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Operations on Functions

You can add, subtract, multiply, and divide numbers. You can add, subtract, multiply, and divide polynomials. You can also add, subtract, multiply, and divide functions. And performing these operations on functions is no more complicated than the notation itself. For instance, when they give you the formulas for two functions and tell you to find the sum, all they're telling you to do is add the two formulas. There's nothing more to this topic than that, other than perhaps some simplification of the expressions involved.

• Given f(x) = 3x + 2 and g(x) = 4 – 5x,
    
  find (f + g)(x), (f – g)(x), (f×g)(x), and (f / g)(x).

To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order they tell me to.

(f + g)(x) = f(x) + g(x) = [3x + 2] + [4 – 5x] = 3x – 5x + 2 + 4 = –2x + 6

(f – g)(x) = f(x) – g(x) = [3x + 2] – [4 – 5x] = 3x + 5x + 2 – 4 = 8x – 2

(f×g)(x) = [f(x)][g(x)] = (3x + 2)(4 – 5x) = 12x + 8 – 15x² – 10x = –15x² + 2x + 8

• Given f(x) = 2x, g(x) = x + 4, and h(x) = 5 – x³,
    
  find (f + g)(2), (h – g)(2), (f × h)(2), and (h / g)(2).

To find the answers, I can either work symbolically (like in the previous example) and then evaluate, or I can find the values of the functions at x = 2 and then work from there. It's probably simpler to evaluate first, so:

f(2) = 2(2) = 4

g(2) = (2) + 4 = 6

h(2) = 5 – (2)³ = 5 – 8 = –3

Now I can evaluate the listed expressions:

(f + g)(2) = f(2) + g(2) = 4 + 6 = 10

(h – g)(2) = h(2) – g(2) = –3 – 6 = –9

(f × h)(2) = f(2) × h(2) = (4)(–3) ;= –12

(h / g)(2) = h(2) ÷ g(2) = –3 ÷ 6 = –0.5

If you work symbolically first, and plug in the x-value only at the end, you'll still get the same results. Either way will work. Evaluating first is usually easier, but the choice is up to you.

• Given f(x) = 3x² – x + 4, find the simplified form of the following expression, and evaluate at h = 0:
     
          Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

This isn't really a functions-operations question, but something like this often arises in the functions-operations context. The simplest way for me to proceed with this exercise is to work in pieces, simplifying as I go; then I'll put everything together and simplify at the end.

f(x + h) = 3(x + h)² – (x + h) + 4
             = 3(x² + 2xh + h²) – x – h + 4
             = 3x² + 6xh + 3h² – x – h + 4

f(x) = 3x² – x + 4

f(x + h) – f(x) = [3x² + 6xh + 3h2² – x – h + 4] – [3x² – x + 4]  
                       = 3x² – 3x² + 6xh + 3h² – x + x – h + 4 – 4  
                       = 6xh + 3h² – h  



Now I'm supposed to evaluate at h = 0, so:

6x + 3(0) – 1 = 6x – 1

Simplified form:  6x + 3h – 1

Value at h = 0:  6x – 1

That's pretty much all there is to "operations on functions", until you get to function composition. Don't let the notation for this topic worry you: it means exactly what it says: add, subtract, multiply, or divide; then simplify and evaluate as necessary. Don't overthink this. It really is this simple.

Oh, and that last example? They put that in there so you can "practice" stuff you'll be doing in calculus. You likely won't remember this by the time you get to calculus, but you'll follow a very similar process for finding something called "derivatives".

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