1.Let functions f and g be defined by f(x)=2x+1, and
g(x)=x^2-2, respectively. Find
a)(gof)(a+2)
b)(fog)(a+2)
2.Let A={x:x≠2) and define f: A→R by f(x)=4x/(2x-1) . Is f is one- to- one?
Find the range of f . Then find f^(-1) and hence determine the domain and range of f^(-1)
(g◦f) = g(f) = f^2-2 = (2x+1)^2-2
(f◦g) = f(g) = 2g+1 = 2(x^2-2)+1
x=4f/(2f-1)
2xf-x = 4f
f(2x-4) = x
f = x/(2x-4)
So, the range of f is all reals except x=2. The domain is all reals except 2 and 1/2.
The domain of f^-1 is the range of f. The range is all reals except x=1/2.
1. To find (gof)(a+2) and (fog)(a+2), we need to substitute (a+2) into the respective functions and perform the composition of functions.
a) (gof)(a+2):
First, we need to find the value of f(a+2) by substituting (a+2) into f(x):
f(a+2) = 2(a+2) + 1 = 2a + 4 + 1 = 2a + 5
Then, we substitute this result into g(x):
(gof)(a+2) = g(f(a+2)) = g(2a + 5)
To find the value of g(x) for any given x, we substitute x into g(x):
g(x) = x^2 - 2
Therefore, substituting (2a + 5) into g(x), we get:
(gof)(a+2) = (2a + 5)^2 - 2
b) (fog)(a+2):
First, we need to find the value of g(a+2) by substituting (a+2) into g(x):
g(a+2) = (a+2)^2 - 2 = a^2 + 4a + 4 - 2 = a^2 + 4a + 2
Then, we substitute this result into f(x):
(fog)(a+2) = f(g(a+2)) = f(a^2 + 4a + 2)
To find the value of f(x) for any given x, we substitute x into f(x):
f(x) = 4x / (2x - 1)
Therefore, substituting (a^2 + 4a + 2) into f(x), we get:
(fog)(a+2) = 4(a^2 + 4a + 2) / (2(a^2 + 4a + 2) - 1)
2. To determine if f(x) is one-to-one, we need to check if it satisfies the horizontal line test. If every horizontal line intersects the graph of f(x) at most once, then f(x) is one-to-one.
Finding the range of f(x):
To find the range of f(x), we need to determine all possible output values of f(x) for every input value in its domain.
Given f(x) = 4x / (2x - 1), we can observe that the denominator (2x - 1) cannot be equal to 0, as it would result in division by zero. Therefore, we need to find the values of x that make (2x - 1) not equal to 0:
2x - 1 ≠ 0
2x ≠ 1
x ≠ 1/2
Thus, the domain of f(x) is all real numbers except x = 1/2.
To find the range, we analyze the behavior of f(x) as x approaches infinity (positive and negative):
As x approaches infinity, 4x / (2x - 1) approaches 2, and as x approaches negative infinity, it also approaches 2.
Hence, the range of f(x) is all real numbers except 2.
Finding f^(-1)(x):
To find the inverse function f^(-1)(x), we interchange the roles of x and f(x) in the equation f(x) = 4x / (2x - 1).
Let y = f(x):
y = 4x / (2x - 1)
Swap x and y:
x = 4y / (2y - 1)
Solve for y:
2xy - x = 4y
2xy - 4y = x
y(2x - 4) = x
y = x / (2x - 4)
Therefore, f^(-1)(x) = x / (2x - 4).
Determining the domain and range of f^(-1)(x):
The domain of f^(-1)(x) is the range of f(x), which we previously determined to be all real numbers except 2.
The range of f^(-1)(x) is the domain of f(x), which is all real numbers except x = 1/2.