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 analyze its symmetry.

1) Odd functions satisfy the property f(-x) = -f(x), meaning that if we replace x with -x, the function's value changes sign.
2) Even functions satisfy the property f(-x) = f(x), meaning that if we replace x with -x, the function's value remains the same.

Let's analyze the function f(x) = x + 1:
- Replacing x with -x, we get f(-x) = -x + 1.
- Comparing f(-x) with -f(x), we have -x + 1 = -(x + 1) = -x - 1.

Since f(-x) = -f(x) does not hold for all values of x, the function f(x) = x + 1 is neither odd nor even.

Now let's move on to finding the least integral upper bound of the zeros of the function f(x) = x^3 - x^2 + 1.

An upper bound is the highest value that a number can reach without going over the desired condition. In this case, we want to find the upper bound for the zeros of the function, which means finding the largest possible value for x where f(x) is zero.

To find the zeros of the function, we can set f(x) equal to zero and solve for x:
x^3 - x^2 + 1 = 0

Unfortunately, there is no straightforward way to find the exact values of the zeros of a cubic function. However, we can determine an upper bound for the zeros by testing different values of x.

Typically, we start by plugging in positive integers and negative integers into the function to see if they result in f(x) = 0. In this case, we can try a few positive and negative values to get an idea of the range where the zeros might lie.

Let's try plugging in x = 0, x = 1, and x = -1 into the function:
f(0) = 0^3 - 0^2 + 1 = 1 ≠ 0
f(1) = 1^3 - 1^2 + 1 = 1 ≠ 0
f(-1) = (-1)^3 - (-1)^2 + 1 = -1 - 1 + 1 = -1 ≠ 0

Since none of these values give us f(x) = 0, we can conclude that the zeros lie outside the range we've tested so far.

To find an upper bound for the zeros, we can keep testing larger positive values of x until we find a value where f(x) becomes positive. Similarly, we can test smaller negative values until we find a value where f(x) becomes negative.

By trial and error, we can find that for x = 1, f(x) = 1 > 0, and for x = -1, f(x) = -1 < 0.

Since f(x) changes sign between x = -1 and x = 1, we can conclude that the zeros of the function f(x) = x^3 - x^2 + 1 lie between -1 and 1.

Therefore, the least integral upper bound of the zeros of the function is 1.