Math 222
Assignment 4
April, 7 2013
Mat222
In this week of class we have been taught how to evaluate, combine and find inverses of working relations or functions. We will be computing, composing, transforming and finding the inverse of some functions. We are working with the following functions:
f(x) = 2x+5 g(x) = x2 -3 h(x) = (7-3x)/3 We have been asked to compute (f – h)(4). (f – h)(4) = f(4) – h(4) So we can evaluate each separately and then subtract.
f(4) = 2*4 +5=13 h(4) = (7-4)/3=1
(f – h)(4) = 13 – 1 = 12
(f-h)(4)=12 This is our answer.
Next we are to compose two pairs of the functions into each other. First we will work out: (f °g)(x)= f ( g ( x )) This means the rule of f will work on g.
= f(x2 -3) In this step f in now going to work on the rule of g.
= 2(x2 -3)+5 The rule of f is applied to g.
= 2x2 -6+5 In this step we simplify.
(f °g)(x)= 2x2 -1 Final results.
Now we will compose the second of the two functions.
(h° g)(x) = h(g(x)) The rule of h will work on g.
= h(x2 – 3) = ((7-(x^2-3)))/3 The rule of h is applied to g. (h ° g)(x) = (10-x^2)/3 The final results.
Next we are asked to transform the g(x) function so that the graph is moved 6 units to the right and 7 units downward from where it would be right now.
Six units to the right means to put -6 in with x to be squared.
Seven units downward means to put -7 outside of the squaring.
The new function will look like this:
G(x) = (x - 6)2 -3 - 7
G(x) = (x - 6)2 -10
Our last job is to find the inverse of two of our functions, f-1 (x) and h-1(x) . To find the inverse we will replace x by y and f(x) with x in f(x) = 2x+5 and for finding h-1(x) we will replace x with y and h(x) with x in h(x) = (7-3x)/3 and solve for y. Here are the functions:
x = 2y + 5 Here we replace f(x) with y:
2y = x - 5 Here we switch the y and the x:
y = ((x-5))/2 Now we solve for y:
April, 7 2013
Mat222
In this week of class we have been taught how to evaluate, combine and find inverses of working relations or functions. We will be computing, composing, transforming and finding the inverse of some functions. We are working with the following functions:
f(x) = 2x+5 g(x) = x2 -3 h(x) = (7-3x)/3 We have been asked to compute (f – h)(4). (f – h)(4) = f(4) – h(4) So we can evaluate each separately and then subtract.
f(4) = 2*4 +5=13 h(4) = (7-4)/3=1
(f – h)(4) = 13 – 1 = 12
(f-h)(4)=12 This is our answer.
Next we are to compose two pairs of the functions into each other. First we will work out: (f °g)(x)= f ( g ( x )) This means the rule of f will work on g.
= f(x2 -3) In this step f in now going to work on the rule of g.
= 2(x2 -3)+5 The rule of f is applied to g.
= 2x2 -6+5 In this step we simplify.
(f °g)(x)= 2x2 -1 Final results.
Now we will compose the second of the two functions.
(h° g)(x) = h(g(x)) The rule of h will work on g.
= h(x2 – 3) = ((7-(x^2-3)))/3 The rule of h is applied to g. (h ° g)(x) = (10-x^2)/3 The final results.
Next we are asked to transform the g(x) function so that the graph is moved 6 units to the right and 7 units downward from where it would be right now.
Six units to the right means to put -6 in with x to be squared.
Seven units downward means to put -7 outside of the squaring.
The new function will look like this:
G(x) = (x - 6)2 -3 - 7
G(x) = (x - 6)2 -10
Our last job is to find the inverse of two of our functions, f-1 (x) and h-1(x) . To find the inverse we will replace x by y and f(x) with x in f(x) = 2x+5 and for finding h-1(x) we will replace x with y and h(x) with x in h(x) = (7-3x)/3 and solve for y. Here are the functions:
x = 2y + 5 Here we replace f(x) with y:
2y = x - 5 Here we switch the y and the x:
y = ((x-5))/2 Now we solve for y: