Question Video: Determining the Domain of a Function Involving Square Roots | Nagwa Question Video: Determining the Domain of a Function Involving Square Roots | Nagwa

Question Video: Determining the Domain of a Function Involving Square Roots

Find the domain of the function ๐‘“(๐‘ฅ) = 6/โˆš(9 โˆ’ 25๐‘ฅยฒ) in โ„.

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Video Transcript

Find the domain of the function ๐‘“ of ๐‘ฅ equals six over the square root of nine minus 25๐‘ฅ squared in the set of real numbers.

Remember, the domain of a function is the complete set of possible values of the independent variable. Here, thatโ€™s ๐‘ฅ. We essentially say that it is the set of all possible ๐‘ฅ values, which makes the function work and outputs real ๐‘ฆ values. When finding the domain, we need to consider that the denominator of some fraction or rational function canโ€™t be equal to zero. Thatโ€™s because when we divide a real number by zero, we say that we get a result thatโ€™s undefined. We also know that the number underneath a square root sign cannot be negative, absolutely must be positive. So, letโ€™s consider how this impacts on our function.

We have a rational function, so the denominator, the square root of nine minus 25๐‘ฅ squared, cannot be equal to zero. So, letโ€™s assume it is equal to zero and solve for ๐‘ฅ. They will tell us the values of ๐‘ฅ that we cannot use. We begin by squaring both sides of this equation. And we get nine minus 25๐‘ฅ squared equals zero. We then factor nine minus 25๐‘ฅ squared, and we could use a difference of two squares. We get three minus five ๐‘ฅ times three plus five ๐‘ฅ. Now, for the product of these two expressions to itself be equal to zero, either three minus five ๐‘ฅ must be zero or three plus five ๐‘ฅ must be zero. Letโ€™s solve the first equation for ๐‘ฅ. Weโ€™re gonna add five ๐‘ฅ to both sides. And we get three equals five ๐‘ฅ. We then divide through by five, and we get ๐‘ฅ equals three-fifths.

To solve the second equation, we subtract three from both sides to get five ๐‘ฅ equals negative three. We then divide through by five, and we found the second solution to the equation. Itโ€™s ๐‘ฅ equals negative three-fifths. Now of course, these are the values for which the denominator of the fraction is equal to zero. So, we say that part of our domain is the fact that ๐‘ฅ cannot be equal to either positive or negative three-fifths.

Weโ€™ll now consider the second part, the number inside the square root must itself be positive. That is, nine minus 25๐‘ฅ squared is greater than zero. Now, we just solved the equation nine minus 25๐‘ฅ squared is equal to zero. And we got ๐‘ฅ equals three-fifths and ๐‘ฅ equals negative three-fifths. The graph of ๐‘ฆ equals nine minus 25๐‘ฅ squared is a parabola. Itโ€™s an inverted parabola since the coefficient of ๐‘ฅ squared is negative. And it has ๐‘ฅ intercepts at three-fifths and negative three-fifths.

Now, the part of the curve that is greater than zero and therefore nine minus 25๐‘ฅ squared is greater than zero is the bit that Iโ€™ve highlighted pink. These are all values of ๐‘ฅ in the open interval negative three-fifths to three-fifths. In other words, values of ๐‘ฅ greater than negative three-fifths and less than three-fifths. Now, to find the domain of our function, weโ€™re looking for the intersection of these two results. Now, in fact, values of ๐‘ฅ in the open interval negative three-fifths to three-fifths are also not equal to positive or negative three-fifths. And so, the intersection of our results and therefore the domain of our function is the open interval from negative three-fifths to three-fifths.

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