User Portlet
| Discussions |
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| I am creating a user defined function f[t_] := where I am evaluating an integral in the body of the function without any issues. I can subsequently evaluate the user function f[t] and obtain the correct answer. However, when I plot the expression... |
| My goal is to display the inverse trigonometric function of ArcCos and Cos through a specific x range of 0 to Pi for Cos and -1 to 1 for the ArcCos since those domain ranges will pass the horizontal line test as being one-to-one and thus inverses... |
| Hi; In calculating an integral (using both the icon from the Classroom Pallett and the Integrate[] function in Mathematica) , it appears that Mathematica is placing an additional value in the results that is both unexpected and unwanted (see... |
| 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... |
| In order to get the exact answer using the Solve[] function, it would be useful to enter the specific interval. For example in calculating the values of x in the Solve function Solve[2 Sin[x]^2 - Sin[x] - 1 == 0, x, Reals] I would like the... |
| How would you go about solving the function: x^2+y^2- 2x + 6y + 9 = 0 in Mathematica? This function has the solution of: (x-1)^2+(y+ 3)^2 = 1, which is a circle of radius 1 centered at (1,-3). When using the Solve[] function of: ... |
| When I use the command "Graphics3D[Arrow[{{1, 1, 2}, {2, 2, 8}}]]" embedded at the end of a Show[] command, the arrow head seems to overwhelm the rest of the plot - see last graphic at end of notebook. Actually, I have more examples that are even... |