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 that πβ(π₯) = (π₯ β 5)/(π₯ + 5) and πβ(π₯) = (π₯Β² β 5π₯)/(π₯Β² + 5π₯), find the largest set on which the functions πβ and πβ are equal.

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### Video Transcript

Given that π one of π₯ is equal to π₯ minus five over π₯ plus five and π two of π₯ is equal to π₯ squared minus five π₯ over π₯ squared plus five π₯, find the largest set on which the functions π one and π two are equal.

Weβre given two rational functions π one and π two, where a rational function is a function of the form π of π₯ over π of π₯, where π and π are polynomials in π₯. And the domain of the function is the set of inputs π₯ for which the denominator π of π₯ is nonzero. Now, we want to find the largest set on which our two rational functions π one and π two are equal. And this means finding the domain that covers both π one and π two.

If we look first at our function π one of π₯, we need to exclude any values for which the denominator, thatβs π₯ plus five, is equal to zero. So we need to solve π₯ plus five is equal to zero. And subtracting five from both sides, we have π₯ equal to negative five. The domain of π one of π₯ is therefore the set of real numbers not including the set containing the number negative five.

Now, if we look at π two of π₯, in order for our denominator π₯ squared plus five π₯ to be nonzero, we need to solve the equation π₯ squared plus five π₯ is equal to zero. We have a common factor of π₯ in both terms, which we can take outside some parentheses. And so we can see that to satisfy our equation, either π₯ is equal to zero or π₯ plus five is equal to zero. And we already know that for π₯ plus five to be equal to zero, π₯ must be negative five. Our solutions are therefore π₯ is equal to zero or π₯ is equal to negative five.

Now making some space, we have the domain of π two of π₯ is the set of real numbers not including the set containing negative five and zero. We have to exclude these values from our domain since if the denominator of the function is equal to zero, the function itself is undefined. Now, letβs take a closer look at our function π two of π₯. We see that both our numerator and denominator have a common factor of π₯. If we then divide both numerator and denominator by π₯, our function π two is π₯ minus five over π₯ plus five. And this is equal to π one of π₯, which is our first function. So we have π two of π₯ is equal to π one of π₯ with the domain the set of real numbers not including the set containing negative five and zero, which is the domain of π two. And thatβs because the domain must exclude any values of π₯ for which the denominator of either function is equal to zero.

Hence, the largest set on which the functions π one and π two are equal is the set of real numbers not including the set containing the numbers negative five and zero.

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