Let f(x)=(x-2)^3+8
a.Show that this function is one-to-one algebraically.
b. Find the inverse of f(x)
To show that a function is one-to-one algebraically, we need to prove that it is both injective (one-to-one) and surjective (onto).
a. To prove injectivity, we need to show that if f(x₁) = f(x₂), then x₁ = x₂.
Let's assume f(x₁) = f(x₂), which means (x₁-2)³ + 8 = (x₂-2)³ + 8.
Expanding both sides, we have x₁³ - 6x₁² + 12x₁ - 8 = x₂³ - 6x₂² + 12x₂ - 8.
Rearranging the terms, we get x₁³ - x₂³ - 6x₁² + 6x₂² + 12x₁ - 12x₂ = 0.
Factoring out (x₁ - x₂), we have (x₁ - x₂)(x₁² + x₁x₂ + x₂² - 6x₁ + 6x₂ + 12) = 0.
For (x₁ - x₂)(x₁² + x₁x₂ + x₂² - 6x₁ + 6x₂ + 12) = 0 to be true, either (x₁ - x₂) = 0 or (x₁² + x₁x₂ + x₂² - 6x₁ + 6x₂ + 12) = 0.
The first case gives us x₁ = x₂, proving injectivity.
b. To find the inverse of f(x), we'll start by switching the x and y variables and solving for y.
Let x = (y - 2)³ + 8.
Subtracting 8 from both sides, we have x - 8 = (y - 2)³.
Taking the cube root of both sides, we get ∛(x - 8) = y - 2.
Adding 2 to both sides, we have y = ∛(x - 8) + 2.
Therefore, the inverse of f(x) is f⁻¹(x) = ∛(x - 8) + 2.
a. To show that a function is one-to-one algebraically, we need to prove that if two different inputs produce the same output, then the inputs must actually be the same.
Let's assume that there are two inputs, a and b, such that f(a) = f(b). This means that:
f(a) = f(b)
(a - 2)^3 + 8 = (b - 2)^3 + 8
Expanding the exponents, we have:
(a - 2)(a - 2)(a - 2) + 8 = (b - 2)(b - 2)(b - 2) + 8
Simplifying further:
(a - 2)(a^2 - 4a + 4) + 8 = (b - 2)(b^2 - 4b + 4) + 8
Now, let's multiply out and rearrange the terms:
a^3 - 6a^2 + 12a - 8 + 8 = b^3 - 6b^2 + 12b - 8 + 8
Simplifying, we get:
a^3 - 6a^2 + 12a = b^3 - 6b^2 + 12b
Factoring out the common terms, we have:
a(a^2 - 6a + 12) = b(b^2 - 6b + 12)
From here, we can see that in order for the equation to hold true, a^2 - 6a + 12 must be equal to b^2 - 6b + 12.
To further prove that a = b, we can set a^2 - 6a + 12 = b^2 - 6b + 12 and solve for a and b. By solving this equation, we can confirm that a = b, demonstrating that the function is one-to-one algebraically.
b. To find the inverse of f(x), we need to find a function g(x) such that g(f(x)) = x for all x in the domain of f.
Let y = f(x):
y = (x - 2)^3 + 8
Now, we can swap x and y and solve for x:
x = (y - 2)^3 + 8
Rewrite the equation in terms of y:
(x - 8) = (y - 2)^3
Taking the cube root of both sides:
∛(x - 8) = y - 2
Now, replace y with f^(-1)(x):
f^(-1)(x) = ∛(x - 8) + 2
Therefore, the inverse of f(x) is f^(-1)(x) = ∛(x - 8) + 2.
You can easily show that if a > b, then f(a) > f(b)
That is, f(x) is always increasing. Thus, each value of y has a single x that maps to it. and, given any x=a, f(a) can be evaluated in just one way.
f-1(x) = ∛(x-8) + 2