Question 1191739
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Here is some graphs:<br>
(1) The basic function |x|
{{{graph(400,400,-5,5,-2,8,abs(x))}}}<br>
(2) The graph of |x-2|: the basic graph shifted 2 to the right
{{{graph(400,400,-5,5,-2,8,abs(x-2))}}}<br>
(3) The graph of 4-|x|, which is the same as -|x|+4: the basic graph flipped vertically and translated up 4
{{{graph(400,400,-5,5,-2,8,4-abs(x))}}}<br>
The two given functions, graphed together:
{{{graph(400,400,-5,5,-2,8,abs(x-2),4-abs(x))}}}<br>
The coefficients of x in both functions are 1, so all the line segments in the graph of each function have slopes of 1 or -1; so the area between the graphs is a rectangle.<br>
Clearly two of the vertices of that rectangle are (0,4) and (2,0).<br>
It appears from the graph that the other two corners are (-1,3) and (3,1); substituting those coordinates in the two functions verifies that those are the other two corners.<br>
However, it is useful to go through the formal algebra to show that the points of intersection of the two graphs are (-1,3) and (3,1).<br>
We need to solve<br>
{{{abs(x-2)=4-abs(x)}}}<br>
The behavior of the two functions change at x=0 and x=2, so we need to check for solutions on three intervals: (1) x less than 0, (2) x between 0 and 2, and (3) x greater than 2.<br>
(1) for x negative, {{{abs(x-2)=2-x}}} and {{{4-abs(x)=4+x}}}<br>
2-x=4+x
2x=-2
x=-1 --> y=3<br>
(2) for x between 0 and 2, {{{abs(x-2)=2-x}}}and {{{4-abs(x)=4-x}}}<br>
2-x = 4-x
2=4
there is clearly no intersection of the two graphs between x=0 and x=2.<br>
(3) for x greater than 2, {{{abs(x-2)=x-2}}} and {{{4-abs(x)=4-x}}}<br>
x-2=4-x
2x=6
x=3 --> y=1<br>
So indeed the four corners of the rectangle are (2,0), (-1,3), (0,4), and (3,1).<br>
The width of the rectangle is the distance between (2,0) and (3,1), which is {{{sqrt(1+1)=sqrt(2)}}}.<br>
The length of the rectangle is the distance between (2,0) and (-1,3), which is {{{sqrt(9+9)=sqrt(18)=3*sqrt(2)}}}.<br>
The area of the rectangle is then length times width: {{{(sqrt(2))*(3*sqrt(2))=6}}}<br>