Square root and cube root functions

Question

What is the end behavior of the function ?

 As x → –∞, f(x) → –∞, and as x → ∞, f(x) → ∞.

 As x → –∞, f(x) → ∞, and as x → ∞, f(x) → –∞.

 As x → –∞, f(x) → 0, and as x → ∞, f(x) → 0.

 As x → 0, f(x) → –∞, and as x → ∞, f(x) → 0.

Answer 1

-4x^1/3 approaches negative infinity as x approaches negative infinity

and as x approaches infinity =4x^1/3 approaches negative infinity

the 2nd choice is correct.

1st, and 4th choices half right

4th choice, totally wrong

Answer 2

Cube root functions are odd, since f(-x) = - f(x)

A quick sketch will show you that the correct answer is the second one:

As x → –∞, f(x) → ∞, and as x → ∞, f(x) → –∞.

Answer 3

If x is positive, its cube root gets larger, but the neg sign in front means that as x approaches +∞, f(x) gets more and more negative (approaches -∞)

.The opposite is true as x approaches -∞. The cube root of a negative is also negative. Think -27^1/3 or the cube rt of -27. That would be -3, because -3 x -3 x -3 is -27. As x gets more negative, its cube root gets more and more negative. Since the function has a negative sign in front, and since neg x neg is pos, the function as as whole gets bigger and bigger (approaches +∞) as x approaches -∞..

.

Looks like the 2nd option is the one to go with.