Let f(x)= (x-2)^3+8
a. Show that this function is one-to-one algebraically.
b. Find the inverse of f(x)
a. To show that a function is one-to-one algebraically, we need to demonstrate that if f(x₁) = f(x₂), then x₁ = x₂.
Let's say f(x₁) = f(x₂). This means that (x₁ - 2)^3 + 8 = (x₂ - 2)^3 + 8.
Expanding both sides, we have:
x₁^3 - 6x₁^2 + 12x₁ - 8 = x₂^3 - 6x₂^2 + 12x₂ - 8.
Subtracting like terms, we get:
x₁^3 - x₂^3 - 6x₁^2 + 6x₂^2 + 12x₁ - 12x₂ = 0.
Factoring out the difference of cubes, we have:
(x₁ - x₂)(x₁^2 + x₁x₂ + x₂^2) - 6(x₁^2 - x₂^2) + 12(x₁ - x₂) = 0.
Factoring further, we get:
(x₁ - x₂)((x₁^2 + x₁x₂ + x₂^2) - 6(x₁ + x₂) + 12) = 0.
For this expression to be true, we need either (x₁ - x₂) = 0, or ((x₁^2 + x₁x₂ + x₂^2) - 6(x₁ + x₂) + 12) = 0.
If we consider the case (x₁ - x₂) = 0, then x₁ = x₂, proving that the function is one-to-one.
Now let's consider the second case:
(x₁^2 + x₁x₂ + x₂^2) - 6(x₁ + x₂) + 12 = 0.
This is a quadratic equation in terms of x₁, but we know it has no real solutions because there is no x₁ that makes the function output the same value for two different inputs, as we already proved.
Therefore, we conclude that the function f(x) = (x - 2)^3 + 8 is one-to-one algebraically.
b. To find the inverse of f(x), let's interchange x and y and solve for y.
x = (y - 2)^3 + 8.
Now, let's solve for y:
(x - 8) = (y - 2)^3.
Extracting the cube root:
∛(x - 8) = y - 2.
Finally, adding 2 to both sides:
y = ∛(x - 8) + 2.
Thus, 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 demonstrate that for any two distinct elements in the domain, the corresponding elements in the range are also distinct.
Let's consider two values in the domain, x₁ and x₂, where x₁ ≠ x₂, and assume that f(x₁) = f(x₂). We can then compare the values of f(x₁) and f(x₂) as follows:
f(x₁) = (x₁ - 2)^3 + 8
f(x₂) = (x₂ - 2)^3 + 8
If f(x₁) = f(x₂), then we have the following equality:
(x₁ - 2)^3 + 8 = (x₂ - 2)^3 + 8
Now, let's simplify and solve for x₁ and x₂:
(x₁ - 2)^3 = (x₂ - 2)^3
Taking the cube root of both sides:
x₁ - 2 = x₂ - 2
x₁ = x₂
Since we have obtained x₁ = x₂, this contradicts our initial assumption that x₁ ≠ x₂. Thus, we can conclude that f(x) is one-to-one algebraically.
b. To find the inverse of f(x), let's first replace f(x) with y:
y = (x - 2)^3 + 8
Next, we'll swap x and y and solve for y:
x = (y - 2)^3 + 8
Now, let's rearrange the equation to solve for y:
x - 8 = (y - 2)^3
Taking the cube root of both sides:
(x - 8)^(1/3) = y - 2
Adding 2 to both sides:
(x - 8)^(1/3) + 2 = y
Therefore, the inverse function of f(x) is:
f^(-1)(x) = (x - 8)^(1/3) + 2
a. To show that the function f(x) = (x - 2)^3 + 8 is one-to-one algebraically, we need to prove that for any two distinct values of x, the corresponding function values f(x) are also distinct.
To do this, we can assume that f(x₁) = f(x₂) and then prove that x₁ must be equal to x₂.
Let's start by assuming f(x₁) = f(x₂):
(x₁ - 2)^3 + 8 = (x₂ - 2)^3 + 8
By expanding both sides of the equation, we get:
x₁^3 - 6x₁^2 + 12x₁ - 8 = x₂^3 - 6x₂^2 + 12x₂ - 8
Rearranging the equation, we have:
x₁^3 - 6x₁^2 + 12x₁ = x₂^3 - 6x₂^2 + 12x₂
Next, we factor out (x₁ - x₂):
(x₁ - x₂)(x₁^2 + x₁x₂ + x₂^2 - 6x₁ - 6x₂ + 12) = 0
Since we can't divide by zero, we can conclude that either (x₁ - x₂) = 0 or (x₁^2 + x₁x₂ + x₂^2 - 6x₁ - 6x₂ + 12) = 0.
If (x₁ - x₂) = 0, it implies x₁ = x₂, which means f(x) is one-to-one.
If (x₁^2 + x₁x₂ + x₂^2 - 6x₁ - 6x₂ + 12) = 0, we can manipulate this equation to show that it is always greater than or equal to zero, and therefore can't equal zero. Hence, it doesn't affect the one-to-one property.
Therefore, we have proven algebraically that f(x) = (x - 2)^3 + 8 is one-to-one.
b. To find the inverse of f(x), we need to interchange x and y, and then solve for y.
Let's start by interchanging x and y:
x = (y - 2)^3 + 8
Now, we can solve for y:
x - 8 = (y - 2)^3
Taking the cube root of both sides, we have:
∛(x - 8) = y - 2
Finally, we isolate y by adding 2 to both sides:
y = ∛(x - 8) + 2
Therefore, the inverse of f(x) = (x - 2)^3 + 8 is given by:
f^(-1)(x) = ∛(x - 8) + 2
see
https://www.jiskha.com/questions/1800159/Let-f-x-x-2-3-8-a-Show-that-this-function-is-one-to-one-algebraically-b-Find