Question

find any points at which the function is discontinuous:
y=x^(2/3)-x^(-2/3)

Answer 1

this function has a nonremovable discontinuity at x=0; there is a vertical asymptote there

Answer 2

how do you know that?

Answer 3

if you put the function into a fractional form you can see that there is a real zero in the denominator which is the location of a VA\[y=x^{2/3}-\frac{1}{x^{2/3}}\]\[y=\frac{x^{4/3}-1}{x^{2/3}}\]

Answer 4

the real zero is, of course, x=0

Answer 5

when you can write a function in fractional form, any real zero of the denominator are the locations of VA; any zeros of the denominator produce removable discontinuities (holes)

Answer 6

so i can take the x^(2/3) out of the denominator and the numerator?

Answer 7

x^(-2/3) means reciprocal; i just got common denominators and made a single fraction out of it; btw here's the graph fyi

Answer 8

ok i think i understand.

Answer 9

i have another problem i cant do though...

Answer 10

y=(sqaure root of (x^4+1)/(1+sin^2x)