Chapter 7 Section 3 Power Functions and Function

Chapter 7 Section 3 Power Functions and Function Operations

Operations on Functions: for any two functions f(x) & g(x) 1. 2. 3. Addition h(x) = f(x) + g(x) Subtraction h(x) = f(x) – g(x) Multiplication h(x) = f(x)*g(x) OR f(x)g(x)

Operations on Functions: for any two functions f(x) & g(x) 1. Division h(x) = f(x)/g(x) OR f(x) ÷ g(x) 2. Composition h(x) = f(g(x)) OR g(f(x)) ** Domain – all real x-values that “make sense” (i. e. can’t have a zero in the denominator, can’t take the even nth root of a negative number, etc. )

Power Functions •

Ex: Let f(x)=3 x 1/3 & g(x)=2 x 1/3. Find (a) the sum, (b) the difference, and (c) the domain for each. (a) 3 x 1/3 + 2 x 1/3 = 5 x 1/3 (b) 3 x 1/3 – 2 x 1/3 = x 1/3 (c) Domain of (a) all real numbers Domain of (b) all real numbers

Let f (x) = 4 x 1/2 and g(x) = – 9 x 1/2. Find the following. a. f(x) + g(x) SOLUTION f (x) + g(x) = [4 + (– 9)]x 1/2 = – 5 x 1/2 b. f(x) – g(x) SOLUTION = [4 – (– 9)]x 1/2 f (x) – g(x) = 4 x 1/2 – (– 9 x 1/2) c. the domains of f + g and f – g = 13 x 1/2 The functions f and g each have the same domain: all nonnegative real numbers. So, the domains of f + g and f – g also consist of all nonnegative real numbers.

Types Domains • All real numbers – if you can use positive numbers, negative numbers, or zero in the beginning functions and the result of combining functions. • All non-negative numbers -- if you can use positive numbers and zero in the beginning functions and the result of combining functions. • All positive numbers -- if you can use only positive numbers in the beginning function and the result of combining functions

Ex: Let f(x)=4 x 1/3 & g(x)=x 1/2. Find (a) the product, (b) the quotient, and (c) the domain for each. (a) 4 x 1/3 * x 1/2 = 4 x 1/3+1/2 = 4 x 5/6 (b) = 4 x 1/3 -1/2 = 4 x-1/6 = (c) Domain of (a) all reals ≥ 0, because you can’t take the 6 th root of a negative number (Non-neg #’s). Domain of (b) all reals > 0, because you can’t take the 6 th root of a negative number and you can’t have a denominator of zero (Positive #’s).

Let f (x) = 6 x and g(x) = x 3/4. Find the following. f (x) g(x) SOLUTION 6 x = x 3/4 = 6 x(1 – 3/4) = 6 x 1/4 the domain of The domain of f consists of all real numbers, and the domain of g consists of all nonnegative real numbers. Because g(0) = 0, the domain of is restricted to all positive real numbers.

Try it… Let f (x) = – 2 x 2/3 and g(x) = 7 x 2/3. Find the following. 1. f (x) + g(x) SOLUTION f (x) + g(x) = – 2 x 2/3 + 7 x 2/3 = (– 2 + 7)x 2/3 = 5 x 2/3 2. f (x) – g(x) SOLUTION f (x) – g(x) = – 2 x 2/3 – 7 x 2/3 = [– 2 + ( – 7)]x 2/3 = – 9 x 2/3 The domains of f and g have the same domain: all non -negative real numbers. So , the domain of f + g and f – g also consist of all non-negative real numbers.

Composition of Functions •

Ex: Let f(x)=2 x-1 & g(x)=x 2 -1. Find (a) f(g(x)), (b) g(f(x)), (c) f(f(x)), and (d) the domain of each. (a) 2(x 2 -1)-1 = (c) 2(2 x-1)-1 = 2(2 -1 x) (b) (2 x-1)2 -1 = 22 x-2 -1 = = (d) Domain of (a) all reals except x=± 1. Domain of (b) all reals except x=0. Domain of (c) all reals except x=0, because 2 x-1 can’t have x=0.

Let f(x) = 3 x – 8 and g(x) = 2 x 2. Find the following. 8. g(f(5)) SOLUTION To evaluate g(f(5)), you first must find f(5) = 3(5) – 8 = 7 Then g( f(3)) = g(7) = 2(7)2 = 2(49) = 98. ANSWER So, the value of g(f(5)) is 98.

Let f(x) = 3 x – 8 and g(x) = 2 x 2. Find the following. 9. f(g(5)) SOLUTION To evaluate f(g(5)), you first must find g(5). g (5) = 2(5)2 = 2(25) = 50 Then f( g(5)) = f(50) = 3(50) – 8 = 150 – 8 = 142. ANSWER So, the value of f( g(5)) is 142.

Let f(x) = 2 x– 1 and g(x) = 2 x + 7. Find f(g(x)), g(f(x)), and f(f(x)). Then state the domain of each composition. SOLUTION f(g(x)) =f(2 x + 7) = 2(2 x + 7)– 1 = 2 2 x + 7 g(f(x)) =f(2 x– 1) = 2(2 x– 1) + 7 = 4 x– 1 + 7 = 4 + 7 x f(f(x)) =f(2 x– 1) = 2(2 x– 1)– 1 = x The domain of f(g(x )) consists of all real numbers except x = – 3. 5. The domain of g(f(x)) consists of all real numbers except x = 0.