It’s been a while since I’ve posted a challenge question. There are two main reasons: my own schoolwork, and the fact that most of my PWN time in the past month or so was dedicated to the Essay Guide. Now that I’m done with that, it’s time to get back in the swing of things with some pull-your-hair-out, way-harder-than-the-SAT-but-work-the-same-muscles math questions!

As always, the first non-anonymous commenter to get this right will win a copy of my Math Guide. Note that I’ll only count your first answer, which is emailed to me when you submit your comment—editing later doesn’t work. Don’t submit until you’re sure! (Full contest rules.)

Let’s do this.

The figure above shows the graph of

f(x). Ifais an integer such that

–6 <a< 6 andf(a) =a,what isf(f(f(a –3)) +f(f(f(f(2)))))?

MUAHAHAHAHA. Good luck!

UPDATE: It’s over! This one fooled a bunch of people. Solution below the cut.

There are a few steps here. I don’t think many of you had trouble with the first one, which is to figure out *a*. Here’s one an important thing to remember about function notation: you can translate function notation expressions like f(*p*) = *q* into ordered pairs (*p*, *q*). So when the question says *f*(*a*) = *a*, you’re looking for the ordered pair (*a*, *a*). There’s only one point on the graph where that occurs: (–2, –2). So *a* = –2, and the expression becomes:

*f*(

*f*(

*f*(–2

*–*3)) +

*f*(

*f*(

*f*(

*f*(2)))))

or…

*f*(

*f*(

*f*(–5)) +

*f*(

*f*(

*f*(

*f*(2)))))

From there, this question is really just an exercise in reading the expression carefully, and finding points on the graph. I think where a lot of people got turned around was that they didn’t work from the inside out, and therefore misread how the parentheses grouped the expression. Let’s go through it together, one step at a time.

*f*(

*f*(

*f*

**(**–5

**)**) +

*f*(

*f*(

*f*(

*f*

**(**2

**)**))))

*f*(–5) and

*f*(2). To do this, follow the

*x*-axis over to –5 and 2, and find the

*y*-coordinate of the graph at those points:

*f*(–5) = 2 and

*f*(2) = 3. Now just repeat that same process a few more times, working from the inside out.

*f*(

*f*

**(**2

**)**+

*f*(

*f*(

*f*

**(**3

**)**)))

*f*(2) on the left there, we run out of closed parentheses on the left.

*This does not mean I made a typo writing the question.*Rather, it means we have to do more work on the right to close off the first parentheses. Like so:

*f*(3 + *f*(*f***(**0**)**))

*f*(3 + *f***(**–3**)**)

*f***(**3 + 1**)**

*f*(4**)**

1

0 +1 = 1?

Well the answer is 1, but not because of 0 + 1. See below…

wait so i didn’t win?? what!!!??!?!

EDIT** Got the email… THANKS!!!!! 🙂

You did win. On the SAT, when you get the right answer for the wrong reason, you still get credit. Same deal here. 🙂

Oh well, guess I’ll have to buy the book…been meaning to check it out. I’ve heard good things

Is the answer 1?

yes!

The answer is 1. btw, f(f(f(a – 3)) is missing a parentheses and f(f(f(f(2))))) has an extra parentheses.

Explanation:

f(f(f(a – 3))): Since f(a)=a, we need to find a coordinate that has equal x and y values. The coordinate is (-2,-2). Since we found a, all we need to do is plug it into f(f(f(a – 3))), and find the result. f(f(f(a – 3))=f(f(f(-5)))=f(f(2))=f(3)=0.

f(f(f(f(2)))): We simply find it with our graph. f(f(f(f(2)))))=f(f(f(3))))=f(f(0))=f(-3)=1.

Hence, our answer is 0+1=1.

Nope…parenthesis are where they’re meant to be.

Actually, you are missing them.

http://www.wolframalpha.com/input/?i=f%28f%28f%28a+%E2%80%93+3%29%29&dataset=

http://www.wolframalpha.com/input/?i=f%28f%28f%28f%282%29%29%29%29%29+

No. The first f( contains everything else. See Tutor Delphia’s comment below.

O. that’s what you meant. ok then.f(3+1)=f(4) so 1

-2 + -2 = – 4

That was a typo.

Never mind; it’s -2

I’m getting 1 but because f(3+1)=f(4)=1

3+1

(-2,-2)

f(f(f(a – 3)) + f(f(f(f(2)))))

f(f(f(-2 – 3)) + f(f(f(3))))

f(f(f(-5)) + f(f(0)))

f(f(2) + f(-3))

f(3 + 1)

f(4)

1

Never mind! I just saw that “a is an integer” !