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Question Video: Finding an Unknown in a Rational Function given Its Domain Mathematics • Third Year of Preparatory School

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If the domain of the function 𝑓(π‘₯) = βˆ’5/(π‘₯Β² βˆ’ 8π‘₯ + π‘˜) is ℝ βˆ’ {4}, determine the value of π‘˜.

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Video Transcript

If the domain of the function 𝑓 of π‘₯ equals negative five over π‘₯ squared minus eight π‘₯ plus π‘˜ is the set of real numbers minus the set containing four, determine the value of π‘˜.

To answer this question, let’s begin by inspecting the function 𝑓 of π‘₯. 𝑓 of π‘₯ is actually a rational function. Assuming π‘˜ is a constant, it’s the quotient of a pair of polynomials. We know that the domain of a rational function is the set of all real numbers. But we must exclude any values of π‘₯ that make the denominator equal to zero. Comparing this to the domain of 𝑓 of π‘₯, the set of real numbers minus the set containing the element four, we can deduce that this element four, this value of π‘₯, must make the denominator zero. In other words, if π‘₯ is equal to four, π‘₯ squared minus eight π‘₯ plus π‘˜ must be equal to zero.

With this in mind, we can substitute π‘₯ equals four into this expression, set it equal to zero, and solve for π‘˜. Substituting π‘₯ equals four, and we get four squared minus eight times four plus π‘˜ equals zero. That’s 16 minus 32 plus π‘˜ equals zero or negative 16 plus π‘˜ equals zero. To solve for π‘˜, we need to add 16 to both sides of our equation. And when we do, we find that π‘˜ is equal to 16. And so, given information about the domain of 𝑓 of π‘₯, we can determine the value of π‘˜ is equal to 16.

Now, in fact, by substituting π‘˜ equals 16 back into the expression for the denominator, we can check our answer. Let’s set it equal to zero. So π‘₯ squared minus eight π‘₯ plus 16 equals zero. To solve this equation for π‘₯ and to find any values of π‘₯ that we have to disregard from the domain of our function, we factor the left-hand side. So π‘₯ minus four times π‘₯ minus four equals zero. And this must mean there is only one solution. It’s the solution to the equation π‘₯ minus four equals zero, which is π‘₯ equals four. We said this was the value of π‘₯ we’d have to disregard from the domain of 𝑓 of π‘₯. So we’ve satisfied our criteria. And so π‘˜ is equal to 16.

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