Wednesday, April 8, 2009

A Recurring Problem for SATs (Functions)

SAT "Grid-In" Type
Level of Difficulty: 5 (High)
Content: Algebra 2, precalculus

The function F satisfies the condition
F(N + 6) = F(N) + 8, for all integers N.
If F(7) = -2, what is the value of F(25)?

Click on Read more to see the answer, solution, discussion.

Answer: 22

Suggested Solution:

Replace N by 7 since F(7) is known:
Therefore, F(7 + 6) = F(7) + 8 or F(13) = -2 + 8 = 6

Next, replace N by 13 since F(13) is known:
F(19) = F(13) + 8 = 6 + 8 = 14

Finally, F(25) = F(19) + 8 = 14 + 8 = 22.

(1) Too difficult for the SATs? Not really! A similar problem recently appeared. There aren't that many "hard" questions (Level 5) on the SAT but, if a student wants to score over 700 they will need exposure to these types in practice.

(2) Consider writing some variations of these function-type problems for additional practice. At first, change the constants, then consider changing the operations (from addition to multiplication for example). One could raise the bar even higher by asking the question in reverse:
If F(25) = 22, what is the value of F(7)?

(3) There is considerable advanced theory in functional equations and recurrence relations underlying these problems. However, the student needs only to feel comfortable with the function symbolism (or should I call it "Math Notations!"). Starting by "plugging in' N = 7 seems simple in retrospect but most students are too intimidated to consider it. Even the precalculus student may be able to get started, but, without experience, they will often get lost. This is all about exposure, but isn't it always?

(4) One could rewrite this problem using sequence notation:
aN+6 = aN + 8. By expressing the problem in the context of the Nth term of a sequence, students may grasp it a bit better, but, in the end, it's all about interpreting function notation.

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