Answer:
x < -5 or x = 1 or 2 < x < 3 or x > 3
Step-by-step explanation:
Given rational inequality:
[tex]\dfrac{(x-1)^2(x-2)^3}{(x^2-5x+6)^2(x+5)}\geq 0[/tex]
[tex]\textsf{Factor }(x^2-5x+6):[/tex]
[tex]\implies x^2-2x-3x+6[/tex]
[tex]\implies x(x-2)-3(x-2)[/tex]
[tex]\implies (x-3)(x-2)[/tex]
Therefore:
[tex]\dfrac{(x-1)^2(x-2)^3}{(x-3)^2(x-2)^2(x+5)}\geq 0[/tex]
Find the roots by solving f(x) = 0 (set the numerator to zero):
[tex]\implies (x-1)^2(x-2)^3=0[/tex]
[tex]\implies (x-1)^2=0\implies x=1[/tex]
[tex]\implies (x-2)^3=0 \implies x=2[/tex]
Find the restrictions by solving f(x) = undefined (set the denominator to zero):
[tex]\implies (x-3)^2(x-2)^2(x+5)=0[/tex]
[tex]\implies (x-3)^2=0 \implies x=3[/tex]
[tex]\implies (x-2)^2=0 \implies x=2[/tex]
[tex]\implies (x+5)=0 \implies x=-5[/tex]
Create a sign chart, using closed dots for the roots and open dots for the restrictions (see attached).
Choose a test value for each region, including one to the left of all the critical values and one to the right of all the critical values.
Test values: -6, 0, 1.5, 2.5, 4
For each test value, determine if the function is positive or negative:
[tex]f(-6)=\dfrac{(-6-1)^2(-6-2)^3}{(-6-3)^2(-6-2)^2(-6+5)}=\dfrac{(+)(-)}{(+)(+)(-)}=+[/tex]
[tex]f(0)=\dfrac{(0-1)^2(0-2)^3}{(0-3)^2(0-2)^2(0+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-[/tex]
[tex]f(1.5)=\dfrac{(1.5-1)^2(1.5-2)^3}{(1.5-3)^2(1.5-2)^2(1.5+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-[/tex]
[tex]f(2.5)=\dfrac{(2.5-1)^2(2.5-2)^3}{(2.5-3)^2(2.5-2)^2(2.5+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+[/tex]
[tex]f(4)=\dfrac{(4-1)^2(4-2)^3}{(4-3)^2(4-2)^2(4+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+[/tex]
Record the results on the sign chart for each region (see attached).
As we need to find the values for which f(x) ≥ 0, shade the appropriate regions (zero or positive) on the sign chart (see attached).
Therefore, the solution set is:
x < -5 or x = 1 or 2 < x < 3 or x > 3
As interval notation:
[tex](- \infty,-5) \cup x=1 \cup (2,3) \cup(3,\infty)[/tex]
