Question Video: Simplifying and Determining the Domain of a Quotient of Two Rational Functions Mathematics

Simplify the function 𝑓(π‘₯) = (π‘₯Β² βˆ’ 12π‘₯ + 36)/(π‘₯Β³ βˆ’ 216) Γ· (7π‘₯ βˆ’ 42)/(π‘₯Β² + 6π‘₯ + 36), and determine its domain.

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Video Transcript

Simplify the function 𝑓 of π‘₯ equals π‘₯ squared minus 12π‘₯ plus 36 over π‘₯ cubed minus 216 divided by seven π‘₯ minus 42 over π‘₯ squared plus six π‘₯ plus 36, and determine its domain.

To simplify 𝑓 of π‘₯, let’s begin by factorizing each of the terms in the rational expressions, starting with the numerator of the left-hand term, π‘₯ squared minus 12π‘₯ plus 36. This readily factorizes to π‘₯ minus six times π‘₯ minus six, which is π‘₯ minus six all squared. For the denominator of the left-hand term, we have a difference of two cubes. Recall that when we have a difference of two cubes, π‘₯ cubed minus 𝑐 cubed, we can factorize this as π‘₯ minus 𝑐 times π‘₯ squared plus 𝑐π‘₯ plus 𝑐 squared. 216 will be something cubed, and it so happens that it’s a perfect cube, six cubed. Therefore, we can factorize this as π‘₯ minus six times π‘₯ squared plus six π‘₯ plus six squared, which is 36.

We could try to factorize this quadratic expression further. By taking the discriminant 𝑏 squared minus four π‘Žπ‘, we get a result of negative 108, which is less than zero. Therefore, this quadratic expression has no real roots and cannot be factorized further. For the numerator of the second rational function, we have seven π‘₯ minus 42. We can take out a common factor of seven to give seven times π‘₯ minus six. And finally, for the denominator of the second rational function, we have π‘₯ squared plus six π‘₯ plus 36. We have just shown that this exact expression has no real roots and cannot be factorized further.

So far then, we can simplify 𝑓 of π‘₯ to π‘₯ minus six all squared over π‘₯ minus six times π‘₯ squared plus six π‘₯ plus 36 divided by seven times π‘₯ minus six over π‘₯ squared plus six π‘₯ plus 36. Recall that if we have a function 𝑓 of π‘₯ that is the quotient of two rational functions, 𝑝 of π‘₯ over π‘ž of π‘₯ and π‘Ÿ of π‘₯ over 𝑠 of π‘₯, the domain of 𝑓 is the set of real numbers ℝ minus 𝑍 of π‘ž of π‘₯ minus 𝑍 of π‘Ÿ of π‘₯ minus 𝑍 of 𝑠 of π‘₯, where 𝑍 denotes the zeros of a function. In our case, π‘ž of π‘₯ is the denominator of the first rational function, π‘₯ minus six times π‘₯ squared plus six π‘₯ plus 36. π‘Ÿ of π‘₯ is the numerator of the second rational function, seven times π‘₯ minus six. And 𝑠 of π‘₯ is the denominator of the second rational function, π‘₯ squared plus six π‘₯ plus 36.

We need to find all the values of π‘₯ that solve the equations π‘ž of π‘₯ equals zero, π‘Ÿ of π‘₯ equals zero, and 𝑠 of π‘₯ equals zero and subtract them from the set of real numbers to find the domain of 𝑓. For the first equation, either π‘₯ minus six must equal zero or π‘₯ squared plus six π‘₯ plus 36 must equal zero. If π‘₯ minus six equals zero, π‘₯ is equal to six.

For this quadratic expression, we already saw earlier that the discriminant, 𝑏 squared minus four π‘Žπ‘, was less than zero. Therefore, this expression has no real roots and therefore cannot equal zero for any real value of π‘₯. The second equation solves easily to give π‘₯ equals six once again. And for the third equation, we have exactly the same quadratic expression with the discriminant less than zero. Therefore, there are no real values of π‘₯ that solve this equation. Therefore, the only value of π‘₯ for which any of these functions are equal to zero is six. Therefore, the domain of 𝑓 is the set of real numbers ℝ minus the single element six.

Now, we can safely proceed with taking the quotient and simplifying. To take the quotient, we reciprocate the dividing rational function. So we swap the numerator and denominator and then we change the division to multiplication. Taking the multiplication by multiplying the numerators together and the denominators together, we get π‘₯ minus six all squared times π‘₯ squared plus six π‘₯ plus 36 all over π‘₯ minus six times π‘₯ squared plus six π‘₯ plus 36 times seven times π‘₯ minus six. The π‘₯ minus six all squared on the numerator will cancel with both π‘₯ minus six terms on the denominator. And the π‘₯ squared plus six π‘₯ plus 36 will cancel with the same on the denominator. This leaves us with just one on the numerator and seven on the denominator, so 𝑓 of π‘₯ equals one-seventh.

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