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Thread: Calculating Delta Δ

  1. #1
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    Angry Calculating Delta Δ

    For both questions: Let f(x) and g(x) be real-valued functions defined on R (All real numbers).

    i.) Calculate Δ(f(x)g(x)

    ii.) Let f(x) = x^2*2^x. Use (i) to calculate Δf(x)
    Read: "Let f of x equal x squared times 2 to the x. Use part i to calculate delta f of x.

    For i, I know that the derivative is 1, but I am unsure if that is the answer or how I can use that to get my answer, and obviously if I don't have part i correct I cannot attempt part ii.
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  2. #2
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    Re: Calculating Delta Δ

    I really do not understand what you mean here. Normally \Delta f(x) means the small change in f(x) corresponding to some small change in x but you have not said what small change in x is to be used. Do you intend that small change to be 1?

    That is, do you mean \Delta f(x)= f(x+ 1)- f(x)

    The you say "For i, I know that the derivative is 1". Did you intend " \Delta x to be the derivative? No, the derivative of f(x)g(x) is not "1"! Since f(x) and g(x) can be any real-valued functions. (Are you not even requiring f and g to be continuous?) I don't see how you can get a specific number for i. If \Delta f is f(x+1)- f(x), then \Delta f(x)g(x)= f(x+1)g(x+ 1)- f(x)f(x) which can be written as f(x+1)g(x+1)- f(x)g(x+ 1)+ f(x)g(x+ 1)- f(x)g(x)= (f(x+1)- f(x))g(x+1)- f(x)(g(x+1)- g(x))= g(x+1)\Delta f(x)+ f(x)\Delta g(x).

    If you mean the derivative, normally written as d(f(x)g(x)/dx, then it is [df(x)/dx]g(x)+ f(x)[dg(x)/dx].
    Last edited by HallsofIvy; Apr 3rd 2017 at 02:12 PM.
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    Re: Calculating Delta Δ

    I wish I could be more clear. Our prof simply assigned this set of questions and was not at all helpful so the question I wrote out is exactly what he is asking us.
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    Re: Calculating Delta Δ

    Quote Originally Posted by azollner95 View Post
    For both questions: Let f(x) and g(x) be real-valued functions defined on R (All real numbers).
    ii.) Let f(x) = x^2*2^x. Use (i) to calculate Δf(x)
    Read: "Let f of x equal x squared times 2 to the x. Use part i to calculate delta f of x.
    $\Delta f(x)=\dfrac{(x+h)^2\cdot 2^{x+h}-x^2\cdot 2^x}{h}$
    Only your instructor knows how much more he/she wants. We do not.
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  5. #5
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    Re: Calculating Delta Δ

    Quote Originally Posted by azollner95 View Post
    I wish I could be more clear. Our prof simply assigned this set of questions and was not at all helpful so the question I wrote out is exactly what he is asking us.
    For both questions: Let f(x) and g(x) be real-valued functions defined on R (All real numbers).

    i.) Calculate Δ(f(x)g(x)

    ii.) Let f(x) = x^2*2^x. Use (i) to calculate Δf(x)
    Read: "Let f of x equal x squared times 2 to the x. Use part i to calculate delta f of x.

    For i, I know that the derivative is 1, but I am unsure if that is the answer or how I can use that to get my answer, and obviously if I don't have part i correct I cannot attempt part ii.
    just so you know, $\Delta (whatever)$ normally indicates a change in $whatever$

    $\Delta x = x_2 - x_1$

    $\Delta f(x) = f(x_2)-f(x_1)$

    $\Delta[f(x) \cdot g(x)] = f(x_2) \cdot g(x_2) - f(x_1) \cdot g(x_1)$


    (i) IF this notation (missing a closing parenthsis) ...

    Δ(f(x)g(x)
    ... is meant to be the derivative of the product of two functions, then using a more standard notation ...

    $\dfrac{d}{dx}\bigg[f(x) \cdot g(x)\bigg] = f(x) \cdot g'(x) + f'(x) \cdot g(x)$


    I have no idea how you came up with this conclusion ...
    For i, I know that the derivative is 1
    -----------------------------------------------------------------------------------------------------------------------------------------------------

    (ii) $f(x) = x^2 \cdot 2^x \implies f'(x) = x^2 \cdot 2^x \cdot \ln{2} + 2x \cdot 2^x = x \cdot 2^x(x\ln{2} + 2)$


    ... of course, all this could be BS because what is being asked is unclear.
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