβre 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, isnβt the interval [2,6], so it couldnβt be the absolute

minimum of the function in this interval.

3.

Determine the first derivative of the function:

π β² (π₯) = (5π₯ β 6)β²

Use the linearity property:

π β² (π₯) = 5(π₯)β² β 6(1)β²

Use the power rule:

π β² (π₯) = 5(1) β 6(0)

π β² (π₯) = 5

Since the first derivative is a non-zero constant, the function wonβt have local extrema.

The second derivative, from the constant rule, is π β²β² (π₯) = 0.

The function π(π₯) is a linear function, so, itβs differentiable for any real values of π₯.

Evaluate the function at the boundaries of the closed intervals:

π(0) = 5 β 0 β 6 = β6

π(9) = 5 β 9 β 6 = 39

π(β2) = 5 β β2 β 6 = β16

π(3) = 5 β 3 β 6 = 9

On the interval, [0,9], the absolute minimum is β6 and it occurs at π₯ = 0; the absolute

maximum is 39 and it occurs at π₯ = 9.

On the interval, [β2,3], the absolute minimum is β16 and it occurs at π₯ = β2; the

absolute maximum is 9 and it occurs at π₯ = 3.

4.

Determine the first derivative of the function:

π β² (π₯) = (π₯ 2 β 2π₯ β 9)β²

Use the linearity property:

π β² (π₯) = (π₯ 2 )β² β 2(π₯)β² β 9(1)β²

Use the power rule:

π β² (π₯) = (2π₯) β 2(1) β 9(0)

π β² (π₯) = 2π₯ β 2

Determine the x-coordinates of the local extrema:

π β² (π₯) = 0

2π₯ β 2 = 0

π₯=1

Determine the second derivative of the function:

π β²β² (π₯) = (2π₯ β 2)β²

Use the linearity property:

π β²β² (π₯) = 2(π₯)β² β 2(1)β²

Use the power rule:

π β²β² (π₯) = 2(1) β 2(0)

π β²β² (π₯) = 2

Since π β²β² (1) > 0, the point at π₯ = 1 is a local minimum.

The function π(π₯) is a quadratic function, so, itβs differentiable for any real values of π₯.

Evaluate the function at the boundaries of the closed interval and at the local minimum:

π(β1) = (β1)2 β 2(β1) β 9 = β6

π(2) = (2)2 β 2(2) β 9 = β9

π(1) = (1)2 β 2(1) β 9 = β10

On the interval, [β1,2], the absolute minimum is β10 and it occurs at π₯ = 1; the

absolute maximum is β6 and it occurs at π₯ = β1.

5.

On an open interval, a function can only have an absolute minimum if it doesnβt go to

negative infinity in this interval, similarly, it can only have an absolute maximum if it doesnβt

go to positive infinity in this interval.

For this function, on its…

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