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 examine the symmetry of the function graph.
1) For a function to be odd, it needs to satisfy the condition f(-x) = -f(x) for all x.
Similarly, for a function to be even, it needs to satisfy the condition f(-x) = f(x) for all x.
Considering the function f(x) = x + 1, we can determine its symmetry:
- To check for odd symmetry, we substitute -x for x and compare the results:
f(-x) = -x + 1
However, -f(x) = -(x + 1) = -x - 1
Since f(-x) is not equal to -f(x), the function does not satisfy the condition for odd symmetry.
- To check for even symmetry, we substitute -x for x and compare the results:
f(-x) = -x + 1
f(x) = x + 1
Since f(-x) is not equal to f(x), the function also does not satisfy the condition for even symmetry.
Therefore, the function f(x) = x + 1 is neither odd nor even.
2) Finding the least integral upper bound of the zeros of a function involves determining the smallest integer value above which all the zeros lie.
Considering the function f(x) = x^3 - x^2 + 1, we need to find the least integral upper bound of its zeros.
- One approach is to use calculus to analyze the function's behavior and find critical points. However, in this case, we can rely on a simpler approach:
Since the highest power of x is x^3, the function is a cubic function.
Cubic functions, by definition, have either one or three real zeros.
To find the least integral upper bound, we can test a few positive values of x and determine when the function becomes positive.
- For x = 1, f(1) = 1^3 - 1^2 + 1 = 1 - 1 + 1 = 1
Since f(1) is positive, we can conclude that all the zeros lie above x = 1.
Therefore, the least integral upper bound of the zeros of the function f(x) = x^3 - x^2 + 1 is 1.