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)=√x, g(x)=1/(x+5)

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 the square root of X and G of X is equal to one divided by the quantity of X plus 64. And 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. There are two possible solutions solutions for the expression of F of G of X. One of which is one divided by the square root of the quantity of X plus 64. And the other possible expression is one divided by the quantity of the square root of X plus 64. And then there are multiple different possibilities for the domain of F of G. So we can begin solving this problem by first identifying the domain of our first given function F of X. So if we recall for the domain of F, we need to recall the, the given F of X function which is F of X is equal to the square root of X. So here we see that X is underneath the square root, which means that X must always be positive or that X must be greater than, or equal to zero. And now, with this inequality, we can write the domain of F and interval notation going from zero to positive infinity where zero is included due to our nonstick inequality and therefore closed by a bracket and infinity is always excluded. So it gets closed by a parentheses. So now we have found the domain of F. So we can move on to identify, find the domain of our other function G. So if we recall G of XX, our given function for G is one divided by the quantity of X plus 64. Now, here we see X is in the denominator of this expression. So this whole expression in the denominator cannot be equal to zero. Therefore, X plus 64 cannot be equal to zero. In solving for X. We see that X cannot be equal to negative 64 because this value results in a zero in the denominator. So now that we have identified both of our domains for both functions, we can move on to finding the expression for F of G of X. And now we can begin by rewriting this F of G of X as F of the function of G of X or G FX in its own set of parentheses. Now, so now we just substitute in the given expression for G FX. So we actually have F of one divided by the quantity of X plus 64. So now we will just need to substitute in this expression for X from F of X. So instead of having just the square root of X four, F of X, we have actually F of one divided by the quantity of X plus 64 which gives us the square root of the quantity of one divided by the quantity of X plus 64. And now just simplifying since we can just take the square root of one, which is just one, we actually have the expression for F of G of X, simplified to one divided by the square root of the quantity of X plus 64. So now that we have identified or found our expression for F of G of X, we can move on to the second pro part of the problem which asks us to find the domain of this expression. So we see that we have an expression X plus underneath a square root, which means this whole expression cannot be equal, cannot be less than zero since it must be infinity. So we know it must be greater than, or equal to zero. And we also know that this whole expression in the denominator where we have the square root of the quantity of X plus 64 cannot be equal to zero because it is underneath or it is, it is a denominator and we cannot divide by zero. So now, just combining these two facts, we see we cannot have our, our expression be equal to zero. But our first inequality says we can have it equal to zero. So again, just combining these two facts, we actually have that, that X must be greater than negative 64. And here we see that G of X is also in the domain of F. And so we can finish up by writing the domain of F of G FX using this inequality that we came up with where X is greater than negative 64. So our domain is from negative 64 to positive infinity where infinity is always excluded. So it gets closed by a parentheses and negative 64 is also excluded as shown in our strict inequality. So it is also closed by a set of parentheses. So now with our domain identified and we found our expression for F of G of X, we have two solutions which leaves us with answer choice C where again, F of G of X is one divided by the square root of the quantity of X plus 64 in the domain of F of G is from negative 64 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.

Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See Examples 6 and 7. ƒ(x)=√x, g(x)=1/(x+5)

Key Concepts

Function Composition

Domain of a Function

Square Root Function

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