Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See ...

Question

Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See Examples 6 and 7.

$ƒ (x) = \frac{2}{x}, g (x) = x + 1$

Accepted Answer

Hey, everyone here we are asked to determine F of G of X for the given functions F of X is equal to 10 divided by X and G F X is equal to X plus seven. And we are also asked to find the domain of F of G of X. So here for this problem, we are given for answer choice options, A B C and D and each answer choice has two solutions. One solution is the expression found for F of G of X. And the other solution is the domain of F of G in interval notation. So we can begin this problem by first identifying the domain of F. Well, the domain of f can be found by assessing the expression given for F of X, which we see is 10 divided by X. Well, we know we cannot divide by zero or we do not want to divide by zero. So we know that X cannot be equal to zero. So with this, we see that X is within or an element of all real numbers are minus or excluding the set of zero. So now with this statement, we need to note that the domain of G is also our or all real numbers. So the domain of G is all real numbers. So with this, we can move on to finding F of G of X F of G of X and we can begin by rewriting this expression as F of the quantity of G of X. So now we just need to substitute in our given expression for G of X. So we actually have F of the F of X plus seven. And so now, instead of having F of X as we are, we were originally given in our problem statement, we have F of X plus seven. So all we need to do is substitute in this expression for X from the original function F of X. So F of X plus seven gives us 10 divided by the quantity of X plus seven. So now we have identified the expression for F of G of X and now we can move on to finding the domain of F of G of X. So the domain of F of G of X can be found the same as we found for F. So we need to assess the expression that we just found and we see that X is in the denominator. Well, we know we cannot divide by zero. So we know that X plus seven cannot be equal to zero or the denominator of our expression cannot be equal to zero. So now solving for X by subtracting seven from both sides, we, we see that X cannot be equal to negative seven since we will result in a zero in the denominator. Now, here we also know that G F X should be in the domain of F since it is the inside function in our expression. So we know that G F X is within or an element of all real numbers minus or excluding the set of zero as it needs to match the domain of F. And so now we also have found that G F X or X itself cannot be equal to negative seven from our domain that we found from F of G of X. So now with all of this, we can rewrite the domain of F of G of X using this information in interval notation by going from negative infinity to our excluded value which is negative seven. And then we go from negative seven to positive infinity. And now we need to note that infinity is always excluded. So both negative and positive infinity get closed by parentheses. And next we know that negative seven is excluded. So both negative sevens also get closed by parentheses. So with this, we have found our domain of F of G of X in the interval notation going from negative infinity to negative seven and from negative seven to positive infinity. And we also found our expression for F of G of X as divided by the quantity of X plus seven. So with these two solutions, we are left here with answer choice D. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See Examples 6 and 7.
$ƒ (x) = \frac{2}{x}, g (x) = x + 1$

Key Concepts

Function Composition

Domain of a Function

Rational Functions and Restrictions

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