Let f and g be the functions defined by
f : R^2 −→ R^2
(x, y) −→ (−y, −x)
and
g : R^2 −→ R^2
(x, y) −→ (x + 1, y − 1).
(a) Describe the geometric effect of each of f and g
(b) Determine the composite function g◦f.
(c) Show that g is one-to-one and onto, and determine its inverse
function g−1.
To describe the geometric effect of each function, we need to understand what each function does to a point in the Cartesian plane.
(a) Geometric effect of f:
The function f takes a point (x, y) in the Cartesian plane and transforms it to (-y, -x). This means that f reflects the point across the x-axis and then the y-axis. In other words, it swaps the x and y coordinates and negates them. So, if you imagine a point (x, y) in the plane, f will map it to the point (-y, -x).
Geometric effect of g:
The function g takes a point (x, y) in the Cartesian plane and transforms it to (x + 1, y - 1). This means that g shifts the point one unit to the right along the x-axis and one unit downwards along the y-axis. So, if you imagine a point (x, y) in the plane, g will map it to the point (x + 1, y - 1).
(b) To determine the composite function g◦f, we need to apply g to the result of applying f. Let's start with an arbitrary point (x, y).
First, apply f to (x, y):
f(x, y) = (-y, -x)
Next, apply g to (-y, -x):
g(-y, -x) = ((-y) + 1, (-x) - 1) = (-y + 1, -x - 1)
So, the composite function g◦f maps the point (x, y) to (-y + 1, -x - 1).
(c) To show that g is one-to-one (injective) and onto (surjective), we need to prove both conditions.
Injective proof:
For g to be injective, we need to show that different inputs produce different outputs. Let's assume g(x1, y1) = g(x2, y2) and prove that it implies (x1, y1) = (x2, y2).
Using the definition of g, we have:
(x1 + 1, y1 - 1) = (x2 + 1, y2 - 1)
From this equation, we can see that x1 = x2 and y1 = y2. Therefore, we have proven that g is one-to-one (injective).
Surjective proof:
For g to be surjective, every point in the target space (R^2) needs to have a pre-image in the domain (R^2). Let's take an arbitrary point (a, b) in R^2 and show that there exists (x, y) in R^2 such that g(x, y) = (a, b).
Using the definition of g, we can set up the following equations:
a = x + 1
b = y - 1
Solving these equations, we find x = a - 1 and y = b + 1. Therefore, for any (a, b) in R^2, we can find a pre-image (x, y) in R^2 such that g(x, y) = (a, b). Thus, g is onto (surjective).
Inverse function g⁻¹:
To find the inverse function of g, we need to interchange the roles of the input and output variables. Let's solve the equations in terms of x and y:
a = x + 1
b = y - 1
Rearranging, we get:
x = a - 1
y = b + 1
So, the inverse function g⁻¹ takes a point (a, b) in the Cartesian plane and maps it to (a - 1, b + 1). Therefore, the inverse function of g is g⁻¹(x, y) = (x - 1, y + 1).
(a) Geometric effect of f: The function f reflects a point across the line y = x. In other words, it swaps the x-coordinate with the negative y-coordinate and swaps the y-coordinate with the negative x-coordinate.
Geometric effect of g: The function g translates every point one unit to the right in the x-direction and one unit down in the y-direction. So, it shifts the points to the right and downwards.
(b) To find the composite function g◦f, we apply the function f first and then apply the function g to the result.
(g◦f)(x, y) = g(f(x, y))
= g(-y, -x) (using the definition of f)
Now, applying the function g to (-y, -x), we get:
(g◦f)(x, y) = (-y + 1, -x - 1)
So, the composite function g◦f is given by (-y + 1, -x - 1).
(c) To show that g is one-to-one, we need to prove that for any two distinct points (x1, y1) and (x2, y2) in the domain of g, the images under g are also distinct.
Let (x1, y1) and (x2, y2) be two distinct points in the domain of g. Then, applying g to these points:
g(x1, y1) = (x1 + 1, y1 - 1)
g(x2, y2) = (x2 + 1, y2 - 1)
If g(x1, y1) = g(x2, y2), then we must have:
x1 + 1 = x2 + 1
y1 - 1 = y2 - 1
From the first equation, we can conclude that x1 = x2. And from the second equation, we can conclude that y1 = y2. Therefore, (x1, y1) = (x2, y2), which implies that g is one-to-one.
To show that g is onto, we need to prove that for any point (x, y) in the codomain of g, there exists a point (a, b) in the domain of g such that g(a, b) = (x, y).
Let (x, y) be any point in the codomain of g. To find the pre-image (a, b), we can work backwards from the definition of g:
x + 1 = a
y - 1 = b
Therefore, g^{-1}(x, y) = (x + 1, y - 1).
Thus, the inverse function of g is g^{-1}(x, y) = (x + 1, y - 1).