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+--- Thread: Try to plot x^(2/3).. what happen for x<0 ? (/thread-13489.html)
Try to plot x^(2/3).. what happen for x<0 ? - josephec - 08-22-2019 03:45 AM
Hi Friends
Why In HP PRIME is different
(X^2)^(1/3) and (x^(1/3))^2 ?
Pd. The first one plots for x<0
Thanks for Your answers.
RE: Try to plot x^(2/3).. what happen for x<0 ? - rkf - 08-22-2019 06:45 AM
I assume the Prime does simpy follow the basic rules of Mathematics:
In the first expression, we start with squaring x, which yields for all reals a positive result - thus the cubic root is defined everywhere.
In the second expression, the cubic root is calculated first - but for reals is only defined for x >= 0. Thus the plot is left for x < 0.
RE: Try to plot x^(2/3).. what happen for x<0 ? - ijabbott - 08-22-2019 08:08 AM
You can redefine the second one as \( \big( \sqrt[3]{x}\big)^2 \) to square the real cube root of \( x\).
Note that \(x^\frac{1}{3}\) and \(\sqrt[3]{x}\) are treated differently, and that \(x^\frac{1}{3}\) usually produces a complex number when \( x < 0 \), and its square is also complex, so it cannot be plotted by the Function app.