Show that the function f(x)= x^(3) +3/(x^2) +2 has exactly one zero on the interval (-infinity, 0).

So far this is what I have:

0=x^3 + 3/(x^2) +2
-2= (1/x^2)(x^5 + 3)
-2x^2= x^5 +3

But now I'm stuck. I also am not sure if this is how I'm supposed to be solving the problem. We had been learning Rolle's Theorem and the Mean Value Theorem, but none of those seem applicable here because they require that all points on the interval be differentiable, and if the function has a zero on the interval then it is not differentiable at that point. Am I going about this the right way? And how would I solve it from here?

To prove that the function f(x) = x^3 + 3/(x^2) + 2 has exactly one zero on the interval (-∞, 0), you can use the Intermediate Value Theorem. Here's how:

1. Start by rewriting the equation f(x) = 0:
x^3 + 3/(x^2) + 2 = 0

2. Notice that the function f(x) is continuous on the interval (-∞, 0) since it is a polynomial and rational function.

3. Evaluate the values of f(x) at the endpoints of the interval:
f(0) = 0^3 + 3/(0^2) + 2 = 2
f(-1) = (-1)^3 + 3/((-1)^2) + 2 = -2 + 3 + 2 = 3

4. Since f(0) = 2 and f(-1) = 3, the function passes through the values 2 and 3 on the interval (-∞, 0).

5. According to the Intermediate Value Theorem, if a function is continuous on an interval and takes on two distinct values at the endpoints, then it must take on every value in between.

6. Since f(x) goes from positive (2) to negative (3) as x goes from 0 to -1, it must pass through zero at some point in between.

Therefore, the function f(x) = x^3 + 3/(x^2) + 2 has exactly one zero on the interval (-∞, 0).

You were on the right track by setting f(x) = 0, but solving the resulting equation analytically can be difficult. Instead, using the Intermediate Value Theorem is a more appropriate method for proving the existence of a zero in this case.