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Question

Find $f+g$ , $f-g$ , $fg$ , and $\frac{f}{g}$ . Determine the domain for each function.

$f\left(x\right)=6-\frac{1}{x}$ , $g\left(x\right)=\frac{1}{x}$

Accepted Answer

Hello. In this video, we are going to be using the given F of X and G of X function to find the composite function F divided by G. We will also be identifying the domain of F divided by G. So the F of X function is given to us as 14 minus three divided by X. And the G of X function is given to us as three divided by X. How can we use these two functions to solve for F divided by G? Well, F divided by G is telling us that we want to take the F of X function and divide it by the G of X function. So we will write the F of X function which is 14 minus three divided by X and divided by the G of X function which is three divided by X. Now, what we want to do is we want to simplify the composite function. Currently, this function is in the form of a complex number. In order to simplify the complex fraction, we want to go ahead and turn the domain to the number one. In order to do this, we can multiply the denominator by its reciprocal, the reciprocal of the denominator will be X divided by three. And if we multiply the denominator by the reciprocal, we must multiply the entirety of the numerator by the reciprocal as well. This will allow us to reduce the denominator of the complex fraction to one. And that will leave us with the numerator of X divided by three multiplied by 14 minus three divided by X. In order to simplify, we will distribute X divided by 3 to 14 and negative three divided by X. That will leave us with 14 X divided by three minus X divided by three multiplied by three divided by X. If we take a look at the right hand term, we can cross reduce the product to just be one. And if we rewrite further, we can rewrite the function to be 14 divided by three multiplied by X minus one. This is going to be the simplest form of F divided by G of X. Now, what about its domain? While F divided by G is in the form of a polynomial function. But the one thing we need to understand is that the original F of X function and G of X function were in the form of rational functions. If we take a look at F of XF of X was given to us as 14 minus three divided by X. The only thing we don't want in a rational function is a denominator of zero And since we have an individual X in this denominator, X cannot be zero, the same can be said for the G of X function G of X was given to us as three divided by X. And the only value that makes this undefined is the value of zero. Since F divided by G is a composite function made up of F and G of X, we must include these restrictions in its domain. We can have all numbers of X beside zero. So we will represent the domain of F divided by G as all real numbers negative infinity, going to zero, not including the value of zero, going to positive infinity. So the domain is negative infinity to zero union zero to infinity. So just to recap F divided by G is equal to 14 divided by three multiplied by X minus one and its domain is all real numbers not including the value of zero. With that being said, the solution to this problem is going to be B. So I hope this video helps you in understanding on how to approach this problem. And I'll go ahead and see you all in the next video.

Find ﻿f+gf+gf+g﻿, ﻿f−gf-gf−g﻿, ﻿fgfgfg﻿, and ﻿fg\frac{f}{g}gf​﻿. Determine the domain for each function.﻿f(x)=6−1xf\left(x\right)=6-\frac{1}{x}f(x)=6−x1​﻿, ﻿g(x)=1xg\left(x\right)=\frac{1}{x}g(x)=x1​﻿

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Find $f+g$ , $f-g$ , $fg$ , and $\frac{f}{g}$ . Determine the domain for each function.
$f\left(x\right)=6-\frac{1}{x}$ , $g\left(x\right)=\frac{1}{x}$