User Portlet
| Discussions |
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| Hi; I am having difficulty creating a plottable user function using the Integrate[] function - please see attached. Given, I am using the Integrate[] function to create an antiderivative with one variable and assuming that the constant is 0 in my... |
| Hi; When I solve the function f[x] = 3x^(2/3)-2x at the x values of [-1,0,1], the independent value of -1 gives a complex number and I cannot understand why - see attached. When I treat the statement of x^(2/3) as the cubed root of x squared... |
| Hi; I am having difficulty displaying my two entering functions in the ContourPlot[] function (Please see attached). As my attachment shows, I can enter my two functions individually in two separate ContourPlot[] functions and combine them using... |
| Hi; In the function that I am using x is undefined at value 3 (see attached). To circumvent the problem of x being undefined, I created a piecewise function defining x at value 3 to be zero and subsequently created a table using the piecewise... |
| Thanks so much. I also found that Reduce[] works real well also. Additionally you do not need to suffix // DeleteDuplicates to the Reduce[] expression. Thanks again, Mitch Sandlin |
| Hi Murray; I tried duplicating your example, but did not get the same results. Please see attached notebook and give me your thoughts. Thanks so much, Mitch Sandlin |
| The one attached below - GradWork01. Not sure what happened to the one that I attached earlier. Thanks so much for reviewing this. Mitch Sandlin |
| Hi; I really like using the "D" Command because of its format and ease of use, however the command seems to cause me calculation problems - See Attached. I defined the function f[x_,y_] and used the function to generate derivatives for both x... |
| Thanks so much for your reply. I was under the impression that D was a shortcut for Derivative, which is apparently not the case. However there seems to still be a problem with D[f[1, 1], x] and D[f[1, 1], y], since it still gave me an answer... |
| Hi Christos; Knowing what the function is equal to is part of the problem. It is actually equal to 5 but you don't know that until you complete the squares. Once you know the function equals 5, you can plug the function into Solve[] and get the... |