First of all, I'm new to multivariable calculus... in a multivariable function, by assuming that its

Donald Johnson 2022-01-04 Answered
First of all, I'm new to multivariable calculus... in a multivariable function, by assuming that its domain is going to be \(\displaystyle{R}^{{2}}\) and its image is going to be all real numbers, the graph of that function is defined as a subset of \(\displaystyle{R}^{{3}}\) in which the x and y axis are going to receive the inputs, and the output is going to be in \(\displaystyle{z}{\left({x},{y},{f{{\left({x},{y}\right)}}}\right)}\) ? Is that correct? Will its graph, in this example, be some kind of surface?

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Edward Patten
Answered 2022-01-05 Author has 1499 answers
Yes. If such a f(x,y) is defined for all real pairs (x,y), for each such pair corresponds a single real value, which could be taken as the z coordinate of a point. This interpretation leads to a surface, as you suppose.
Using Wolfram Alpha, for instance, you can get the plots of the following functions. Guessing the surfaces before looking them up would be an interesting exercise.
\(\displaystyle{f{{\left({x},{y}\right)}}}={x}\)
\(\displaystyle{f{{\left({x},{y}\right)}}}={y}\)
\(\displaystyle{f{{\left({x},{y}\right)}}}={x}+{y}\)
\(\displaystyle{f{{\left({x},{y}\right)}}}={x}−{y}\)
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