Pls. help me with this function?

Question

domain range intercepts asymtotes

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Answer 1

y = (x^3 - 1)^(1/3)
Domain: what values can you put in? Answer: anything, domain is all reals.

Range: what values can you get out? For what a is there no solution to a = (x^3 - 1)^(1/3)?
a^3 = x^3 - 1
a^3 + 1 = x^3
(a^3 + 1)^(1/3) = x, so for any a there is an x with f(x) = a; Range: all reals

Y intercept: x = 0, y = -1
X intercept(s): y = 0 = (x^3 - 1)^(1/3) x^3 - 1 = 0, x^3 = 1, x = 1 is the only solution.
Asymptotes: as x gets big the 1 becomes less and less important, the function starts to look like y = (x^3)^(1/3) = x, so the asymptote is y = x.

This is all restricted to reals. If you allow complex numbers the answers are different, but usually when you refer to x and y axes it means reals.

Answer 2

Cube roots are defined for negative arguments, so that's not a problem. The domain appears to be any real value.

Range is any real value.

x=0 -> y=-1 is the y intercept

y=0 -> x = 1 is the x intercept

It does not have any asymptotes.

I cheated by graphing it with my TI-83. It does have an undefined slope at x=1.

Note that for large |x| the -1 can be neglected and it looks like the function y=x.

Answer 3

....... _______
y = 3/(x^3 - 1) -----------------(1)

Resolving it comes to
....... ________
x = 3/(y^3 + 1) -----------------(2)

By trial,
putting x = 1 in equ.(1), y = 0
putting y = -1 in equ.(2), x = 0

The asymptote is (1,0), (0,-1)

Answer 4

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