on a separate sheet of paper, graph f(x)=x³ - 3x + 1 by using a table of values. then, estimate the x…

Explanation:

Step1: Create a table of values

Choose some values of (x) such as (x=- 3,-2,-1,0,1,2,3) and calculate (y = f(x)=x^{3}-3x + 1). When (x=-3), (y=(-3)^{3}-3\times(-3)+1=-27 + 9+1=-17). When (x=-2), (y=(-2)^{3}-3\times(-2)+1=-8 + 6+1=-1). When (x=-1), (y=(-1)^{3}-3\times(-1)+1=-1 + 3+1=3). When (x = 0), (y=0^{3}-3\times0+1=1). When (x = 1), (y=1^{3}-3\times1+1=1 - 3+1=-1). When (x = 2), (y=2^{3}-3\times2+1=8 - 6+1=3). When (x = 3), (y=3^{3}-3\times3+1=27-9 + 1=19).

Step2: Find zeros

A zero of the function (y = f(x)) is a value of (x) for which (y = 0). By looking at the sign - change of (y) values in the table, the function changes sign between (x=-2) and (x=-1), between (x = 0) and (x = 1), and between (x = 1) and (x = 2).

Step3: Find the derivative

The derivative of (y=f(x)=x^{3}-3x + 1) is (y'=f'(x)=3x^{2}-3). Set (y' = 0), then (3x^{2}-3=0), which simplifies to (x^{2}-1=0), so (x=\pm1).

Step4: Determine relative extrema

Take the second - derivative (y''=f''(x)=6x). When (x=-1), (y''=6\times(-1)=-6<0), so there is a relative maximum at (x=-1). When (x = 1), (y''=6\times1=6>0), so there is a relative minimum at (x = 1).

Step5: Determine domain and range

The domain of a polynomial function (y=x^{3}-3x + 1) is all real numbers, (D=(-\infty,\infty)). Since the function is a cubic function with a positive leading coefficient ((a = 1) in (y=ax^{3}+bx^{2}+cx + d)), the range is also all real numbers, (R=(-\infty,\infty)).

Answer:

Zeros between (x=-2) and (x=-1), between (x = 0) and (x = 1), and between (x = 1) and (x = 2). Relative maximum at (x=-1). Relative minimum at (x = 1). (D=(-\infty,\infty)), (R=(-\infty,\infty))