Question Video: Finding an Unknown in a Rational Function given the Common Domain of Two Rational Functions Mathematics

If the common domain of the two functions 𝑛₁ (π‘₯) = π‘₯/(π‘₯Β² + 64) and 𝑛₂ (π‘₯) = βˆ’5/(π‘₯Β² + 11π‘₯ βˆ’ 𝑏) is ℝ βˆ’ {βˆ’7, βˆ’4}, find the value of 𝑏.

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Video Transcript

If the common domain of the two functions 𝑛 sub one of π‘₯ equals π‘₯ over π‘₯ squared plus 64 and 𝑛 sub two of π‘₯ equals negative five over π‘₯ squared plus 11π‘₯ minus 𝑏 is the set of real numbers minus the set containing negative seven, negative four, find the value of 𝑏.

So we’re given the common domain of our two functions 𝑛 sub one of π‘₯ and 𝑛 sub two of π‘₯. So before we go any further, let’s inspect each of these functions. 𝑛 sub one of π‘₯ and 𝑛 sub two of π‘₯ are rational functions. That is, they are the quotient of a pair of polynomials. And we know that the domain of any rational function is the set of real numbers, not including any values of π‘₯ that make the denominator zero.

So what we’re going to do is actually identify the respective domains of 𝑛 sub one and 𝑛 sub two. To do so, we’re going to set the denominator equal to zero in each case. And those are the values of π‘₯ we can disregard from the domain of those functions. Then, we can use this information to identify and compare the common domain.

Let’s begin by setting π‘₯ squared plus 64 equal to zero. To solve for π‘₯, we would begin by subtracting 64 from both sides. And that gives us the equation π‘₯ squared equals negative 64. Our next step would be to take both the positive and negative square root of negative 64. But we know that does not give us a real value. So there are no real solutions to the equation π‘₯ squared plus 64 equals zero. And so the domain of 𝑛 sub one of π‘₯ is in fact just the set of real numbers. There are no values that make the denominator equal to zero, so no values of π‘₯ that we have to exclude from our domain.

So what about the domain of 𝑛 sub two of π‘₯? Once again, it’s the set of real numbers, but we need to exclude any values of π‘₯ that make the expression π‘₯ squared plus 11π‘₯ minus 𝑏 equal to zero. Now, normally, we would try to factor the left-hand side here. But 𝑏 is an unknown constant, so we can’t do that.

So instead, let’s consider the information we’ve been given about the common domain. It’s the set of real values minus the set containing negative seven and negative four. We know that there were no exclusions for 𝑛 sub one of π‘₯. So this must mean that if π‘₯ is negative seven or if π‘₯ is equal to negative four, that makes the denominator π‘₯ squared plus 11π‘₯ minus 𝑏 equal to zero.

So to find the value of 𝑏, that means that π‘₯ squared plus 11π‘₯ minus 𝑏 is equal to zero when π‘₯ is equal to negative seven, let’s substitute π‘₯ equals negative seven in. When we do, we get negative seven squared plus 11 times negative seven minus 𝑏 equals zero. The left-hand side simplifies to negative 28 minus 𝑏. And then adding 𝑏 to both sides of the equation, we get 𝑏 equals negative 28.

So we assume the value of 𝑏 that makes the expression π‘₯ squared plus 11π‘₯ minus 𝑏 equal to zero, where the domain also excludes π‘₯ equals negative seven and π‘₯ equals negative four, is negative 28. But let’s check by substituting π‘₯ equals negative four. That’s negative four squared plus 11 times negative four minus 𝑏 equals zero, which once again simplifies to negative 28 minus 𝑏 equals zero, so 𝑏 equals negative 28.

Given information then about the common domain of our functions, 𝑏 is negative 28.

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