Answer:
[tex]\displaystyle -\frac{1}{2} \leq x < 1[/tex]
Step-by-step explanation:
Inequalities
They relate one or more variables with comparison operators other than the equality.
We must find the set of values for x that make the expression stand
[tex]\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0[/tex]
The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.
The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as
[tex]\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0[/tex]
Simplifying by x-1 and taking x=1 out of the possible solutions:
[tex]\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0[/tex]
We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions
[tex](6x^3+14x^2+10x+12)[/tex]
is always positive and doesn't affect the result. It can be neglected. The expression
[tex](x-\frac{1}{2})^2[/tex]
can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression
[tex]\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0[/tex]
For the expression to be negative, both signs must be opposite, that is
[tex](x+\frac{1}{2})\geq 0, (x-1)<0[/tex]
Or
[tex](x+\frac{1}{2})\leq 0, (x-1)>0[/tex]
Note we have excluded x=1 from the solution.
The first inequality gives us the solution
[tex]\displaystyle -\frac{1}{2} \leq x < 1[/tex]
The second inequality gives no solution because it's impossible to comply with both conditions.
Thus, the solution for the given inequality is
[tex]\boxed{\displaystyle -\frac{1}{2} \leq x < 1 }[/tex]
