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 𝑓(𝑥) = 6/√(9 − 25𝑥²) in ℝ.

02:46

### Video Transcript

Find the domain of the function 𝑓 of 𝑥 equals six over the square root of nine minus 25𝑥 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, that’s 𝑥. We essentially say that it is the set of all possible 𝑥 values, which makes the function work and outputs real 𝑦 values. When finding the domain, we need to consider that the denominator of some fraction or rational function can’t be equal to zero. That’s because when we divide a real number by zero, we say that we get a result that’s undefined. We also know that the number underneath a square root sign cannot be negative, absolutely must be positive. So, let’s consider how this impacts on our function.

We have a rational function, so the denominator, the square root of nine minus 25𝑥 squared, cannot be equal to zero. So, let’s assume it is equal to zero and solve for 𝑥. They will tell us the values of 𝑥 that we cannot use. We begin by squaring both sides of this equation. And we get nine minus 25𝑥 squared equals zero. We then factor nine minus 25𝑥 squared, and we could use a difference of two squares. We get three minus five 𝑥 times three plus five 𝑥. Now, for the product of these two expressions to itself be equal to zero, either three minus five 𝑥 must be zero or three plus five 𝑥 must be zero. Let’s solve the first equation for 𝑥. We’re gonna add five 𝑥 to both sides. And we get three equals five 𝑥. We then divide through by five, and we get 𝑥 equals three-fifths.

To solve the second equation, we subtract three from both sides to get five 𝑥 equals negative three. We then divide through by five, and we found the second solution to the equation. It’s 𝑥 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 𝑥 cannot be equal to either positive or negative three-fifths.

We’ll now consider the second part, the number inside the square root must itself be positive. That is, nine minus 25𝑥 squared is greater than zero. Now, we just solved the equation nine minus 25𝑥 squared is equal to zero. And we got 𝑥 equals three-fifths and 𝑥 equals negative three-fifths. The graph of 𝑦 equals nine minus 25𝑥 squared is a parabola. It’s an inverted parabola since the coefficient of 𝑥 squared is negative. And it has 𝑥 intercepts at three-fifths and negative three-fifths.

Now, the part of the curve that is greater than zero and therefore nine minus 25𝑥 squared is greater than zero is the bit that I’ve highlighted pink. These are all values of 𝑥 in the open interval negative three-fifths to three-fifths. In other words, values of 𝑥 greater than negative three-fifths and less than three-fifths. Now, to find the domain of our function, we’re looking for the intersection of these two results. Now, in fact, values of 𝑥 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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