Function and Inverse Cindy Dean
Composition and Inverse
Cindy Dean
Intermediate Algebra: MAT 222
Instructor:
February 14, 2014
Composition and Inverse
In this week’s assignment, I will be solving functions with different values and variables. Many companies and businesses, use these methods to either make progress or to change something that will benefit their success. The first function is:
(f – h)(4) f(4) – h(4) I multiplied 4 with each variable. f(4) = 2(4) + 5 The x is replaced with 4. f(4) = 13 I used the order of operation to evaluate this function. h(4) = (7 – 3)/3 I will repeat the steps that I used in the previous problem for the variable, h. h(4) = 3/3 I used the order of operation to simplify. h(4) = 1 This is the solution for h.
(f – h)(4) = 13 + 1 I combined the values of f and h.
(f – h)(4) = 14 This is my solution. The next problem, I will be finding the solution for g(x). I will replace the x in the f function with the g function. My function will look like this:
(f° g)(x) = f(g(x))
(f° g)(x) = f (x^2 – 3) My g function is replacing the x.
(f° g)(x) = 2(x^2 – 3) + 5
(f° g)(x) = 2x^2 – 6 + 5 I used the order of operation to simplify my answer.
(fg)(x) = 2x^2 – 1 This is my solution.
This problem, I will be composing for (hg)(x).
(h° g)(x) = h(g(x)) I will repeat the steps that I have been using.
(h° g)(x) = h (x^2 + 5)
(h g)(x) = 5 (x^2 – 3)/-1
(h g)(x) = 2 + x^2/-1 This is my final answer. The next step is to graph the g(x) function and transform it to graph 6 units to the right and 7 units down. g(x) = x^2 – 3 This is the original function. g(x) = (x – 6)^2 – 3 – 7 I plugged in the -6 and -7. g(x) = (x-6)^2 – 10 This is my solution after using the order of operation and simplifying. My last problem is to find the inverse of functions of f and h. The function is going to be y instead of its own name. The x and y will
Cindy Dean
Intermediate Algebra: MAT 222
Instructor:
February 14, 2014
Composition and Inverse
In this week’s assignment, I will be solving functions with different values and variables. Many companies and businesses, use these methods to either make progress or to change something that will benefit their success. The first function is:
(f – h)(4) f(4) – h(4) I multiplied 4 with each variable. f(4) = 2(4) + 5 The x is replaced with 4. f(4) = 13 I used the order of operation to evaluate this function. h(4) = (7 – 3)/3 I will repeat the steps that I used in the previous problem for the variable, h. h(4) = 3/3 I used the order of operation to simplify. h(4) = 1 This is the solution for h.
(f – h)(4) = 13 + 1 I combined the values of f and h.
(f – h)(4) = 14 This is my solution. The next problem, I will be finding the solution for g(x). I will replace the x in the f function with the g function. My function will look like this:
(f° g)(x) = f(g(x))
(f° g)(x) = f (x^2 – 3) My g function is replacing the x.
(f° g)(x) = 2(x^2 – 3) + 5
(f° g)(x) = 2x^2 – 6 + 5 I used the order of operation to simplify my answer.
(fg)(x) = 2x^2 – 1 This is my solution.
This problem, I will be composing for (hg)(x).
(h° g)(x) = h(g(x)) I will repeat the steps that I have been using.
(h° g)(x) = h (x^2 + 5)
(h g)(x) = 5 (x^2 – 3)/-1
(h g)(x) = 2 + x^2/-1 This is my final answer. The next step is to graph the g(x) function and transform it to graph 6 units to the right and 7 units down. g(x) = x^2 – 3 This is the original function. g(x) = (x – 6)^2 – 3 – 7 I plugged in the -6 and -7. g(x) = (x-6)^2 – 10 This is my solution after using the order of operation and simplifying. My last problem is to find the inverse of functions of f and h. The function is going to be y instead of its own name. The x and y will