In Exercises 67-74, finda. (fog) (x)b. the domain of f o g. f(x) ...

Question

In Exercises 67-74, finda. (fog) (x)b. the domain of f o g. f(x) = x² + 4, g(x) = √(1 − x)

Accepted Answer

So this problem says to solve F of G of X. Also final domain of F of G of X where F X is three X minus two and G X equals the square root of X. So let's first write out what F G F X meets F G X. This is the same thing as saying F on the outside G X on the inside of the premises, we're pretty much saying we're gonna take what we have for G M X and plug it into the X for our F of X. So if our G X is a scroll of X, R F G X is then F discredit effects. Now this we can plug in the squared of X everywhere. We have an X in the F X equation. So we have three squared of X minus two. Now we can't simplify this any further. So that is the equation for F of G of X. Now we want to find what the domain of this equation is. Let's see if we have any restrictions on our equation. We notice here, we have a squared of X. So we'll look at the domain of the square defects. Well squared of X. We cannot have a negative number under the square root, We do that. So the -1. We do not have a real answer. We have a complex answer. So we cannot have any negative numbers in the domain of squared events. Let's see what happens if you have zero squared of zero. Well, that is just zero. So our domain will start from zero. And now we want to check to see where it goes to not a squared of X. It only all positive numbers and zero. So this is going to go to infinity, meaning the answer to the equation x minus two as a domain from zero to infinity. This makes our answer answer B. Okay, I hope that will solve the problem. Thank you for watching. Goodbye.

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In Exercises 67-74, finda. (fog) (x)b. the domain of f o g. f(x) = x² + 4, g(x) = √(1 − x)

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