#Algebra 2 Help Please?!
If f = (1, 2), (2, 3), (3, 4), (4, 5),
g = (1, -2), (3, -3), (5, -5), and
h = (1, 0), (2, 1), (3, 2),
find the following and state the domain:
2a.f+g
2b.f-g
2c.f(times)g
2d.f/h
2e.g o f o h
Let f(x)=|x-5| and g(x)=1/x .
Write the given function as a composition function
using f and g.
3a. H(x)=|1/x-5|
Well, aren't you in for a treat with Algebra 2! Let's dive right in and find the answers you're looking for.
2a. To find 2a.f+g, we need to "add" the functions f and g using scalar multiplication. So, let's see:
f = (1, 2), (2, 3), (3, 4), (4, 5)
g = (1, -2), (3, -3), (5, -5)
Using scalar multiplication, we multiply each point of f by 2, and then add it to the corresponding point of g:
2a.f+g = (2(1), 2(2)+(-2)), (2(2), 2(3)+(-3)), (2(3), 2(4)+(-5)), (2(4), 2(5))
Simplifying, we get:
2a.f+g = (2, 2), (4, 3), (6, 3), (8, 5)
The domain of this function would be the x-values of f and g, which is {1, 2, 3, 4, 5}.
2b. For 2b.f-g, we follow a similar process. We multiply each point of f by 2, and then subtract g from it:
2b.f-g = (2(1), 2(2)-(-2)), (2(2), 2(3)-(-3)), (2(3), 2(4)-(-5)), (2(4), 2(5))
Simplifying, we get:
2b.f-g = (2, 6), (4, 9), (6, 13), (8, 15)
The domain of this function would also be {1, 2, 3, 4, 5}.
2c. To find 2c.f times g, we multiply each point of f by 2 and then multiply it by g:
2c.f x g = (2(1), 2(2) * (1/1)), (2(2), 2(3) * (1/3)), (2(3), 2(4) * (1/5)), (2(4), 2(5) * (1/1))
Simplifying, we get:
2c.f x g = (2, 4), (4, 2), (6, 1.6), (8, 8)
The domain of this function would be the x-values of f and g, which is {1, 2, 3, 4, 5}.
2d. For 2d.f/h, we divide each point of f by the corresponding point of h:
2d.f/h = (1/1, 2/0), (2/2, 3/1), (3/3, 4/2), (4/1, 5/0)
Uh-oh! We have an issue here because division by zero is undefined. So the domain of this function would be the x-values of f, which is {1, 2, 3, 4}.
2e. Now, onto g o f o h! This means we need to find g(f(h(x))). So, let's break it down step by step:
f(x) = |x-5| -> Subtract 5 from x, then take the absolute value
g(x) = 1/x -> Take the reciprocal of x
h(x) = |1/x-5| -> Substitute h(x) = f(g(x)), which means we need to find f(g(x-5)) since g(x) = 1/x
Substituting g(x) into h(x), we get:
h(x) = |1/(x-5)-5| -> Simplifying, we get:
h(x) = |1/(x-5)-25/(x-5)| -> Further simplifying, we get:
h(x) = |((-25) + 1)/(x-5)| -> Combining terms, we have:
h(x) = |-24/(x-5)|
So, the given function as a composition function using f and g is H(x) = |-24/(x-5)|.
Phew, that was quite a ride! I hope my clownish explanations made algebra a bit more enjoyable for you. Let me know if you have any more questions or if there's anything else I can help you with!