Hello, function fans. In this encore to our function series, I am going to cover two more types of function problems that I did not cover in the series. Without further ado, these are…

1.) Nested Function Problems

These problems are the ones that define two different functions for you and then ask you to make computations involving numbers or variables based on combinations of those functions. For these, the most crucial thing to keep in mind is that you should always do the inner function first! Just like in any calculation where you follow PEMDAS, parentheses should be dealt with first – and this is no exception.

For example:

The function f(x) is defined for all real values of x as f(x) = 5x + 4 . The function g(x) is defined for all real values of x as g(x) = 3x² – 2. What is the value of f(g(2))?

Let’s evaluate the inner function first! g(2) = 3(2)² – 2 = 3 * 4 – 2 = 12 – 2 = 10.

Then, we can plug that into our outer function. Since g(2) = 10, then f(g(2)) must equal f(10). So, here we go! f(10) = 5(10) + 4 = 50 + 4 = 54.

So, f(g(2)) = 54. Problem solved!

You might have to evaluate this kind of problem, but in terms of variables. For example, using the functions f and g as defined above, you might be asked, which of the following answer choices is equivalent to f(g(x + 1)) ?

This looks scary but the exact same process that we used above can be applied here. Fist, we’ll evaluate the inner function, then plug that into the outer function. The algebra will be a little hairy since we’re keeping it in terms of variables, but we’ll deal with it.

The inner function: g(x + 1) = 3(x + 1) ² – 2 = 3(x² + 2x + 1) – 2 = 3x² + 6x + 3 – 2 = 3x² + 6x + 1.

The outer function: f(3x² + 6x + 1) = 5(3x² + 6x + 1) + 4 = 15x² + 30x + 5 + 4 = 15x² + 30x + 9. Problem solved!
(In a multiple choice situation, you would then simply see which answer matches this one and you’d be good to go!).

Sometimes, in a particularly tricky variation on this, the SAT might throw a nested symbol function problem at you. For example…
The function @a, b @ is defined as a+2ab + b and the function %(x) is defined by 3x/2 + 3. What is the value of %(@4, 8@) ?

Here again, we need to figure out the inner and outer functions and then follow the same process as above. The inner function here is the @a, b@ function and the outer one is the %(x) function.
So, we first need to evaluate @4, 8@ , which we are told must equal 4 + 2(4*8) + 8 = 4 + 2(32) + 8 = 4 + 64 + 8 = 76.
Then, we need to plug that into the %(x) function, so %(76) = 3(76)/2 + 3 = 228/2 + 3 = 114 + 3 = 117. Problem solved!

2.) Function Graph Transformation Questions

These questions involve various graphs of functions – sometimes recognizable ones like lines or parabolas, but often randomly-shaped function graphs – and ask you something about a possible transformation that could be applied to that graph.

These are pretty doable if you’re familiar with transformation rules – but can be very tricky if you aren’t. The main rules that you need to know are:

Vertical shifts: if a constant is added to the function – i.e. f(x) becomes f(x) + a or y = 3x becomes y = 3x – 2, then the graph gets shifted vertically up or down by the appropriate number of units. A positive constant means an upward shift and a negative one means a downward shift.

Horizontal shifts : if a constant is added to or subtracted from the argument or x term – i.e. f(x) becomes f(x+2) or g(x) becomes g(x-3) – then the graph gets shifted to the left or right by the appropriate number of units. A positive constant means a shift to the left and a negative one means a shift to the right (this is often counterintuitive for people, so be aware!).

Flipping across axes: if the x term is multiplied by -1, the graph is flipped across the y-axis. If the y or f(x) term is multiplied by -1, the graph is flipped across the x-axis.

Steepness: If the function is multiplied by a nonzero constant, its steepness will change accordingly. If that constant is greater than 1, the graph will become steeper/vertically stretched. If that constant is a fraction, the graph will become less steep/vertically compressed.

And that’s about all we’ve go to say about functions. Now that you have such a great background in functions, go ahead and practice them with confidence! You are now officially a function rockstar!

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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.