<p> MA 11B Tech Math I Credit Arcadia Valley Career Technology Center Last Update: March 2008 Topic: Relations and Functions Focus: Performing Operations with Functions. </p><p>Show-Me Standards: MA3, G1-8,MA4 MO Grade Level Expectations: D1C10, NCTM Standards: 14C, 14D D3A9 </p><p>OBJECTIVE: The students will be able to find the sum, difference, product, and quotient of two functions and find the composition of two functions in order to create new functions.</p><p>Operations with Functions:</p><p>Let’s say f(x) and g(x) are any two functions. You can create a new function by adding, subtracting, multiplying, or dividing the functions according to the following rules:</p><p>(f + g)(x) = f(x) + g(x) (f – g)(x) = f(x) – g(x)</p><p>(f ● g)(x) = f(x) ● g(x) (f/g)(x) = f(x) g(x) </p><p>Examples:</p><p>EX1. Given f(x) = x2 + 3x – 1 and g(x) = 8x + 2 , find each of the following.</p><p> a. (f + g)(x) </p><p>(f + g)(x) = f(x) + g(x) = (x2 + 3x – 1) + (8x + 2) = x2 + 11x + 1</p><p> b. (f – g)(x)</p><p>(f – g)(x) = f(x) – g(x) = (x2 + 3x – 1) – (8x + 2) = x2 – 5x – 3 </p><p>EX2. Given h(x) = 3x and r(x) = 7x2 – 5 , find each of the following.</p><p> a. (h ● r)(x) </p><p>(h ● r)(x) = h(x) ● r(x) = (3x)(7x2- 5) = 21x3 – 15x b. (r ● h)(x)</p><p>(r ● h)(x) = r(x) ● h(x) = (7x2 – 5)(3x) = 21x3 – 15x</p><p> notice, that (h ● r)(x) and (r ● h)(x) give you the same function. </p><p>EX.3 Given t(x) = 26x – 10 and m(x) = 2x , find each of the following.</p><p> a. (t/m)(x)</p><p>(t/m)(x) = t(x) = 26x – 10 = 13 – 5/x m(x) 2x</p><p> b. (m/t)(x)</p><p>(m/t)(x) = m(x) = 2x____ = x___ t(x) 26x – 10 13x – 5 </p><p>Composition of Functions:</p><p>Functions can also be combined to create new functions in a way called “composition of functions”. In this way, one function is performed and its outcome is then performed by the second function. It can denoted by “ f ᵒ g ” and is read “ f of g ”. It can be described in equation form as: [f ᵒ g](x) = f[g(x)]</p><p>When you work with composite functions, the performance of the operations in one of the functions on its domain numbers leads you to a set of range numbers that are then used as the domain numbers for the second function. In the above case, function g’s range numbers are used as the domain numbers for f . Examples:</p><p>EX1. If f(x) = {(-2,3),(1,0),(6,5)} and g(x) = {(2,6),(-1,1),(5,-2)} , find f ᵒ g .</p><p> f[g(2)] = f(6) = 5 because, in the g function, 2 gets paired with 6 and in the f function, 6 gets paired with 5. So, [f ᵒ g](2) = 5 .</p><p> f[g(-1)] = f(1) = 0 so, [f ᵒ g](-1) = 0 .</p><p> f[g(5)] = f(-2) = 3 so, [f ᵒ g](5) = 3 .</p><p>Therefore, [f ᵒ g](x) = {(2,5),(-1,0),(5,3)} and we have created a new function.</p><p>EX2. If f(x) = {(2,6),(9,4),(6,5),(7,-7)} and g(x) = {(6,-3),(4,0),(5,14),(8,-1)}, find g ᵒ f .</p><p> g[f(2)] = g(6) = -3</p><p> g[f(9)] = g(4) = 0</p><p> g[f(6)] = g(5) = 14</p><p> g[f(7)] = g(-7) = is undefined, there is no range number that is paired with -7 in the g function.</p><p>Therefore, [g ᵒ f](x) = {(2,-3),(9,0),(6,14)}</p><p>Simplifying Compositions in Equation Form:</p><p>Suppose f(x) = x + 3 and g(x) = x2 + x – 1 . We can simplify the composition f ᵒ g into a single equation by substituting/replacing the whole g equation for the domain of f .</p><p>[f ᵒ g](x) = f[g(x)] = f(x2 + x – 1) = (x2 + x – 1) + 3 = x2 + x + 2 </p><p> the x in the f(x) was replaced by the whole g function. That meant that wherever we saw an x in the f function, we replaced it with x2 + x – 1 . Examples:</p><p>EX1. Suppose f(x)= 3x – 1 and g(x)= 7x + 5. Find f ᵒ g and g ᵒ f .</p><p>[f ᵒ g](x) = f[g(x)] = 3(7x + 5) – 1 = 21x + 15 – 1 = 21x + 14</p><p>[g ᵒ f](x) = g[f(x)] = 7(3x – 1) + 5 = 21x – 7 + 5 = 21x – 2 </p><p>Now go back and find [f ᵒ g](2) and [g ᵒ f](-9) .</p><p>[f ᵒ g](2) = 21(2) + 14 = 42 + 14 = 56 so, [f ᵒ g](2) = 56</p><p>[g ᵒ f](-9) = 21(-9) – 2 = -189 – 2 = -191 so, [g ᵒ f](-9) = -191</p><p>EX2. If h(x) = x2 – 2 and m(x) = 6x , find h ᵒ m and m ᵒ h . Afterwards, evaluate [h ᵒ m](-4) and [m ᵒ h](0) .</p><p>[h ᵒ m](x) = (6x)2 – 2 = 36x2 – 2 </p><p>[m ᵒ h](x) = 6(x2 – 2) = 6x2 – 12 </p><p>[h ᵒ m](-4) = 36(-4)2 – 2 = 36 ● 16 – 2 = 574</p><p>[m ᵒ h](0) = 6(0)2 – 12 = 6 ● 0 – 12 = -12 PROBLEMS:</p><p>Given f(x) = 2x – 9 , g(x) = 4x + 9 , h(x) = 2x2 , k(x) = x2 + 6x + 9 , p(x) = 5 – 20x , and t(x) = 10x . Find the following:</p><p>1. (f + t)(x)</p><p>2. (g – f)(x)</p><p>3. (h ● g((x)</p><p>4. (t ● k)(x)</p><p>5. (p/t)(x)</p><p>6. (t/h)(x) 7. (h + k)(x)</p><p>8. (k – f)(x)</p><p>9. [f ᵒ g](x)</p><p>10. [p ᵒ k](x)</p><p>11. [g ᵒ h](x)</p><p>12. [k ᵒ t](x) </p><p>Use the new functions you found in problems #1-12 to evaluate the following:</p><p>13. (h + k)(4) 14. (g – f)(-2)</p><p>15. (t ● k)(-1)</p><p>16. (t/h)(15)</p><p>17. [f ᵒ g](-3)</p><p>18. [k ᵒ t](1)</p><p>19. [g ᵒ h](-4)</p><p>20. If f(x) = {(-1,6),(0,-7),(-2½,3)} and g(x) = {(2,0),(4,-2½),(-10,-1)} , find f ᵒ g .</p>