QUERY 23

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course MTH 271

11/18,1550

assignment #023023. `query 23

Applied Calculus I

11-17-2009

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01:21:16

3.2.10 all relative extrema of x^4 - 32x + 4

Give the x and y coordinates of all the relative extrema of x^4 - 32x + 4.

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RESPONSE -->

f(x)=x^4 - 32x + 4

f'(x) = 4x^3 -32

0=4x^3 - 32

0=4(x^3 - 8)

0=4(x - 2)^3

Critical Number=2

(-infinity, 2)

test value: -5

f(-5)=Negative

(2, infinity)

test value: 10

f(10)=Positive

f(2)=-44

Relative Minimum at (2,-44)

confidence rating:

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01:23:19

The procedure is to find the critical numbers, where the derivative is zero, since at a 'peak' or a 'valley' the function levels off and the derivative is for that one instant zero.

The derivative of this function is 4 x^3 - 32.

4 x^3 - 32 = 4 ( x^3 - 8) = 4 ( x-2)^3 has a zero at x = 2. This is the only value for which the derivative is zero and hence the only critical point.

For x < 2, x - 2 is negative and hence (x-2)^3 is negative.

For x > 2, x-2 is positive and hence (x-2)^3 is positive.

So the derivative changes from negative to positive at this zero. This means that the function goes from decreasing to increasing at x = -2, so x = -2 is a relative minimum of x^4 - 32x + 4.

The value of the function at the relative minimum is -44. That is the function has its minimum at (2, -44). **

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RESPONSE -->

self critique rating:: 3

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01:27:57

3.2.30 abs extrema of 4(1+1/x+1/x^2) on [-4,5]

What are the absolute extrema of the given function on the interval?

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RESPONSE -->

h(x)=4(1 + 1/x + 1/x^2)

h(x)=4 + 4x^-1 + 4x^-2

h'(x)=-4x^-2 - 8x^-3

-4/x^2 - 8/x^3

(-4x - 8 )/x^3

-4(x + 2) / x^3 = 0

Critical Number:

-2, f(-2)= 3, Minimum:(-2,3)

Endpoints:

-4, f(-4)=3.25

5, f(5)=4.96, Maximum: (5, 4.96)

confidence rating:

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01:28:34

the derivative of the function is -4/x^2 - 8 / x^3.

Multiplying through by the common denominator x^3 we see that -4/x^2 - 8 / x^3 = 0 when x^3( -4/x^2 - 8 / x^3 ) = 0, x not 0. This simplified to -4 x - 8 = 0, which occurs when x = -2.

At x = -2 we have y = 4 ( 1 + 1 / (-2) + 1 / (-2)^2 ) = 4 ( 1 - .5 + .25) = 4(.75) = 3.

Thus (-2, 3) is a critical point.

Since large negative x yields a negative derivative the derivative for all x < -2 is negative, and since as x -> 0 from 'below' the derivative approaches +infinity the derivative between x=-2 and x = 0 is positive. Thus the derivative goes from negative to positive at x = 2, and the point is a relative minimum. A second-derivative test could also be used to show that the point is a relative minimum.

We also need to test the endpoints of the interval for absolute extrema.

Testing the endpoints -4 and 5 yields 4(1+1/(-4)+1/(-4)^2) = 3.25 and 4(1+1/5+1/25) = 4(1.24) = 4.96 at x = 5. However these values aren't necessarily the absolute extrema.

Recall that the derivative approaches infinity at x = 0. This reminds us to check the graph for vertical asymptotes, and we find that x = 0 is a vertical asymptote of the function. Since as x -> 0 the 1 / x^2 terms dominates, the vertical asymptote will approach positive infinity on both sides of zero, and there is no absolute max; rather the function approaches positive infinity.

However the min at (-2, 3), being lower than either endpoint, is the global min for this function. **

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RESPONSE -->

self critique rating:: 3

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01:28:44

3.2.44 demand x inversely proportional to cube of price p>1; price $10/unit -> demand 8 units; init cost $100, cost per unit $4. Price for maximum profit?

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RESPONSE -->

x=y/p^3

8=y/(10^3)

8=y/1000

8000=y

x=8000/ p^3

R=xp

x=8000/p^3

p^3=8000/x

p=(8000)^(1/3) / x^(1/3)

p= 20 / x^(1/3)

R=xp

R=x( 20 / x^(1/3))

R=20x/x^(1/3)

R=20x^(2/3)

P=R-C

R=20x^(2/3)

C=100 + 4x

P=(20x^(2/3)) - (100 + 4x)

P=(20x^(2/3)) - 100 - 4x

P'= (20(2/3)x^(-1/3)) -4

P'=40/3 x^(-1/3) - 4

0=40/3 x^(-1/3) - 4

4= 40/3 x^(-1/3)

12=40 x^(-1/3)

(3/10)=x^(-1/3)

x=37.04

(-infinity, 37.04), P'(x)=Positive

(37.04, infinity), P'(x)=Negative

P(37.04)= (20)(37.04^(2/3)) - 100 - (4)(37.04)

P(37.04)= -25.9

Maximum at (37.04, -25.9)

confidence rating:

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self critique rating:: 3"

&#Very good responses. Let me know if you have questions. &#