Hi mike! This question is from the May 2015 SAT.

(will post photo)

In the xy plane above, f and g are functions defined by f(x)=abs[x] and g(x)=-abs[x] + 3 for all values x. What is the area of the shaded region bounded by the graphs of the two functions?

You don’t even need to post the picture—I can graph those functions:

abs val graphs

First, note that the bounded region is a square. You know this if you remember that the absolute value graphs will create 45º angles with the axes, just like the graph of yx does.

Once you’ve realized that, all you need to do is find the length of one of its sides. Best way to go there is to recognize that the diagonal of the square goes from (0, 0) to (0, 3), so it has a length of 3. A square with a diagonal of length 3 will have sides of length \dfrac{3\sqrt{2}}{2}. (Brush up on your 45º-45º-90º triangles if you’re not clear why.)

So the area of the square is \dfrac{3\sqrt{2}}{2}\times\dfrac{3\sqrt{2}}{2}=\dfrac{9}{2}.

Comments (2)

Wow, I seriously over-complicated this…I kept plugging in values for x ad then i set the lines equal to other b/c they intersect.

How did you realize the second point for (0,3) though?
EDIT: nvm, I looked at my old algebra notes and realized. The graph (on the sat) had no intervals so I had a hard time.

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