Video: Finding the Domain of a Given Vector-Valued Function

Find the domain of the vector-valued function π‘Ÿ(𝑑) = (2𝑑²) 𝑖 + (βˆšπ‘‘ βˆ’ 1) 𝑗 + (5/(2𝑑 + 4)) π‘˜.

02:16

Video Transcript

Find the domain of the vector-valued function π‘Ÿ of 𝑑 equals two 𝑑 squared 𝑖 plus root 𝑑 minus one 𝑗 plus five over two 𝑑 plus four π‘˜.

We’re looking to find the domain of our vector-valued function. Now, each component function in our vector-valued function will have its own domain, the set of values it can take. The domain of our vector-valued function π‘Ÿ will be the intersection of these three domains. So our job is to begin by identifying the domain of each component function. Let’s begin by finding the domain of the horizontal component two 𝑑 squared.

This is a polynomial. Now, we know that the domain of a polynomial is simply the set of all real numbers. So the domain of this component function is indeed the set of real numbers. And what about the domain of our 𝑗 component, the domain of the square root of 𝑑 minus one? Well, we’re interested in the values the square root of 𝑑 minus one can take. And we know that the square root of a negative number is not a real number. This means we need the expression inside our square root 𝑑 minus one to be either equal to zero or greater than zero. Solving for 𝑑 by adding one to both sides and we obtain that 𝑑 must be greater than or equal to one. And we found the domain of this component function.

Our final component function is five over two 𝑑 plus four. We recall that the domain of the quotient of two functions is equal to the intersection of the domain of each, but where the denominator is not equal to zero. Our numerator and denominator are polynomials. So the domain is all real numbers. But we must ensure that two 𝑑 plus four cannot be equal to zero. By subtracting four from both sides, we see that two 𝑑 cannot be equal to negative four. And then dividing by two, we find that 𝑑 cannot be equal to negative two.

So now, we have the domain of each of our component functions. Remember, the domain of our vector-valued function is the intersection of these. So it’s real numbers greater than or equal to one, but not including negative two. Of course, negative two is less than one. So our domain is simply real numbers greater than or equal to one. And we can write this using interval notation, as shown.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.