Evaluating a piecewise defined function

Given : if x belongs to the interval -2 ≤ x < -1, then f(x) = -2

As x = -2 belongs to the interval -2 ≤ x < -1 ⇒ f(-2) = -2

As x = -1.5 belongs to the interval -2 ≤ x < -1 ⇒ f(-1.5) = -2

Given : if x belongs to the interval 1 ≤ x < 2, then f(x) = 1

As x = 1 belongs to the interval 1 ≤ x < 2 ⇒ f(1) = 1

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.