What would the graph f of x equals x raised to the xth power look like when you allow x to be all real numbers?

Answer 1

For positive x, it will start near 1, for x near zero, then will dip down before increasing up to f(1) = 1, then it will increase more than exponentially for x > 1.

For negative values of x, most values of f(x) will be complex numbers, so you would only be able to plot scattered points on a real number graph.

For example (-1/2)^(-1/2) = 1/sqrt(-1/2) which is imaginary, but (-1/3)^(-1/3) is 1/cube_root(-1/3) which is a real number. Rational negative values of x should produce either pure real or pure imaginary values of f(x). Irrational negative values of x will produce f(x) being a complex number.

All negative integer values of x will result in f(x) real. As x becomes more negative, the values of f(x) will be exponentially closer to zero, but jumping between positive and negative values. Example (-2)^(-2) = (-1/2)^2 = 1/4, but (-3)^(-3) = (-1/3)^3 = -1/27

If you could plot an additional imaginary axis (perpendicular to the x and to the y axis, in 3-dimensions), f(x) would be seen spiraling around the x-axis, approaching the x-axis but never quite getting there, as x gets more negative. For values of x where f(x) is real, this is where the spiral intersects the 'real' x-y plane.