express the function in the form f * g

g(x) = cube-root (x/1+x)

i know f(x) has to be something
and g(x) has to be something

i think that f(x) = x/1+x
and g(x) = x^(1/3)
but im not sure how to do this....thanks for the help!

Yours is a correct solution. There are several others

thanks, what are the others?

To express the function g(x) = cube-root (x/1+x) in the form f * g, we need to find f(x) such that f(x) * g(x) = g(x). In other words, we need to find a function f(x) that when multiplied by g(x) will result in g(x) itself.

To do this, let's start by expressing g(x) as the product of two functions, f(x) and g(x).

g(x) = f(x) * g(x)

Now, let's substitute the given g(x) in terms of its components:

cube-root (x/1+x) = f(x) * x^(1/3)

To isolate f(x), we can divide both sides of the equation by x^(1/3):

cube-root (x/1+x) / x^(1/3) = f(x)

Simplifying the left side, we can rewrite x/1+x as 1/(1/x + 1):

cube-root (1/(1/x + 1)) / x^(1/3) = f(x)

Further, we can simplify the cube root:

(1/(1/x + 1))^(1/3) / x^(1/3) = f(x)

Finally, we can simplify further by applying the power rule for division:

(1 / (x+1)^3)^(1/3) / x^(1/3) = f(x)

Therefore, the function f(x) in the form f * g is:

f(x) = (1 / (x+1)^3)^(1/3) / x^(1/3)

To express the function g(x) = cube-root(x/1+x) in the form f * g, you first need to find the functions f(x) and g(x) such that g(x) = f(x) * g(x).

Let's start with g(x) = cube-root(x/1+x).

To express g(x) as the product of two functions, we can rewrite g(x) as follows:

g(x) = (x/1+x)^(1/3)

Now, let's take a closer look at the expression (x/1+x)^(1/3). Notice that the denominator 1+x can be factored as 1 * (x + 1).

So, we can rewrite the expression as:

g(x) = (x/(1 * (x + 1)))^(1/3)

Now, we can split g(x) into two functions, f(x) and g(x):

Let f(x) = x
And g(x) = (1/(1+x))^(1/3)

Now, if we multiply f(x) * g(x), we get back g(x):

f(x) * g(x) = x * (1/(1+x))^(1/3)

So, we have expressed the function g(x) in the form f * g, where f(x) = x and g(x) = (1/(1+x))^(1/3).

Note that there could be multiple ways to express a function as the product of two functions, but this is one possible solution for this specific function.