Let f(x)= Cube root(x + 2) and g(x) = x^3 - 2. Which of the following are true?

I. g(x) = f^-1(x) for all the real values of x.
II. (f£»g) (x) = 1 for all reak values of x.
III. The Function is one-to-one.

Lets try the inverse of f(x)

x=cubrt(y+2)
x^3=(y+2
y= x^3-2 so g(x)=inverse of f(x)

lets try f(g(x))=cubrt(x^3-2+2)=x

Now, is it one to one? I will leave that to you to examine.

To determine which of the statements are true, let's analyze each one and explain the reasoning behind it.

I. g(x) = f^-1(x) for all real values of x.

To determine if two functions are inverses, we need to check whether applying one function followed by the other will result in the original input for all possible x-values.

To find the inverse of f(x), we can switch the roles of x and y in the equation and then solve for y:

x = (y + 2)^(1/3)
x^3 = y + 2
y = x^3 - 2

The equation y = x^3 - 2 is the formula for g(x). Therefore, g(x) is indeed the inverse of f(x). So, statement I is true.

II. (f£»g) (x) = 1 for all real values of x.

To evaluate (f£»g) (x), we need to substitute g(x) into f(x) and simplify:

f(g(x)) = f(x^3 - 2) = Cube root((x^3 - 2) + 2) = Cube root(x^3) = x

We can see that (f£»g) (x) simplifies to x. Therefore, (f£»g) (x) is equal to 1 for all real values of x, so statement II is false.

III. The function is one-to-one.

A function is one-to-one if it satisfies the horizontal line test, meaning that no two different x-values map to the same y-value.

Let's analyze f(x) = Cube root(x + 2):

For any two distinct x-values, say x1 and x2, if f(x1) = f(x2), then Cube root(x1 + 2) = Cube root(x2 + 2). To find if this is possible, we raise both sides of the equation to the power of 3:

(x1 + 2) = (x2 + 2)
x1 = x2

Since the equation above implies that x1 = x2, we can conclude that f(x) = Cube root(x + 2) is a one-to-one function. Therefore, statement III is true.

In summary:

I. True
II. False
III. True