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Definitions: Sum, Difference, Product, and Quotient of Functions.
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Definitions: Sum, Difference, Product, and Quotient of Functions • Let f and g be two functions. The sum of f + g, the differencef – g, the productfg, and the quotientf /g are functions whose domains are the set of all real numbers common to the domains of f and 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) 0 /
Example: Combinations of Functions Let f(x) = x2 – 3 and g(x)= 4x + 5. Find • (f + g) (x) • (f + g)(3) • (f – g)(x) • (f * g)(x) • (f/g)(x) What is the domain of each combination?
Example: Combinations of Functions If f(x) = 2x – 1 and g(x) = x2 + x – 2, find: • (f-g)(x) • (fg)(x) • (f/g)(x) What is the domain of each combination?
Review For Test 1 • Equations of Lines • Slope of a line, parallel & perpendicular • Graph with intercepts, graph using slope-intercept method • Formulae for distance and midpoint • Graph circle with standard and general form (complete the square) • Difference Quotient • Piecewise-defined functions • Domain & range of a function • Intervals where the function increases, decreases, and/or is constant • Be able to determine when a relation is a function – ordered pairs or graph • Relative Max and Min • Average Rate of change • Even and odd functions, symmetry • Transformations • Algebra of functions, domain
The Composition of Functions • The composition of the function f with g is denoted by fog and is defined by the equation • (fog)(x) = f (g(x)). • The domain of the composition function fog is the set of all x such that • x is in the domain of g and • g(x) is in the domain of f.
f (x) = 3x – 4 This is the given equation for f. (fog)(x) = f (g(x)) = 3g(x) – 4 = 3(x2 + 6) – 4 = 3x2 + 18 – 4 = 3x2 + 14 Replace g(x) with x2 + 6. Use the distributive property. Simplify. Example: Forming Composite Functions Given f (x) = 3x – 4 and g(x) = x2 + 6, find: a. (fog)(x) b. (gof)(x) • Solution • We begin the composition of f with g. Since (fog)(x) = f (g(x)), replace each occurrence of x in the equation for f by g(x). Thus, (fog)(x) = 3x2 + 14.
g(x) = x2 + 6 This is the given equation for g. (gof )(x) = g(f (x)) = (f (x))2 + 6 = (3x – 4)2 + 6 = 9x2 – 24x + 16 + 6 = 9x2 – 24x + 22 Replace f (x) with 3x – 4. Square the binomial, 3x – 4. Simplify. Example: Forming Composite Functions Given f (x) = 3x – 4 and g(x) = x2 + 6, find: a. (fog)(x) b. (gof)(x) Solution b. Next, (gof )(x), the composition of g with f. Since (gof )(x) = g(f (x)), replace each occurrence of x in the equation for g by f (x). Thus, (gof )(x) = 9x2 – 24x + 22. Notice that (fog)(x) is not the same as (gof )(x).