rakave Posted March 9, 2005 Share Posted March 9, 2005 Hi can anyone help me sketch the shape of the graph X^X . Thank you Link to comment Share on other sites More sharing options...
fallen_6666 Posted March 9, 2005 Share Posted March 9, 2005 Umm, do you mean X to the power of X? wouldnt you just make it y = x^x so if x = 0 then y = 0^0, x=1 then y = 1^1, then x =2 y = 2^2, if x = 3 then y = 3^3? Link to comment Share on other sites More sharing options...
rakave Posted March 10, 2005 Author Share Posted March 10, 2005 hey it's easy to just say it. try drawing it. For negative powers it gets really confusing Link to comment Share on other sites More sharing options...
jordan Posted March 10, 2005 Share Posted March 10, 2005 Are you aware that the negative exponents are just the inverses of the corresponding positive exponent (for even x's) and the negative inverse of the corresponding positive exponent (for odd x's)? X=-2 ----- You know that 2^2=4 so -2^-2 is 1/4 X=-3 ----- You know that 3^3=27 so -3^-3 is -1/27 Link to comment Share on other sites More sharing options...
5614 Posted March 10, 2005 Share Posted March 10, 2005 Because: -2^-2 = 1 / -2^2 = 1 / 4 (because -2 x -2 (two negatives make a plus!) so 4) Link to comment Share on other sites More sharing options...
bloodhound Posted March 10, 2005 Share Posted March 10, 2005 the graph of x^x is weird. its alrite for positive values of x. not defined at 0, but tends to 1 as x tends ti 0 from the right. its alrite for negative integers. but for negative numbers like -1/2. we have (-1/2)^(-1/2) . which is is 1 over the square root if a negative number. which is not it ]R Link to comment Share on other sites More sharing options...
jordan Posted March 10, 2005 Share Posted March 10, 2005 Yeah, I forgot about all the fractions and stuff. My initial thoughts were to use complex numbers and graph it with those but then I figured that was propably too complicated so I just stuck to integers. Don't know why I didn't think of fractions. So if we used complex numbers, it would just have an asymptote at y=0, assuming the equation is z=x+iy? Link to comment Share on other sites More sharing options...
rakave Posted March 11, 2005 Author Share Posted March 11, 2005 It seems that even a few programs can't plot the curve.( see attachment) Link to comment Share on other sites More sharing options...
rakave Posted April 6, 2005 Author Share Posted April 6, 2005 so wat is the shape of the graph? atleast a rough sketch Link to comment Share on other sites More sharing options...
Dave Posted April 6, 2005 Share Posted April 6, 2005 ? You can't draw a graph for it for negative x; there's just too many discontinuities. Link to comment Share on other sites More sharing options...
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