Find the range of values of $x$ for which $1-x<(x-1)(5-x)<3$.

Question

Find the range of values of $x$ for which $1-x<(x-1)(5-x)<3$.

First of all, I solved $1-x<(x-1)(5-x)<3$ which gives me $(x-1)(x-6)<0$ and $(x-4)(x-2)<0$.

How to find the range, here I get at least $4$ values of $x$. How do I plot them on a number line?

Answer 1

Hint: A product of two factors is negative if and only if one of the factors is negative and one is positive. When is one of the factors of $(x-1)(x-6)$ negative and one positive (you get two inequalities for $x$)? When is one of the factors of $(x-4)(x-2)$ negative and one positive (you get two more inequalities for $x$)? Which values of $x$ satisfy all conditions?

Answer 2

Here is a graph of the functions. Calculate the intersections and see for what values of x the inequality is true.

Answer 3

I think what you need is an elementary explanation on how to group the values of the two inequalities to arrive at a combined range and to plot them on the line so I am going to write this additional answer hopefully to get you there.

What you have done so far is perfect. You have separated the inequalities $1-x<(x-1)(5-x)$ and $ (x-1)(5-x) <3$, solved them separately and come up with two inequalities $(x-1)(x-6)<0 --- (1)$ and $(x-4)(x-2)<0 --- (2)$. Now each of these inequalities is part of the initial problem. The range of $x$ which satisfies $1-x<(x-1)(5-x)<3$ is that which satisfies both inequalities $(1) \ $ and $(2)$.

Let's start with $(2):$ which is $(x-4)(x-2)<0$

Now the way to do it to build separate cases. The trick is to find $3$ separate ranges for $x$ (since there are two factors) such that the sign of each factor is affected. I pick $x \le 2, 2 \lt x \lt 4$ and $ 4 \ge x$. See that these ranges exhaust all possible values for $x$. When $x \le 2$ both factors in $(2)$ are negative and hence the product is positive or zero so $x$ cannot be in this range. Similar reasoning will rule out $x \ge 4$ since both factors are positive. But if we pick any $x$ such that $ 2 \lt x \lt 4$ the first factor will be negative, the second positive and the product less than zero which says this is the range for which inequality $(2)$ is satisfied.

Similar reasoning will tell you that inequality $(1)$ is satisfied if and only if $1 \lt x \lt 6$. Now the range of $x$ for which our original inequality is satisfied is when $x$ adheres to both $ 2 \lt x \lt 4 ----(a) \ $ and $1 \lt x \lt 6 ---- (b)$. The ranges $2$ to $4$ and $1$ to $6$ are the ones you should plot in the number line.

Now what range for $x$ satisfies both $(a)$ and $(b)$. You should see that it is the section in which the red and blue ranges intersect - which is $2 \lt x \lt 4$. Or you can also take the less than and greater than parts separately and argue thus:

$x \gt 1$ and $x \gt 2 \implies x \gt 2$ AND $ x \lt 6 $ and $x \lt 4 \implies x \lt 4$, So $2 \lt x \lt 4$ woudl work.

Hope I helped.