In Exercises 31–50, find ƒ+g, f−g, fg, and f/g. Determine the dom...

Question

In Exercises 31–50, find ƒ+g, f−g, fg, and f/g. Determine the domain for each function. f(x) = √(x -2), g(x) = √(2-x)

Accepted Answer

for the given functions F of X equals the square root of x minus 11 and G X equals the square of 11 minus x. Find F plus G and write the domain. So we're first gonna add together our equations are fx and vfx together. F plus G. So let's say we have F X plus G X. Now we will add our equations together. You have square x minus Plus, 11 -X. Now these are not the same radical. We have x minus 11 here minus X there. So we actually can't simplify this any further. Sorry, the question is gonna be this now we want to figure out what the domain of this equation is to do that. We have to look at our square roots. So the square root we cannot have a negative value underneath our square root or else we do not have a real answer. So let's go and look at our square roots here, we have the square root of X -11. Let's look at the domain of this. Well, If we were to plug in everything under the square root instead of equal to zero, let's say we take X -11 equal to zero. Soft for X X equals 11. So any number less than 11 is going to give us a negative and a radical. You can see that here. If you plug it in Let's say we want x equals 10, -11. That gives us -1. So domain of squared of X -11 is 11 to infinity. Let's check squared of 11 -6. So you have square 11 minus X. If this domain we're gonna set everything on radical equal to zero, solved. We get 11 minus X equals zero minus 11, negative X equals negative 11 X equals 11. Now I love minus X is a little bit different because the X is now the second number. So this is actually the everything greater than 11 that will cause this to be negative. You can see that here if we say X is 12 You will get the square of -12 which is squared of either one. So the main of this is everything from negative infinity. Two, 11 like so Now you look here the first equation we have, it is undefined if X is less than 11, the second one is undefined, X is greater than 11. So they kind of both cancel each other out. And the only place where both of these functions are not undefined at the same time is at X equals 11, meaning that our domain is X equals 11. With our equation squared of x minus 11 plus squared of minus X. Making this answer answer. D. Okay, I hope so. We solve the problem. Thanks for watching. Goodbye

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