Determine whether the function f(x)[x+1] is odd, even, or neither.

Answer: neither odd nor even

2)Find the least integral upper bound of the zeros of the function f(x)=x^3-x^2+1.

Answer: upper bound, 1

To determine whether a function is odd, even, or neither, we need to evaluate the function at "x" and "-x" and compare the results.

For the function f(x) = x + 1, let's first evaluate f(-x):
f(-x) = (-x) + 1 = -x + 1

Now let's compare f(x) and f(-x):
f(x) = x + 1
f(-x) = -x + 1

If f(x) = f(-x), then the function is even.
If f(x) = -f(-x), then the function is odd.
If neither of the above conditions are met, then the function is neither odd nor even.

Let's substitute x = 1 into the function to check:
f(1) = 1 + 1 = 2
f(-1) = (-1) + 1 = 0

Since f(1) ≠ f(-1) and f(1) ≠ -f(-1), the function f(x) = x + 1 is neither odd nor even.

Moving on to the second question:

To find the least integral upper bound of the zeros of a function, we need to determine the smallest integer value greater than or equal to all the zeros.

For the function f(x) = x^3 - x^2 + 1, we want to find the least integral upper bound for its zeros.

We can use the Rational Root Theorem to check for potential rational zeros. According to the theorem, any rational zero of the form p/q (where p is the factors of the constant term and q is the factors of the leading coefficient) could be a potential zero. In our case, the leading coefficient is 1 and the constant term is 1.

The factors of 1 are ±1, while the factors of 1 are also ±1.

So, the potential rational zeros we need to check are: ±1.

Substituting x = 1 into the function f(x), we get:
f(1) = 1^3 - 1^2 + 1 = 1 - 1 + 1 = 1

Since f(1) ≠ 0, 1 is not a zero of the function.

Substituting x = -1 into the function f(x), we get:
f(-1) = (-1)^3 - (-1)^2 + 1 = -1 - 1 + 1 = -1

Since f(-1) ≠ 0, -1 is not a zero of the function.

Based on our analysis, we cannot find any integral upper bound that is greater than or equal to all the zeros of the function f(x) = x^3 - x^2 + 1. Therefore, the answer is that there is no least integral upper bound for the zeros of this function.