A few months ago we introduced the concept of a Function, which is the relation between a set of inputs and the related outputs. These problems are not difficult, but they require attention to detail and careful calculation. Today we will discuss combining two (or more) functions to produce a third function. This concept appeared 14 times last year, with a median placement at question # 66.
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You will probably remember that a typical function looks like this:
Let's say we want to find g[f(x)], using the two given functions above, where x = 5. It will be written as finding g[f(5)]. According to the order of operations, we'll work from the inside parentheses out, so we will find f(5) first:
- f(5) = 4(5) + 2 = 20 + 2 = 22.
- Now we will plug 22 in for x in the second function: g(22) = 22/2 - 4 = 11 - 4 = 7.
That's really all there is to it--the real "trick" to these is not to get lost in the process.
Example 1: g(x) = 2x + 3 and h(x) = 2 - 3x. g(h(4)) = ___
- Find h(4) first. h(4) = 2 - 3(4) = 2 - 12 = -10.
- Now find g(-10): 2(-10) + 3 = -20 + 3 = -17.
Example 2: g(x) = (x! - 3) and h(x) = x ÷ 7. h(g(4)) = ___
- Find g(4) first. g(4) = (4! - 3) = (24 - 3) = 21.
- Now find f(21): 21 ÷ 7 = 3.
Example 3: If f(x) = 2x - 5, then f[f(3)] = ___
- This problem requires performing the same function twice. f(3) = 2(3) - 5 = 6 - 5 = 1.
- f(1) = 2(1) - 5 = 2 - 5 = -3.
Example 4: f(x) = 2x - 7 and g(x) = 3x. f[g(-1)] = ___
- g(-1) = 3(-1) = -3.
- f(-3) = 2(-3) - 7 = -6 - 7 = -13.
Example 5: If f(x) = 3x + 2, g(x) = 4 - x, and h(x) = x + 1, then h{g[f(-1)]} = ___
- Work from the inside out, so find f(-1) first. f(-1) = 3(-1) + 2 = -3 + 2 = -1.
- g(-1) = 4 - (-1) = 5.
- h(5) = 5 + 1 = 6.
Example 6: If f(x) = 3x + 5, g(x) = x - 2, and h(x) = 3x, then f[g{h(1)}] = ___
- h(1) = 3(1) = 3.
- g(3) = 3 - 2 = 1.
- f(1) = 3(1) + 5 = 8.