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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"