Recent content by eumyang

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    Removable discontinuity solution

    For a function to be continuous at a point c, three conditions must be met: 1) f(c) is defined. The left graph in my attachment shows an example of a graph where f(c) is NOT defined. 2) lim_{x \rightarrow c} f(x) exists. The middle graph shows an example of a graph where f(c) is defined, but...
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    What rule have they used to change the integral?

    It's called "adding zero." :tongue: I see it more as a trick to eventually introduce a factor that can be canceled. For instance, if the integral was this: \int \frac{3x}{(x-2)^2} dx I would subtract and add 6: = \int \frac{3x - 6 + 6}{(x-2)^2} dx = \int \frac{3x - 6}{(x-2)^2} dx + \int...
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    Integral error

    If dv = e^{ax} dx, then v = e^{ax} is wrong. There are also multiple errors on the second line, but you need to fix what I said first.
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    Derivatives with constants

    Perhaps, but I have seen students who do, however, use a quotient rule even if the denominator is a constant. Weird, for sure... (shrugs)
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    Proof by induction sequence

    Best I can do is give you some other examples that use the properties that you may need to continue: 4(2+5^{n-3}) = 8 + 4 \cdot 5^{n-3} -7 \cdot 4^{n-1} = -7 \cdot 4 \cdot 4^{n-2} = -28 \cdot 4^{n-2} 12 \cdot 10^{n-5} - 4 \cdot 10^{n-5} = 8 \cdot 10^{n-5} Study the above and try your...
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    Proof by induction sequence

    5 \cdot 2^{n-2} \ne 10^{n-2} 5 \cdot 3^{n-2} \ne 15^{n-2} ...and so on. I don't know if this will help, but note that 2^{n-2} = 2\cdot 2^{n-3} and 3^{n-2} = 3\cdot 3^{n-3}
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    Function and relations

    Isn't the notation wrong? It looks like you want (f \circ g)(x), (g \circ f)(x) and (f \circ h)(x) (function composition) but it looks more like (fg)(x), (gf)(x) and (fh)(x) (combining functions by multiplication)
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    Algebra/Trigonometey (Precal) books with rigor?

    How about another old one: Principles of Mathematics by Allendoerfer/Oakey?
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    Linearization of a function

    I was told that the linearization is defined this way: L(x) = f(a) + f'(a)(x - a), where f is differentiable at a.
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    Local Extrema

    I knew people who were fooled by this sort of problem. It is possible that a local maximum be "lower" than a local minimum. Look at the graphs of secant or cosecant, for example.
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    Finding the volume using cylindrical shells

    Almost. Switch the order of subtraction.
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    Finding the volume using cylindrical shells

    No, you don't want to add 2 and 1 + (y - 2)2.
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    Finding the volume using cylindrical shells

    (To help visualize what you did earlier, I attached two pics. The red region is what is being rotated around the x-axis. The "Wrong.bmp" file shows what you did, and the "Right.bmp" file shows what the problem is asking.) The way you had set it up, your heights of the representative rectangle...
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    Finding the volume using cylindrical shells

    Here's where you've gone wrong, I think. The representative rectangles need to be inside the parabola, not outside, as you have set the integral up.
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    Emptying a Tank of Beer

    Whew! Glad I'm not totally out of it. Thanks.