a. To show that the function f(x) is one-to-one algebraically, we need to show that for any two distinct values of x, their corresponding y-values will also be distinct.
Let's assume two different values of x, x1 and x2, such that x1 ≠ x2.
f(x1) = (x1 - 2)^3 + 8
f(x2) = (x2 - 2)^3 + 8
Now, we will show that if f(x1) = f(x2), then x1 = x2:
If f(x1) = f(x2), then (x1 - 2)^3 + 8 = (x2 - 2)^3 + 8
Subtracting 8 from both sides,
(x1 - 2)^3 = (x2 - 2)^3
Taking the cube root of both sides,
x1 - 2 = x2 - 2
Cancelling out "2" from both sides,
x1 = x2
Hence, we have shown that for any two distinct values of x, their corresponding y-values will also be distinct. Therefore, the function f(x) is one-to-one algebraically.
b. To find the inverse of f(x), we will exchange the roles of x and y and solve for y.
Let y = (x - 2)^3 + 8
To find the inverse, we need to solve this equation for x.
Step 1: Swap x and y.
x = (y - 2)^3 + 8
Step 2: Solve for y.
x - 8 = (y - 2)^3
Taking the cube root of both sides,
∛(x - 8) = y - 2
Adding 2 to both sides,
y = ∛(x - 8) + 2
Therefore, the inverse of f(x) is given by:
f^(-1)(x) = ∛(x - 8) + 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)
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