# Knowledge Base

## Solution 15629: Graphing Functions with Fractional Powers on Texas Instruments Products Where X < 0.

### Why are the graphs of x^(2/3) and (x^(1/3))2 not the same when x<0 on TI graphing calculators?

This information was originally written and posted to answer the specific question of why the graphs of x^(2/3) and(x^(1/3))^2 were not the same for x<0. It probably is a bit more detailed than the question demands, but in some ways the question raises some fairly complex issues.

To begin, some basic facts will make the discussion clear: (If the user follows along with a calculator, parentheses should be used in expressions such as (-3)^.5, because the negation key is a lower precedence than power.)

If we have x^(1/q) where x is a real number and q is a positive integer, there will be q answers (roots). These multiple values are called branches and the "principal branch" has magnitude (abs(x))^(1/q) and an angle that is (angle of x)/q. The user can get the angle of a complex number on the TI-85, for example, with the angle function. For the function to work right, the number should be entered as a complex data type, i.e. (-3,0). For positive real numbers the angle is zero, and for negative real numbers the angle is Pi. As a consequence, when x is positive, the principal branch is always a real number. (Zero divided by q is zero.) When x is negative, the principal branch is always complex for q>1. (Ð divided by a q>1 is never zero or Ð.) However, for x<0 and q an odd integer, one of the branches is always real. (The branches for q=3 have angles Pi/3, 3* Pi/3 and 5* Pi/3. The first branch is principal, the second branch is real. The branches for q=4 are Pi/4, 3* Pi/4, 5* Pi/4, and 7* Pi/4. When q is even, there is not a real branch.)

If we have x^(p/q) where x is a real number and p and q are positive integers, and compute the result as (x^(1/q))^p, the result is real if x^(1/q) is real. So, in general, x^y (for real x and y) has a real answer only for x>=0 or when y can be expressed as a rational number p/q, with q odd. However, as mentioned above, the principal branch of x^y is not real for x<0 and y<1.

Now to digress for a bit of history. The universal power function is calculated as: x^y = exp(y ln x). The early scientific calculators with a universal power or root function give an error for x^y if x<0 (even (-2)^2, for example). This is because ln(-2) is complex and the early calculators did not have the memory space to compute complex intermediate results or make special tests to reformulate the problem. However, for our more modern scientific calculators, it is our standard practice to make appropriate tests to return 4 for (-2)^2 and, since a root key is featured, to return answers when a real root exists (for instance: cube root of (-8) or (-8)^(1/3) returns -2). It is this feature that seems to cause some confusion when implemented on our graphing calculators.

The graphing calculators (as well as the TI-68) return the principal branch of x^y except when y can be expressed as a rational fraction (1/q) and q is odd. For this exception, the real branch is returned. The capability to return real results for integer roots of negative numbers may lead to some confusion, but have found it to have practical utility.

The principal branch of x^(1/3) and x^(2/3) are not real for x<0, but the graphing calculators plot the real branch for x^(1/3) due to the "odd root" feature. So (x^(1/3))^2 plots as a real branch, because an integer power of a real number is always real. The formulation (x^(1/q))^p or (x^p)^(1/q) plots the real branch of x^(p/q) if a real branch exists.

In the case of derivatives, if the numeric derivative function is used, you get a graph of the derivative over the same range that the function plots. However, the der1 and der2 functions on the TI-85 are exact derivatives and are computed in a manner equivalent to forming the symbolic derivative. So, der1(x^(1/3),x) only plots for x>0 because the derivative is 1/(3x^(2/3)), but der1(x^(2/3),x) plots for all x (except x=0) even though x^(2/3) doesn't plot for x<0, because the derivative is 2/(3x^(1/3)).