Question Video: Finding the Domain for Two Rational Functions to Be Equal | Nagwa Question Video: Finding the Domain for Two Rational Functions to Be Equal | Nagwa

Question Video: Finding the Domain for Two Rational Functions to Be Equal Mathematics • Third Year of Preparatory School

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Given the functions 𝑛₁(π‘₯) = π‘₯/(π‘₯Β² βˆ’ 10π‘₯) and 𝑛₂(π‘₯) = 1/(π‘₯ βˆ’ 10), what is the set of values on which 𝑛₁ = 𝑛₂?

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Video Transcript

Given the functions 𝑛 one of π‘₯ is equal to π‘₯ over π‘₯ squared minus 10π‘₯ and 𝑛 two of π‘₯ is equal to one over π‘₯ minus 10, what is the set of values on which 𝑛 one is equal to 𝑛 two?

We’re given two rational functions 𝑛 one and 𝑛 two, where a rational function is a function of the form 𝑃 of π‘₯ over 𝑄 of π‘₯, such that 𝑃 and 𝑄 are polynomials in π‘₯. And the domain of the function, that’s the input values π‘₯, are such that 𝑄 of π‘₯ is not equal to zero. Now, we’re asked to find the set of values on which 𝑛 one is equal to 𝑛 two, that is, the domain or set of input values satisfying 𝑛 one equals 𝑛 two. This means we need to exclude any values of π‘₯ that make our denominators equal to zero. And so we need to solve π‘₯ squared minus 10π‘₯ is equal to zero and π‘₯ minus 10 is equal to zero.

Let’s look first at 𝑛 one, where we have a common factor of π‘₯ in both terms. And taking this outside some parentheses, we have π‘₯ multiplied by π‘₯ minus 10 is equal to zero. Solutions to this are either π‘₯ is equal to zero or π‘₯ minus 10 is equal to zero. And to solve π‘₯ minus 10 is equal to zero, we add 10 to both sides, giving us π‘₯ is equal to 10. Our solutions are therefore π‘₯ is equal to zero and π‘₯ is equal to 10. The domain of 𝑛 one of π‘₯ is therefore the set of real numbers not including the set containing zero and 10.

And now turning to 𝑛 two of π‘₯, we see that the denominator is equal to zero if π‘₯ is equal to 10. And so the domain of 𝑛 two of π‘₯ is the set of real numbers not including the set containing the number 10.

Now let’s take a closer look at the function 𝑛 one of π‘₯. We’ve seen that we have a common factor of π‘₯ in the denominator, which we can take outside some parentheses. If we then divide both numerator and denominator by π‘₯, 𝑛 one is equal to one over π‘₯ minus 10, which is actually equal to our second function 𝑛 two of π‘₯. This means that the set of values on which 𝑛 one is equal to 𝑛 two is the domain of 𝑛 one. And this is because the domain of 𝑛 one includes the domain of 𝑛 two and our domain must apply to both functions. The set of values on which 𝑛 one is equal to 𝑛 two is therefore the set of real numbers not including the set containing the numbers zero and 10.

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