Recent content by antinerd
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Maximize Profit: Find Optimal Production with C(x) & R(x)
Homework Statement Find the level of production x that will maximize profit. Homework Equations C(x) = 500 + 100x^2 + x^3, where x = units produced. R(x) = 7000x - 80x^2 The Attempt at a Solution Should I use marginal cost and marginal revenue, or is there a way to...- antinerd
- Thread
- Maximization
- Replies: 1
- Forum: Calculus and Beyond Homework Help
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Finding Inflection Points on f(x)=x^4-2x^2-1
Would the inflection points of this equation: f``(x) = 60x^3 - 120x be 60x = 0 and x^2 - 2 = 0 so are the inflectoin points: x = 0, x= +- sqrt(2) ?- antinerd
- Post #4
- Forum: Calculus and Beyond Homework Help
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Finding Inflection Points on f(x)=x^4-2x^2-1
Homework Statement The inflection points are where the function changes its concavity, and can be found through the second derivative of the function... so, I have been given this equation: f(x) = x^4 - 2x^2 - 1 and I have to find the inflection points. Homework Equations The...- antinerd
- Thread
- Points
- Replies: 9
- Forum: Calculus and Beyond Homework Help
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Linear Approximation: Approximating \sqrt{4.1}-\sqrt{3.9}
Holy crap man, nevermind. I've been up for like 3 days straight so I'm buggin' out. Thanks for your insight, guys :)- antinerd
- Post #11
- Forum: Calculus and Beyond Homework Help
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Linear Approximation: Approximating \sqrt{4.1}-\sqrt{3.9}
When I do: e^0 - (h / e^0) = 1 + (0.1/1) = 1.1 It should be 0.9 ... I have the correct approximation for -ln(1.1) = 0.1 but I made a mistake Shouldn't it be 1 PLUS 0.1 to give me 1.1? Cuz h = NEGATIVE 0.1 What's wrong here?- antinerd
- Post #10
- Forum: Calculus and Beyond Homework Help
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Linear Approximation: Approximating \sqrt{4.1}-\sqrt{3.9}
EDIT: Should I use -0.1 for h or can i use 0.1...- antinerd
- Post #9
- Forum: Calculus and Beyond Homework Help
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Linear Approximation: Approximating \sqrt{4.1}-\sqrt{3.9}
So let's say I was doing something like e^(-0.1) - ln (1.1) I can just go about doing it thus: h = -0.1 so that: -0.1 = 0 + h and 1.1 = 1 - h for the first term: f(x) = e^(x) f`(x) = e^(x) but then how do I go about it from there... Could someone help me...- antinerd
- Post #7
- Forum: Calculus and Beyond Homework Help
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Definite Integrals Homework: Evaluate & Feedback
Thanks. So if I did the second correctly, it's: 1/2 e^5 - e/2 Right? And I can leave it like that?- antinerd
- Post #3
- Forum: Calculus and Beyond Homework Help
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Definite Integrals Homework: Evaluate & Feedback
Homework Statement Evaluate the definite integrals. Homework Equations Integral of (t+1)/(t^2+2t+1) dt from 1 to 4 (a=1, b=4) and Integral of (xe^(x^2+1)) dx from 0 to 2 (a=0, b=2) The Attempt at a Solution I have done them out, just wondering if this is the best way to...- antinerd
- Thread
- Definite integrals Integrals
- Replies: 4
- Forum: Calculus and Beyond Homework Help
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Linear Approximation: Approximating \sqrt{4.1}-\sqrt{3.9}
Thanks.- antinerd
- Post #5
- Forum: Calculus and Beyond Homework Help
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Linear Approximation: Approximating \sqrt{4.1}-\sqrt{3.9}
It seems to me that it is a large margin of error, I was just wondering why this is so. Or maybe it's not too large, I'm not sure...- antinerd
- Post #3
- Forum: Calculus and Beyond Homework Help
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What is the derivative for x^2 + y^2 + ln(2) = xy?
Because isn't it: (2x-y) = dy/dx (x-2y) and then you divide to get dy/dx = (2x-y)/(x-2y)- antinerd
- Post #42
- Forum: Calculus and Beyond Homework Help
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What is the derivative for x^2 + y^2 + ln(2) = xy?
Wait I got (2x-y)/(x-2y) = dy/dx ...- antinerd
- Post #41
- Forum: Calculus and Beyond Homework Help
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Linear Approximation: Approximating \sqrt{4.1}-\sqrt{3.9}
Homework Statement OK, I'm doing this linear approximation problem: Approximate \sqrt{4.1} - \sqrt{3.9} Homework Equations f(a + h) ~ f(a) + hf`(a) The Attempt at a Solution This is what I have done so far: I approximated each square root separately... 4.1 = 4 + h h = .1 f(x) =...- antinerd
- Thread
- Approximation Linear
- Replies: 12
- Forum: Calculus and Beyond Homework Help
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Indefinite Integrals: Solving Homework Problems
Because if you take the derivative as such: log(3x+1) dx You will get d/dx f(g(x)) = f`(g(x))(g`(x)) Which means that: d/dx log(3x+1) = (1/(3x+1)) (3) = 3/(3x+1) But so now do I have to place a constant to make the derivative 15? I'm wondering if \int15/(3x+1) = just...- antinerd
- Post #8
- Forum: Calculus and Beyond Homework Help