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Determine two simple functions who composition is $f g(x)$ ).$f(g(x))=\sqrt[3]{x^{4}-7 x^{2}+3 x-2}$

$$f(x)=\sqrt[3]{x}, g(x)=x^{4}-7 x^{2}+3 x-2$$

Algebra

Chapter 1

Functions and their Applications

Section 2

Basic Notions of Functions

Functions

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Lectures

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In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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Determine two simple funct…

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Find two functions $f$ and…

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If $f(x)=3 x, g(x)=\sqrt{x…

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Let $g(x)=x^{2}+3 .$ Find …

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for this problem. We have been given a composite function f of G f X, and our goal is to break this into two simple functions F of X and G of X. Now there are more than one. There's more than one way to do this, but we're gonna find one set of functions that we can break this composite function into. So let's go back and look at what we're actually doing. In a composite function, if I have f of g f X, it means I have an outer function f that I'm evaluating at some point g of X. So I'm gonna take out. I'm gonna go to F take Out X. I'm gonna plug in this g of X. I'm gonna plug something into its place, and I know it's a very high tech way of doing it, but we're actually going to use that word, something to see if he can help us figure out which pieces F in which pieces G because I need an outer function. So I'm gonna be doing something. I'm gonna be doing some function to something, so let's see if we can make sense of that with what we have here. I have a cube root function. I'm taking the cube root of something. So that's something is going to be my G of X. Everything is under that radical, that thing that I'm doing that cube root function that is my f That's my outer function. So let's see if we could go back and get what we started with. If I have f of X equal in the cube root of X, that means if I do f of g of X, I'm gonna take the cube root of something. Well, what is that? Something? I'm going to take G of X and plug it in. That gives me what I started with. So this is indeed a valid decomposition into f of X and G f X.

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