Welcome back to the function series! Today, we’re going to talk about some more complex/potentially confusing variations on the standard function problems we discussed last time. But, don’t stress – we’re going to apply the same principles we used last time, in terms of defining what you know and what you need to know in terms of the function and then plugging in accordingly to solve.

The type of problem we’re looking at today might take several slightly different forms, but generally will be something like this:

If f(x) = 2x + 4 for all real x values, and f(p) = 14, then what is the value of f(2p)?

This might look overwhelming at first. There are so many equations, and variables. Where can we start? How can we use all of this to get an answer to this question? Why, oh why, is the SAT so evil?!

It’s okay. Just breathe. Re-read the question and break it down, so you can figure out what each part of it is telling you and how that’s useful.

“If f(x) = 2x + 4 for all real x values”

This is the part where they’re defining the function for us. For any given x value, the value of the corresponding y value is equal to 2 times that number, plus 4.

“f(p) = 14”

Now we have a number, p, which is a specific x value (remember the function was defined as true for all values of x, so we’re not changing variables here; we’re just picking out a specific x value named p).  We also know that f(p) = 14, which means we can solve for the value of p by using the function as defined and plugging in 14 for f(p).

In other words, if f(x) = 2x + 4 for all real values of x, then f(p) = 2p + 4. If we know that f(p) = 14, then 14 = 2p + 4. That means that 2p = 10 and p = 5.

“what is the value of f(2p)?”

This tells us what we’re looking for: the value of f(2p).  Since we know that p = 5, then we know that 2p = 2 * 5, or 10. So, figuring f(2p) isn’t too bad. We just need to plug 10 into the function as our x value, and see what we get for a y value.  Since f(x) = 2x + 4 for all x values, f(10) must equal 2(10) + 4, or 24.  Problem solved!

There are a couple of possible variations on this problem type. Perhaps the function will include a constant, and you will need to solve for that constant and then plug back in to find whatever you’re asked to find.  (E.g. The function g(x) = 5x + k for all real values of x. If g(6) = 35, what is the value of g(8)?).  Perhaps the function will involve a quadratic relationship, so you will ultimately need to factor it and solve it as a quadratic equation.  (E.g. the function h(x) = x² + 16 for all real values of x.  If h(m) = 10m, what is one possible value of m?).  These are cases that we can look at more in depth in later weeks**, but the general process of solving will remain essentially the same as the one discussed above.

Tune in next time for a discussion of symbol functions, and then the following time for a discussion of table and/or graph-based function questions. Til then, be well, and may the function force be with you!

Here’s a summary of how to solve them, for the inquisitive ones among you.

For the first one, plug your given x and y values (6 and 35, respectively) into the equation to solve for k. So, 35 = 5(6) + k. That means 35 = 30 + k, so k = 5. So, that means g(x) = 5x + 5 for all values of x.  Then to find the value of g(8), simply plug 8 back into the function. g(8) = 5(5) + 5 = 45. Problem solved!

For the second one, you know that h(x) = x² + 16 for all real values of x . So, that means that h(m) = m² + 16. You also know that h(m) = 10m.  So, that means that h(m) = m² + 16 = 10m.  If m² + 16 = 10m, we can solve for m as a quadratic by getting everything on one side of the equals sign, so m² – 10m + 16 = 0, and then factoring and solving for m. In this case, (m -8) (m-2) = 0, so m can equal 8 or 2. Problem solved!

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Alice Rothman-Hicks is a Veritas Prep SAT 2400 instructor. Since graduating from Columbia University (Magna Cum Laude, Phi Beta Kappa), Alice has been teaching and tutoring test prep, helping students achieve their own academic successes. She scored a 2350 on the SAT.