- a.Determine the solution set for the inequal-
ity ≥0.
First, we must make certain that the numera-
tor and denominator are both completely fac-
tored and that all common terms are canceled,
as follows:= =Next determine the x-values that make this
expression equal to zero or undefined. Then,
we assess the sign of the expression on the left
side on each subinterval formed using these
values. To this end, observe that these values
are x= –^12 , 0, and ^23 . Now, we form a number
line, choose a real number in each subinterval,
and record the sign of the expression above
each:Since the inequality includes “equals,” we
include those values from the number line that
make the numerator equal to zero. Since none
of these values make the numerator equal to
zero, we conclude that the solution set is
(–^12 ,0)∪(0,^23 ).- a.Determine the solution set for the inequal-
ity 0.First, we must simplify the complex fraction
on the left side of the inequality:==
–(xx^22 (–x^2 –x 1 +)1)xx (^22) – (x 4 x+–3) 12 =
(–x( 2 x(x––^1 ) 12 )(xx–^2 ( 6 x)(+x^3 +)2)=
- ((xx––1 6 ))((xx++^32 ))
So, the original inequality can be written as- ((xx––1 6 ))((xx++^32 ))≥0, or equivalently (upon multi-
plication by –1 on both sides),((xx––1 6 ))((xx++^32 )) 0.
Next, we determine the x-values that make this
expression equal to zero or undefined, includ-
ing the values that make any factors common
to both numerator and denominator equal to
zero. Then, we assess the sign of the expression
on the left side on each subinterval formed
using these values. To this end, observe that
these values are x= –3, –2, 0, 1, and 6. Now,
we form a number line, choose a real number
in each subinterval, and record the sign of the
expression above each:
The inequality includes “equals,” so we include
those values from the number line that make
the numerator equal to zero. The solution set
is [–3, –2)∪[1, 6).3 2 0 1 6+ – + +
– –
– +
2(xx (^2) (–x^1 –) 1 – )x^2
x 2
x–^2 (^4 x(x++3)3)
x^22 – x–^1 1
x+^1 3 – x^4 ^2
x^22 – x–^1 1
x+^1 3 – x^42
0
- ++–
3 - 2
2 - 3
–(x^2 + 1)
x^2 (2x+1)(3x– 2)
–(x^2 + 1)
x^2 (6x^2 – x– 2)
–x^2 – 1
6 x^4 – x^3 – 2x^2–x^2 – 1
6 x^4 – x^3 – 2x^2ANSWERS & EXPLANATIONS–