To complete the table, we want to choose any value on the given intervals and test its sign on each columns, (t - 2) and (t - 1). If both have + signs, the las column will also have + sign. If both have - sign, the last column will have + sign. If one has + sign and the other has - sign, the last column will have - sign.
The first internal is "less than 1 hour".
One easi pick here is t = 0:
[tex]\begin{gathered} (t-2)=0-2=-2\longrightarrow- \\ (t-1)=0-1=-1\longrightarrow- \end{gathered}[/tex]Since both are negative, we have:
[tex]4(t-2)(t-1)\longrightarrow+[/tex]So, the first row is:
less than 1 hour - - +
The second is not an interval it is "1 hour", so:
[tex]\begin{gathered} (t-2)=1-2=-1\longrightarrow- \\ (t-1)=1-1=0\longrightarrow0 \end{gathered}[/tex]Since any number multiplied by 0 is also 0, the result on the third column will also be 0:
[tex]4(t-2)(t-1)\longrightarrow0[/tex]So, the second row is:
1 hour - 0 0
The third interval is "between 1 and 2 hours", so we can pick 1.5, for example. We have:
[tex]\begin{gathered} (t-2)=1.5-2=-0.5\longrightarrow- \\ (t-1)=1.5-1=0.5\longrightarrow+ \end{gathered}[/tex]Since one is - and the other is +, the third column will be negative:
[tex]4(t-2)(t-2)\longrightarrow-[/tex]So, the third row is:
between 1 and 2 hours - + -
In the fourth, we have a value of 2 hours, so:
[tex]\begin{gathered} (t-2)=2-2=0\longrightarrow0 \\ (t-1)=2-1=1\longrightarrow+ \end{gathered}[/tex]Since one of them is 0, the third column will also be 0, so:
[tex]4(t-2)(t-1)\longrightarrow0[/tex]So, the fourth row is:
2 hours 0 + 0
The last interval is "more than 2 hours", so we can pick 3, for example:
[tex]\begin{gathered} (t-2)=3-2=1\longrightarrow+ \\ (t-1)=3-1=2\longrightarrow+ \end{gathered}[/tex]Since both are +, the third column will also be +:
[tex]4(t-2)(t-1)\longrightarrow+[/tex]So, the fifth row is:
more than 2 hours + + +
So, we have:
interval of time (t-2) (t-1) 4(t-2)(t-1)
less than 1 hour - - +
1 hour - 0 0
between 1 and 2 hours - + -
2 hours 0 + 0
more than 2 hours + + +
