This is one of my favorite functions. x can equal zero. Only when dealing with limits is 0⁰ undefined, otherwise it is 1. x will only be defined in the negatives for integer values because you can’t take the root (fractional exponent) of a negative number. Since we’re working with real numbers, don’t worry about that. Otherwise x is defined for all real numbers, and the function will go off to infinity. The range is going to be the trickiest thing to find. At x = 1 it is obviously 1 and at x = 0 it is 1, but between them it will be smaller than 1 because we know that (1/2)^(1/2) = 1/√2 < 1. Because y(0) = 1 and y(1) = 1 you can conclude by Rolle’s theorem (a specific case of the mean value theorem) that the derivative will equal zero somewhere between x = 0 and x = 1. This will be the minimum. To derive this you much use logarithmic differentiation. ln(y) = xln(x) (dy/dx) / y = ln(x) + (x/x) = ln(x) + 1 dy/dx = y(ln(x) + 1) = (x^x)(ln(x) + 1) 0 = (x^x)(ln(x) + 1) x^x = 0 which doesn’t happen ln(x) + 1 = 0 ln(x) = -1 x = e⁻¹ = 1/e y = (1/e)^(1/e) = (e⁻¹)^(e⁻¹) = e^(-e⁻¹) Domain: {x ∈ ℝ | x ≥ 0} Range: {y ∈ ℝ | y ≥ e^(-e⁻¹)} If you want to include negative numbers and not complex numbers write it as Domain: {x ∈ ℝ | x ≥ 0} ∪ {x ∈ ℤ | x < 0} 0}
The range changes because as x approaches negative infinity the function approaches zero.
Originally posted 2013-11-09 22:47:20.