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

Question

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

Accepted Answer

Hey, everyone here we are asked to determine G of F of X for the given functions F of X is equal to the square root of the quantity of X plus 12 and G of X is equal to divided by X. And we also need to find the domain of G of F of X. So here we have four answer choice options, ABC and D and we have two different possible expressions given for G of F of X in these solutions. And then we have various different possibilities for the domain of G of F. So we can begin this problem by first identifying the domain of our first given function F. So if we recall, we are given F of X, which is equal to the square root of the quantity of X plus 12. In now we see this expression X plus 12 is underneath a square root. So we know that it must always be positive or in other words, X plus 12 must be greater than, or equal to zero. Now, just solving for X, we see that X must be greater than or equal to negative 12. And so now with this inequality, we can write the domain of F in interval notation, going from negative infinity to negative 12 and then from negative 12 to positive infinity. And now our infinities are always excluded. So they both get closed by a, by parentheses. And then our negative 12 value gets closed by a bracket since it is included as shown by our, our no strict inequality. So now we can move on to identifying the domain of our other function G. So if we recall G of X is equal to divided by X and here we have X in the denominator of this expression. And we know we cannot divide by zero. Therefore, X cannot be equal to zero. And now we can rewrite this expression here in interval notation to express the domain of G. So the domain of G is from negative infinity to zero and from zero to positive infinity. And all of these terms are closed by parentheses since infinity is always excluded. And in this case, zero is al also excluded since we cannot divide by zero. So now that we have identified both out of our individual domains for each function, we can move on to finding the expression for G of F of X. And now we can begin by rewriting this G of F of X as G of the function F of X or F of X is now in its own set of parentheses. So now we just want to substitute in the expression given for F of X. So we actually have have G of the square root of the quantity of X plus 12. And now we just need to substitute in this expression into G of X. So we actually have G of F of X is equal to. Now substituting in the expression, we have 14 divided by the square root of the quantity of X plus 12. So here we see in this expression that X is in the denominator and also underneath the square root. So to start finding the domain of this expression that we found, we first need to note that X plus 12 or the expression underneath the square root must be greater than or equal to zero. Or in other words, it must be positive. And solving for X, we see that X must be greater than or equal to negative 12. And now we also see that this whole expression, the square root of the quantity of X plus 12 must not be equal to zero. Since it is in the denominator of our expression in solving for XX, we see that X cannot be equal to negative 12 since this will result in a zero in our denominator. So now combining these two expressions here where X can be greater than or equal to negative 12 and X cannot be equal to negative 12. We see that X must be greater than negative and we see this with this inequality that F is in the domain of G here. So we can finish up by writing the domain of G of F of X in interval notation where it goes from negative 12 to positive infinity where infinity is always excluded and therefore closed by a parentheses. And here as shown in our strict inequality, negative 12 is not included. So it gets closed by a parentheses as well. So with our domain of G F F of X found, and the expression also found we have two solutions which leaves us with answer choice C where again, G F F of X is 14 divided by the square root of the quantity of X plus 12. And the domain of G F F is from negative 12 to positive infinity where both 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, and (b)(g∘ƒ)(x) and its domain. See Examples 6 and 7. ƒ(x)=√(x+2), g(x)=-(1/x)