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

Question

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

$ƒ (x) = \sqrt x, g (x) = \frac{1}{(x + 5)}$

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 X and G FX is equal to one divided by the quantity of X plus 64. And we also need to find the domain of G F F of X. So here we have four answer choice options, ABC and D. And with each of these answer choices, we have two different possibilities for expressions for G F F of X. One of them is one divided by the square root of the quantity of X plus 64. And the other expression could be one divided by the quantity of the square root of X plus 64. And then we have various options for the domains of G F F. So to begin solving this problem, we first want to take a look at our first given function F of X and we want to use this function to find the domain of F. So again, we have F of X is equal to the square root of X. Well, since X is underneath the A root, we know that X must always be positive. So we can see that X can be greater than, or must be greater than, or equal to zero. And so with this statement, this inequality, we can the domain of F in interval notation by going in the interval from zero to positive infinity. Now infinity is always excluded. So it gets closed by a parentheses and zero in this case is included due to our nonstick inequality. So it gets closed by a bracket. Now we can move on to determining the domain of our other function which is G of X. So for the domain of G, we need to recall our given G of X function. So we have G of X is equal to one divided by the quantity of X plus 64. Well, with this expression here, we see X is in the denominator and we know that we do not want to divide by a denominator of zero. So that means that X plus 64 or the expression in the denominator must not be equal to zero. So that means that X cannot be equal to negative 64 since this will result in a denominator of zero. So now with our domains identified, we can move on to finding the expression for G of F of X. And we can begin by rewriting G of F of X as G of the function F of X. And now we can just substitute in the given expression for F of X. So we actually have G of the square root of X. So now we just need to substitute in this expression where we have the square root of X into our original given expression for G of X. So instead of G of X, we have G of the square root of X, which gives us the expression one divided by the quantity of the square root of X plus 64. And now by the property of a square root, we can find the domain of our G of F of X expression here. So, so again, our one divided by the quantity of the square root of X plus is the G of F of X expression that we were asked to find. And now finding the domain again, we use the property of the square root. So we see that X can be greater than or equal to zero since X must always be positive. So with this statement, we also need to notice that X is in the denominator. So the square root of X plus 64 cannot be equal to zero. Well, we know that for when X is greater than or equal to zero, that this expression in the denominator is always greater than zero. So we see that F is in the domain of G and therefore the domain of G of F is from zero to positive infinity. Now again, infinity is excluded. So it gets closed by a parentheses and zero is included since we have this non strict inequality. So it gets closed by a bracket. Therefore, we have found our two solutions here. Our expression for G F F of X and the domain of the expression. So we are left with answer choice B where G F F of X is equal to one divided by the quantity of the square root of X plus 64. And the domain of G F F is from zero included to positive infinity. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Given functions f and g, find (b)(g∘ƒ)(x) and its domain. See Examples 6 and 7.
$ƒ (x) = \sqrt x, g (x) = \frac{1}{(x + 5)}$

Key Concepts

Function Composition

Domain of a Function

Square Root and Rational Function Domains

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