Answer:
Part A)
In interval notation:
[tex](1, 5)[/tex]
Or, as an inequality:
[tex]1<x<5[/tex]
Part B)
[tex]x\approx 3.5[/tex]
Part C)
V(x) is increasing for 1 < x < 3.5 and decreasing for 3.5 < x < 5.
Step-by-step explanation:
The building boxes have lengths represented by (x + 1), widths by (5 - x), and heights by (x - 1).
Part A)
Since they are side lengths, they must be positive. In other words:
[tex]\displaystyle x+1>0, 5-x>0\text{ and } x-1>0[/tex]
Solving for each inequality yields:
[tex]x>-1, x<5,\text{ and } x>1[/tex]
We can eliminate the first inequality. Thus, our compound inequality is:
[tex]1<x<5[/tex]
So, the reasonable domain for V(x) will be all values greater than one and less than five.
In interval notation:
[tex](1, 5)[/tex]
Or as an inequality:
[tex]1<x<5[/tex]
Part B)
We only need to look at the part of the graph within our domain.
For 1 < x < 5, we can see that its maximum is 17.
This occurs around x = 3.5.
Part C)
Again, we only need to look at the part of the graph within our domain.
As we can see, starting from x = 1 to x = 3.5 (approximately), our function is increasing (i.e. sloping upwards).
From x = 3.5 (approximately) and ending at x = 5, our function is decreasing.
Therefore, V(x) is increasing for 1 < x < 3.5 and decreasing for 3.5 < x < 5.
