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

Question

Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See Examples 6 and 7. ƒ(x)=1/(x-2), g(x)=1/x

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 one divided by the quantity of X minus 16 and G of X is equal to one divided by X. Now, we also need to find the domain of F of G of X. So here we have four answer choice options. ABC and D and we have two possible expressions to describe F of G of X. One of them is just X minus 16 and the other is X divided by the quantity of one minus 16 X. And then we have multiple different options or possibilities for the domain of F O G. So we can begin this problem by first finding or identifying the domain of our first given function, which is F and so for the domain of F, we need to recall our given function for F of X which is one divided by the quantity of X minus 16. And now we see here that X is in the denominator of this expression. So we know that we do not want to divide by zero, which means that this expression in the denominator X minus 16 must not be equal to zero. So just adding 16 to both sides, we see that we actually cannot have X be equal to positive 16. Since this will result in our denominator being equal to zero. Now using interval notation, we can use this statement of X being not equal to 16 to write the domain of. So first we go from negative infinity to 16 and then we go from 16 to positive infinity and now infinity is always excluded. So both infinities here get closed by parentheses. And we know that X cannot be 16. So both 16 values are closed by parentheses since they are also excluded. Now that we have found the domain of F, we can move on to identifying the domain of our other function G. So if we recall G of X is equal to one divided by X. Now here again, X is in the denominator but it is all alone. So we just have that X cannot be equal to zero. Now, in interval notation, the domain of G becomes from negative infinity to zero and then from zero to positive infinity. And just like before all of these terms get closed by parentheses, since zero and infinity are excluded. Now, our next step in solving this problem is to determine the expression for F of G of X. And now we can begin by rewriting this expression F of G of X as F of the, the function of G of X. And now we just substitute in the given expression for G F F four G FX. So we actually have F 01 divided by X. And now we just need to substitute in one divided by X four X from our original F of X fun function. So F of one divided by X gives us one divided by the quantity of one divided by X minus 16. And now simplifying, we see that our expression for F of G of X simplifies two X divided by the quantity of one minus 16 X. And now we need to find the domain of F of G FX as stated in the problem. So again, we need to look at our denominator here since we have X in the denominator and we know that this expression one minus 16 X cannot be equal to zero. Now we know solving for X, that X can actually not be equal to 1/16. And now, with this value, we also know that G of X should be in the domain of X or the domain of F, excuse me. So that means that G of X must be greater than 16 as it is in the domain of F that we identified earlier or G FX must also be less than 16. So using these two inequalities, we can find the specifics of our domain of F of G of X. So now we just need to substitute in the given expression for G FX. So we actually have one divided by X is greater than and one divided by X is less than 16. Now solving again for X, we see that X must be greater than 1/16 and X must be less than 1/16. And now with these two inequalities, as well as the domain, we identified for our insight function G we can finish by writing the domain of G and interval notation. So first, we go from negative infinity to zero as shown in our domain of G. So again, go from negative infinity to zero. And both of these terms are still closed by parentheses. And then we go from zero to the next closest value which is 1/16 and both of these terms are excluded. So they also get closed by parentheses. And then lastly, we go from this value of 1/16 all the way to positive infinity. And we close both of these terms with parentheses. So here we have identified the domain of our F of G FX expression and we also found the expression. So with the, our two solutions, we are left with answer choice A where summarizing we have F of G FX equal to X divided by the quantity of one minus 16 X. And the domain of F of G is from negative infinity to zero and from 0 to 1/16. And then from 1/16 to infinity. And all of these terms are closed by parentheses. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

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Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See Examples 6 and 7. ƒ(x)=1/(x-2), g(x)=1/x