Question

Solve the given inequality :

Answer 1

Answer:

x < -5  or  x = 1  or  2 < x < 3  or  x > 3

Step-by-step explanation:

Given rational inequality:

$\dfrac{(x-1)^2(x-2)^3}{(x^2-5x+6)^2(x+5)}\geq 0$

$\textsf{Factor }(x^2-5x+6):$

$\implies x^2-2x-3x+6$

$\implies x(x-2)-3(x-2)$

$\implies (x-3)(x-2)$

Therefore:

$\dfrac{(x-1)^2(x-2)^3}{(x-3)^2(x-2)^2(x+5)}\geq 0$

Find the roots by solving f(x) = 0  (set the numerator to zero):

$\implies (x-1)^2(x-2)^3=0$

$\implies (x-1)^2=0\implies x=1$

$\implies (x-2)^3=0 \implies x=2$

Find the restrictions by solving f(x) = undefined  (set the denominator to zero):

$\implies (x-3)^2(x-2)^2(x+5)=0$

$\implies (x-3)^2=0 \implies x=3$

$\implies (x-2)^2=0 \implies x=2$

$\implies (x+5)=0 \implies x=-5$

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:

$f(-6)=\dfrac{(-6-1)^2(-6-2)^3}{(-6-3)^2(-6-2)^2(-6+5)}=\dfrac{(+)(-)}{(+)(+)(-)}=+$

$f(0)=\dfrac{(0-1)^2(0-2)^3}{(0-3)^2(0-2)^2(0+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-$

$f(1.5)=\dfrac{(1.5-1)^2(1.5-2)^3}{(1.5-3)^2(1.5-2)^2(1.5+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-$

$f(2.5)=\dfrac{(2.5-1)^2(2.5-2)^3}{(2.5-3)^2(2.5-2)^2(2.5+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+$

$f(4)=\dfrac{(4-1)^2(4-2)^3}{(4-3)^2(4-2)^2(4+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+$

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:

$(- \infty,-5) \cup x=1 \cup (2,3) \cup(3,\infty)$