Solving the Inequality:
$\frac{(x-1)(x-2)^2}{x-3} \cdot (x-4)^2 \cdot (x-5)^4 \cdot (x-6)^3 \leq 0$
Step 1: Find the Critical Points
To solve the inequality, we need to find the critical points where the function changes signs. These critical points are the values of x that make the numerator or denominator zero.
- Numerator:
- $x-1=0 \Rightarrow x=1$
- $x-2=0 \Rightarrow x=2$
- Denominator:
- $x-3=0 \Rightarrow x=3$
- Other critical points:
- $x-4=0 \Rightarrow x=4$
- $x-5=0 \Rightarrow x=5$
- $x-6=0 \Rightarrow x=6$
Step 2: Create a Sign Chart
Create a sign chart using the critical points to determine the signs of the function in different intervals.
Interval | $(x-1)$ | $(x-2)^2$ | $(x-3)$ | $(x-4)^2$ | $(x-5)^4$ | $(x-6)^3$ | Function Sign |
---|---|---|---|---|---|---|---|
$(-\infty,1]$ | - | + | - | + | + | + | - |
$(1,2]$ | + | + | - | + | + | + | + |
$(2,3]$ | + | + | - | + | + | + | - |
$(3,4]$ | + | + | + | + | + | + | + |
$(4,5]$ | + | + | + | - | + | + | - |
$(5,6]$ | + | + | + | - | - | + | + |
$[6,\infty)$ | + | + | + | - | - | - | - |
Step 3: Solve the Inequality
Using the sign chart, we can determine the intervals where the function is less than or equal to zero.
- $(-\infty,1]$ : -
- $(2,3]$ : -
- $(4,5]$ : -
- $(5,6]$ : +
The final solution to the inequality is:
$x \in (-\infty, 1] \cup [2,3] \cup [4,5]$
Note that the endpoints are not included in the solution set because the inequality is strict.