# Thread: Calculating Delta Δ

1. ## 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.

2. ## Re: Calculating Delta Δ

I really do not understand what you mean here. Normally $\displaystyle \Delta f(x)$ 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 $\displaystyle \Delta f(x)= f(x+ 1)- f(x)$

The you say "For i, I know that the derivative is 1". Did you intend "$\displaystyle \Delta x$ 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 $\displaystyle \Delta f$ is $\displaystyle f(x+1)- f(x)$, then $\displaystyle \Delta f(x)g(x)= f(x+1)g(x+ 1)- f(x)f(x)$ which can be written as $\displaystyle f(x+1)g(x+1)- f(x)g(x+ 1)+ f(x)g(x+ 1)- f(x)g(x)= (f(x+1)- f(x))g(x+1)- f(x)(g(x+1)- g(x))= g(x+1)\Delta f(x)+ f(x)\Delta g(x)$.

If you mean the derivative, normally written as d(f(x)g(x)/dx, then it is $\displaystyle [df(x)/dx]g(x)+ f(x)[dg(x)/dx]$.

3. ## 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.

4. ## Re: Calculating Delta Δ

Originally Posted by azollner95
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.

5. ## Re: Calculating Delta Δ

Originally Posted by azollner95
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
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(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.