1. C OMBINATIONS OF F UNCTIONS ; C OMPOSITE F UNCTIONS

2. D OMAIN OF A F UNCTION Exclude any number that causes division by zero Domain: (-∞, 3) (3, ∞) Exclude any number that results in the square root of a negative number Domain: [-4, ∞) Do checkpoint #1 page 211

3. D EFINITIONS : S UM, D IFFERENCE, P RODUCT, AND Q UOTIENT OF F UNCTIONS Let f and g be two functions. The sum of f + g, the difference f – g, the product fg, and the quotient f / g are functions whose domains are the set of all real numbers common to the domains of f and g D f ∩ D g, defined as follows: Sum: ( f + g )( x ) = f ( x )+g( x ) Difference:( f – g )( x ) = f ( x ) – g( x ) Product:( f g )( x ) = f ( x ) g( x ) Quotient:( f / g )( x ) = f ( x )/g( x ), provided g( x ) does not equal 0

4. E XAMPLE Let f ( x ) = 2 x + 1 and g ( x ) = x 2 – 2. Find f + g, f - g, fg, and f/g. Solution: f + g = 2 x + 1 + x 2 – 2 = x 2 + 2 x – 1 f – g = (2 x + 1) - ( x 2 – 2)= - x 2 + 2 x + 3 fg = (2 x + 1)( x 2 – 2) = 2 x 3 + x 2 – 4 x - 2 f/g = (2 x + 1)/( x 2 – 2) Do checkpoint 2, page 214

5. E XAMPLE 3 Let f(x) = and g(x) =. a)Find f + g b)The domain of f + g Solution: a)f + g = Domain of f: [-3, ∞) Domain of g: [2, ∞) b) Domain of f + g is [2, ∞) Do checkpoint #3 page 215

6. T HE C OMPOSITION OF F UNCTIONS The composition of the function f with g is denoted by f o g and is defined by the equation ( f o g )( x ) = f ( g ( x )). The domain of the composite function f o g is the set of all x such that x is in the domain of g and g ( x ) is in the domain of f. (see visual on page 216)

7. T EXT E XAMPLE Given f (x) = 3x – 4 and g(x) = x 2 + 6, find: a. (f o g)(x) b. (g o f)(x) Solution a. We begin with (f o g)(x), the composition of f with g. Because (f o g)(x) means f (g(x)), we must replace each occurrence of x in the equation for f by g(x). f (x) = 3x – 4 This is the given equation for f. (f o g)(x) = f (g(x)) = 3g(x) – 4 = 3(x 2 + 6) – 4 = 3x 2 + 18 – 4 = 3x 2 + 14 Replace g(x) with x 2 + 6. Use the distributive property. Simplify. Thus, (f o g)(x) = 3x 2 + 14.

8. Solution b. Next, we find (g o f )(x), the composition of g with f. Because (g o f )(x) means g(f (x)), we must replace each occurrence of x in the equation for g by f (x). g(x) = x 2 + 6 This is the given equation for g. (g o f )(x) = g(f (x)) = (f (x)) 2 + 6 = (3x – 4) 2 + 6 = 9x 2 – 24x + 16 + 6 = 9x 2 – 24x + 22 Replace f (x) with 3x – 4. Square the binomial, 3x – 4. Simplify. Thus, (g o f )(x) = 9x 2 – 24x + 22. Notice that (f o g)(x) is not the same as (g o f )(x). E XAMPLE Given f (x) = 3x – 4 and g(x) = x 2 + 6, find: a. (f o g)(x) b. (g o f)(x)

9. H OMEWORK Page 219 #13 – 71 every other odd # 81 – 92 Bring your textbooks tomorrow