Question

How do you find the end behavior of `f(x) = 4x- 5x^3`?

Answer 1

Take the derivative, look at the signs.

Explanation:

`f'(x) = 4-15x^2`.
This equation shows the rate of change of `f(x)` at certain x value.
From the equation you can see that `f'(x)>=0` when `-2/sqrt(15)<=x<=2/sqrt(15)`. For all other values, `f'(x)<0`.

The end behavior of `f(x)=4x-5x^3` is that `f(x)` approaches `-oo` as `x -> oo` and `oo` as `x -> oo`.

Note: `f(x)` approaches `oo` as `x` decreases because negative `f'(x)` means `f(x)` decreases as `x` increases.