Thread: f(x) and g(x), need to solve (fg)(x) and (gf)(x)

1. f(x) and g(x), need to solve (fg)(x) and (gf)(x)

f(x) = cuberoot(x+1) and g(x) = cosx

how do I do this?

Here is the note I took in class:

Let f and g be two functions having the domain D. The product of f and g is that function which assigns to x is a subset of D the number f(x)g(x)

fg
x ----> (fg)(x) = f(x)g(x)

Did I copy this incorrectly? I thought (fg)(x) = f[g(x)] not f(x)g(x)

help! Thanks

2. Originally Posted by bloodyfinger
f(x) = cuberoot(x+1) and g(x) = cosx

how do I do this?

Here is the note I took in class:

Let f and g be two functions having the domain D. The product of f and g is that function which assigns to x is a subset of D the number f(x)g(x)

fg
x ----> (fg)(x) = f(x)g(x)

Did I copy this incorrectly? I thought (fg)(x) = f[g(x)] not f(x)g(x)

help! Thanks

Hi bloodyfinger,

Usage varies sometimes by textbook and author. When in doubt, ask for clarification.

My opinion...I would look at $(fg)(x)$ as the product $f(x) \cdot g(x)$

It should have been written as $(f \cdot g)(x)$. This way the meaning is clear.

If they had intended to indicate the composition of functions, they would've used $[f \circ g](x)]$ to mean $f[g(x)]$

3. Oh shoot, my mistake. I didn't realize there was a difference between $\circ$ and $\cdot$ .

The original question actually asks me to find $[f \circ g](x)]$ and $[g \circ f] (x)]$.

How do I do this if $f(x) = \sqrt[3](x+1)$ and $g(x) = cosx$

4. Originally Posted by bloodyfinger
Oh shoot, my mistake. I didn't realize there was a difference between $\circ$ and $\cdot$ .

The original question actually asks me to find $[f \circ g](x)]$ and $[g \circ f] (x)]$.

How do I do this if $f(x) = \sqrt[3](x+1)$ and $g(x) = cosx$
$f[g(x)] = f[\cos{x}] = \sqrt[3]{\cos{x} + 1}$

you do $g[f(x)]$

5. Originally Posted by skeeter
$f[g(x)] = f[\cos{x}] = \sqrt[3]{\cos{x} + 1}$

you do $g[f(x)]$
Is there any way to simplify $f[g(x)] = \sqrt[3]{\cos{x} + 1}$ or $g[f(x)] = cos\sqrt[3]{x+1}$ ?

6. Originally Posted by bloodyfinger
Is there any way to simplify $f[g(x)] = \sqrt[3]{\cos{x} + 1}$ or $g[f(x)] = cos\sqrt[3]{x+1}$ ?
looks like both are in "simple" form to me.

7. Originally Posted by skeeter
looks like both are in "simple" form to me.
That was shockingly easy.... I have a hard time believing the question was that simple... haha

Thanks so much though!!