# Having trouble graphing values of x<0

This is prefaced by saying that I have at most 3 hours of experience in Mathematica, so please treat me gently. Anyways, my problem is that I'm trying to plot g(x) from x=-3 to x=3 and it seems to only want to display values of x>0 I've uploaded images of my inputs and what Mathematica is giving me.

I'm assuming the trouble is coming about because g(x) has a non-differentiable point a x=0, but even when I set the domain to x[-3,-1] and avoid zero entirely it still makes a graph with x[0,1] and doesn't even plot anything. Help please?

• Table[g[x], {x, -3, 3}] // N shows that results for x < 0 are imaginary. – Chris Degnen Nov 19 '17 at 0:47

The function in question is not real valued at negative x, so naturally Plot fails to graph the negative portions. If you intended for x^(2/3) to be the square of the real cube root of x, then you can explicitly request that with:

g[x_] := CubeRoot[x]^2/(1+x+x^4);


As an additional note, general real odd roots are provided by the Surd function, e.g. g[x_] := Surd[x,3]^2/(1+x+x^4).

• I'm able to get a graph on my TI-89 of this function for values less than zero, what's it doing differently? – Alex Nov 19 '17 at 2:27
• @Alex - Mathematica generally operates in the complex plane. Your calculator doesn't. – Bob Hanlon Nov 19 '17 at 4:35

x^(2/3) can be real or complex for negative x. Try

Plot[Re[g[x]], {x, -3, 3}]