1. Find the inverse function for f(x)=sqr 2x-6
2. Let f(x)=(x-2)^3+8
a. Show that this function is one-to-one algebraically.
b. Find the inverse of f(x).
3. let f(x)=1/x and g(x)=x^2+5x
a. Find (f · g)(x).
b. Find the domain and range of (f · g)(x)
1. To find the inverse function for f(x)= sqrt(2x-6), we follow the steps below:
- Let y = sqrt(2x-6)
- Swap x and y: x = sqrt(2y-6)
- Solve for y: x^2 = 2y-6
- Add 6 to both sides: x^2 + 6 = 2y
- Divide by 2: (x^2 + 6)/2 = y
- Therefore, the inverse function is f^(-1)(x) = (x^2 + 6)/2
2.a. To show that the function f(x) = (x-2)^3 + 8 is one-to-one algebraically, we need to show that for any two different values of x, the values of f(x) are also different.
- Assume x1 and x2 are two different values of x.
- Then, f(x1) = (x1-2)^3 + 8 and f(x2) = (x2-2)^3 + 8.
- If f(x1) = f(x2), then (x1-2)^3 + 8 = (x2-2)^3 + 8.
- Simplifying the equation gives us (x1-2)^3 = (x2-2)^3.
- Taking the cube root of both sides gives us x1 - 2 = x2 - 2.
- Simplifying further gives us x1 = x2, which contradicts our assumption that x1 and x2 are different.
- Therefore, f(x) = (x-2)^3 + 8 is one-to-one.
b. To find the inverse of f(x) = (x-2)^3 + 8, we follow the steps below:
- Let y = (x-2)^3 + 8.
- Swap x and y: x = (y-2)^3 + 8.
- Solve for y: x-8 = (y-2)^3.
- Take the cube root of both sides: (x-8)^(1/3) = y-2.
- Add 2 to both sides: (x-8)^(1/3) + 2 = y.
- Therefore, the inverse function is f^(-1)(x) = (x-8)^(1/3) + 2.
3.a. To find (f · g)(x), we first need to plug g(x) into f(x):
- (f · g)(x) = f(g(x)) = f(x^2 + 5x).
- Plugging the function g(x) = x^2 + 5x into f(x), we get:
- f(g(x)) = f(x^2 + 5x) = 1/(x^2 + 5x).
b. To find the domain and range of (f · g)(x) = 1/(x^2 + 5x), we consider the restrictions:
- The denominator, x^2 + 5x, cannot be equal to 0 for the function to be defined.
- Solving x^2 + 5x = 0 gives us x(x+5) = 0, which gives us two possible values for x: x = 0 and x = -5.
- Therefore, the domain of (f · g)(x) is all real numbers except x = 0 and x = -5.
- The range of (f · g)(x) is all real numbers except y = 0, since the denominator cannot be 0.
1. To find the inverse function of f(x) = sqrt(2x - 6), we can follow these steps:
Step 1: Start with the given function:
f(x) = sqrt(2x - 6)
Step 2: Replace f(x) with y:
y = sqrt(2x - 6)
Step 3: Swap the variables x and y:
x = sqrt(2y - 6)
Step 4: Solve for y:
x^2 = 2y - 6
x^2 + 6 = 2y
(x^2 + 6)/2 = y
Step 5: Replace y with f^(-1)(x):
f^(-1)(x) = (x^2 + 6)/2
Therefore, the inverse function for f(x) = sqrt(2x - 6) is f^(-1)(x) = (x^2 + 6)/2.
2a. To show that the function f(x) = (x - 2)^3 + 8 is one-to-one algebraically, we need to prove that if f(a) = f(b), then a = b.
Let's assume f(a) = f(b):
(a - 2)^3 + 8 = (b - 2)^3 + 8
Expanding both sides:
(a^3 - 6a^2 + 12a - 8) + 8 = (b^3 - 6b^2 + 12b - 8) + 8
a^3 - 6a^2 + 12a = b^3 - 6b^2 + 12b
Now, let's subtract (b^3 - 6b^2 + 12b) from both sides:
a^3 - 6a^2 + 12a - (b^3 - 6b^2 + 12b) = 0
Using the difference of cubes formula:
(a - b)(a^2 + ab + b^2) - 6(a^2 - ab + b^2) + 12(a - b) = 0
Factoring out (a - b):
(a - b)(a^2 + ab + b^2 - 6a^2 + 6ab - 6b^2 + 12) = 0
Simplifying:
(a - b)(-5a^2 + 7ab - 5b^2 + 12) = 0
For this equation to be true, either (a - b) = 0 or -5a^2 + 7ab - 5b^2 + 12 = 0.
(a - b) = 0 implies that a = b, which proves one-to-one.
2b. To find the inverse of f(x) = (x - 2)^3 + 8, we can follow these steps:
Step 1: Start with the given function:
y = (x - 2)^3 + 8
Step 2: Swap the variables x and y:
x = (y - 2)^3 + 8
Step 3: Solve for y:
x - 8 = (y - 2)^3
(y - 2)^3 = x - 8
Step 4: Take the cube root of both sides:
y - 2 = (∛(x - 8))
Step 5: Add 2 to both sides:
y = (∛(x - 8)) + 2
Therefore, the inverse of f(x) = (x - 2)^3 + 8 is f^(-1)(x) = (∛(x - 8)) + 2.
3a. To find (f · g)(x), we need to find the composition of f(x) and g(x).
Let's first find f(g(x)):
f(g(x)) = f(x^2 + 5x)
Now, substitute f(x) = 1/x into f(g(x)):
f(g(x)) = 1/(x^2 + 5x)
Therefore, (f · g)(x) = 1/(x^2 + 5x).
3b. To find the domain and range of (f · g)(x) = 1/(x^2 + 5x), we need to consider the restrictions on the value of x.
For the domain, since the denominator cannot be equal to zero, we have to exclude values of x that would make the denominator zero. So, we solve the equation x^2 + 5x = 0:
x(x + 5) = 0
x = 0 or x = -5
Therefore, the domain of (f · g)(x) is all real numbers except x = 0 and x = -5.
For the range, since the function (f · g)(x) is a rational function, the range will be all real numbers except any values that would make the denominator equal to zero. Since x^2 + 5x is always greater than zero for all real values of x (except x = 0 and x = -5), the range of (f · g)(x) is all real numbers except zero.
Domain: All real numbers except x = 0 and x = -5.
Range: All real numbers except zero.