Question

How do you solve the inequality $\dfrac{{\left( {x - 3} \right)\left( {x - 4} \right)}}{{\left( {x - 5} \right){{\left( {x - 6} \right)}^2}}} \leqslant 0$?

Answer 1

Hint: First of all, find all the points at which the function takes the value 0 and is undefined. Then, we will see where the function takes positive and negative values and thus we have the answer.

Complete step-by-step solution:
We are given that we are required to solve the inequality $\dfrac{{\left( {x - 3} \right)\left( {x - 4} \right)}}{{\left( {x - 5} \right){{\left( {x - 6} \right)}^2}}} \leqslant 0$.
Let us say $f(x) = \dfrac{{\left( {x - 3} \right)\left( {x - 4} \right)}}{{\left( {x - 5} \right){{\left( {x - 6} \right)}^2}}}$
Let us now find the points where the function can be zero or undefined.
We know that those points are 3, 4, 5 and 6.
Now, we will see how our function behaves before, after and in between these points.
Let us first consider before 3, let us put in x = 2 in the given function, we will then get:-
$ \Rightarrow f(2) = \dfrac{{\left( {2 - 3} \right)\left( {2 - 4} \right)}}{{\left( {2 - 5} \right){{\left( {2 - 6} \right)}^2}}}$
Simplifying the numerator and the denominator on the right hand side of above equation, we will then obtain the following function with us:-
$ \Rightarrow f(2) = \dfrac{{\left( { - 1} \right)\left( { - 2} \right)}}{{\left( { - 3} \right){{\left( { - 4} \right)}^2}}}$
Simplifying the numerator and the denominator on the right hand side of above equation further, we will then obtain the following function with us:-
$ \Rightarrow f(2) = \dfrac{2}{{ - 48}} = - \dfrac{1}{{24}} < 0$
Let us now consider between 3 and 4, let us put in $x = 3.5$ in the given function, we will then get:-
\[ \Rightarrow f\left( {3.5} \right) = \dfrac{{\left( {3.5 - 3} \right)\left( {3.5 - 4} \right)}}{{\left( {3.5 - 5} \right){{\left( {3.5 - 6} \right)}^2}}}\]
Simplifying the numerator and the denominator on the right hand side of above equation, we will then obtain the following function with us:-
$ \Rightarrow f(3.5) = \dfrac{{\left( {0.5} \right)\left( { - 0.5} \right)}}{{ - 1.5{{\left( { - 2.5} \right)}^2}}}$
Simplifying the numerator and the denominator on the right hand side of above equation further, we will then obtain the following function with us:-
$ \Rightarrow f(3.5) = \dfrac{{ - 0.25}}{{( - 1.5)(6.25)}} > 0$
Let us now consider between 4 and 5, let us put in $x = 4.5$ in the given function, we will then get:-
\[ \Rightarrow f\left( {4.5} \right) = \dfrac{{\left( {4.5 - 3} \right)\left( {4.5 - 4} \right)}}{{\left( {4.5 - 5} \right){{\left( {4.5 - 6} \right)}^2}}}\]
Simplifying the numerator and the denominator on the right hand side of above equation, we will then obtain the following function with us:-
$ \Rightarrow f(4.5) = \dfrac{{\left( {1.5} \right)\left( {0.5} \right)}}{{ - 0.5{{\left( { - 1.5} \right)}^2}}}$
Simplifying the numerator and the denominator on the right hand side of above equation further, we will then obtain the following function with us:-
$ \Rightarrow f(4.5) = \dfrac{{(1.5)(0.5)}}{{( - 0.5)(2.25)}} < 0$
Let us now consider between 5 and 6, let us put in $x = 5.5$ in the given function, we will then get:-
\[ \Rightarrow f\left( {5.5} \right) = \dfrac{{\left( {5.5 - 3} \right)\left( {5.5 - 4} \right)}}{{\left( {5.5 - 5} \right){{\left( {5.5 - 6} \right)}^2}}}\]
Simplifying the numerator and the denominator on the right hand side of above equation, we will then obtain the following function with us:-
$ \Rightarrow f(5.5) = \dfrac{{\left( {2.5} \right)\left( {1.5} \right)}}{{0.5{{\left( { - 0.5} \right)}^2}}}$
Simplifying the numerator and the denominator on the right hand side of above equation further, we will then obtain the following function with us:-
$ \Rightarrow f(5.5) = \dfrac{{(2.5)(1.5)}}{{(0.5)(0.25)}} > 0$
Let us now consider after 6, let us put in $x = 7$ in the given function, we will then get:-
\[ \Rightarrow f\left( 7 \right) = \dfrac{{\left( {7 - 3} \right)\left( {7 - 4} \right)}}{{\left( {7 - 5} \right){{\left( {7 - 6} \right)}^2}}}\]
Simplifying the numerator and the denominator on the right hand side of above equation, we will then obtain the following function with us:-
$ \Rightarrow f(7) = \dfrac{{\left( 4 \right)\left( 3 \right)}}{{\left( 2 \right){{\left( 1 \right)}^2}}}$
Simplifying the numerator and the denominator on the right hand side of above equation further, we will then obtain the following function with us:-
$ \Rightarrow f(7) = \dfrac{{12}}{2} = 6 > 0$
We can clearly see that the answer is $x \in \left( { - \infty ,\left. 3 \right]} \right. \cup \left[ {4,\left. 5 \right)} \right.$.
Thus, we have the required answer.

Note: The students must note that we included 3 and 4 in the solution set because the inequality was not strict but less than and equal to sign was given to us. Therefore, we had to consider the possibility of the function being 0 as well. Now, we did not include 5 because if x would have been equal to 5, then the denominator would have become equal to 0 and the function would not be defined at all.