Algebraic Combinations

In Ex...

Question

Algebraic Combinations

In Exercises 3 and 4, find the domains of f, g, f/g and g/f.

f(x) = 1, g(x) = 1 + √x

Accepted Answer

Welcome back, everyone. Consider the functions H and G defined as follows H of X equals 2 and G of X equals 3 minus root of X. Determine the domains of HG H divided by G and G divided by H. So let's begin with the first part. Let's consider the function H. We know that H of X equals 2. It is a constant function which is independent of X, right? So the domain would be all real numbers because we don't have any restrictions on any constant function. This means that X belongs to all real numbers or basically from negative infinity up to positive infinity. Now let's consider our next function, which is J. Our g of X function is 3 minus square root of X, and we have to recall that we cannot take square root of a negative number, meaning when we consider our domain X must be greater than or equal to 0, and basically we can say that in an interval notation this is X belongs to the interval from zero inclusive up to infinity. That's how we get our two domains for H and J respectively. Now, let's consider a composition H divided by J. So, in the numerator, We can write that H of G of X. This is going to be equal to 2 in our numerator and in the in the denominator, we are substituting 3 minus square root of X. So here we have to be very careful. We have two conditions to satisfy. First of all, we want to make sure that X is greater than or equals to 0 because we cannot take square root of a negative number. And also, we want to make sure that our denominator is not equal to 0 because division by zero is undefined, right? So what can we tell? Well, we can say that first of all, X belongs to the interval from zero inclusive up to infinity, and we also want to make sure that X. Is not equal to what value? Well, we can simply analyze and say that 3 is not equal to square root of X, and if we erase, if if we basically square both sides, then we can see that 9 is not equal to X. So we can also say that X is not equal to 9. What we want to do is simply exclude 9 from our innerval, right? So we are starting with 0. We have all real numbers from 0 up to infinity, excluding 9. So we can say that in this case, Our domain would be from 0 up to infinity excluding 9, so this means that we have X belongs to the interval from 0 inclusive up to 9, not inclusive, and from 9. Up to infinity. Well done. And finally, let's analyze G divided by H. So we have a composition G divided by H of X. In this case, we simply have G in the numerator that's 3 minus square root of X divided by in the denominator we simply have 2. So now our denominator is number equal to 0, meaning we only want to satisfy. X being not negative, so X is greater than or equal to 0 because square root of a negative number is not defined for real numbers, which means that the domain. Is Xs belongs to. Zero inclusive up to infinity. And that's it, we have our answers. So the first one is all real numbers, that's the domain for H. For G, the domain is from zero inclusive up to infinity. For J H divided by J, we have a domain. From 0 inclusive up to 9 not inclusive, and from 9 not inclusive up to infinity and for G divided by H our domain is from zero inclusive up to infinity. Thank you for watching.

Algebraic Combinations

In Exercises 3 and 4, find the domains of f, g, f/g and g/f.

f(x) = 1, g(x) = 1 + √x

Key Concepts

Domain of a Function

Algebraic Combinations of Functions

Restrictions on Domains

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