Roots (Zeros)

Show that the ...

Question

Roots (Zeros)

Show that the functions in Exercises 19–26 have exactly one zero in the given interval.

g(t) = 1/(1 − t) + √(1 + t) − 3.1, (−1, 1)

Accepted Answer

Welcome back, everyone. In this problem, we want to test whether the the statement that the function H of T equals 2 divided by 2 minus T plus the square root of 2 + T minus 3.5 has 10 in the interval from -2 to 1 is true or false. Now how can we know if this statement is true? Well, if the function has exactly 10 in the interval, it's going to meet three conditions. First, it's going to be continuous in the domain. Second, it will change sign from one end of the interval to another. And third, the zero will be unique. That is, the first derivative will either be positive, will either be strictly increasing, so that means it's gonna be positive all the time, or it will be strictly decreasing. It will be negative all the time. So if we test those three criteria and they are true, then this statement is true. So let's start with the first part of that statement. Let's first check if our function is continuous. How can we know if a function is this function is continuous? Well, let's look at its natural domain, OK? Now, notice. That the square root, we have the square root of 2 + T in our function, OK? And for this to be true. So let's say 4 for the square root of 2 + T to be true, then 2 + T would have to be greater than or equal to 0, which implies that t would have to be greater than or equal to -2. But remember our interval is from -2 to 1, so this this condition is satisfied. Next, notice that. We also have 2 divided by 2 minus T in our function and thus we know that this is defined, OK, but since it's a quotient, we know that this function is defined when T is not equal to 2. So if T was equal to 2 within our interval, then there would be a hole at that point. However, our interval is from -2 to 1. It doesn't include 2. Therefore, the function is continuous, OK? So we've satisfied the first condition. Do you remember the 2nd? Well, we need to check if the function changes sign, OK. So does the function. In other words, let's say, does HFT change sign? Across endpoints. Because if it changes and it goes from positive to negative, then we know it crosses the X axis and thus the function has a zero. How can we test that? Well, let's look at what's happening at the end points. At T equals -2, then H of -2 is gonna be equal to 2 divided by 2 minus -2. Plus the square root of 2 + -2 minus 3.5, OK? And now, if we think about it here, this when we solve, it should give us -3. At the other end point, that is at T equals 1, then age of 1 is gonna be equal to 2 divided by 2 minus 1. OK, plus. The square root of 2 + 1. -3.5. That's gonna give us 2 + route 3 minus 3.5, OK. And when we work that out, then we should get it to be approximately equal to 0.23205. Now what do you notice about the signs of our results? We'll notice that H of -2 is negative. It's less than 0, while H1 is positive. It's greater than 0. So therefore that means that the sign or signs change between the end points and now by the intermediate value theorem. OK, based on what we said earlier, there is at least 10. At least 10. In 21. So we know that our function is continuous. We know that it has a 0 within the interval. Now we need to check the uniqueness of the 0. So let's try to find our derivative. OK. So let's, the question I'm asking here is the 0, is this 0 that we know exists within the interval unique? To figure that out, let's differentiate our function and then see if it is strictly increasing on our interval or strictly decreasing. Now, for a function here, its derivative will be equal to the derivative of 2 divided by 2 minus T, OK? Which, if we apply the poor rule here, that's gonna be 2 divided by 2 minus T squared, OK? And no For the square root of 2 + t, if we apply the chain rule of differentiation, its derivative will be 1 divided by 2 multiplied by root 2 + T. Now, what do you notice here? Well, notice that both terms are positive and negative to 1. So this means that H of T, OK, is strictly increasing on or interval. And if it's strictly increasing on the interval, that means the 0 is unique. So if we think back to our three criteria, first, we know the function is continuous. Next, we know there is at least 10 in 21. We've confirmed that. And by checking the derivative, we know that the the function is strictly increasing on the interval, so the zero is unique. Therefore, the statement is true. That is, the function H of T has exactly 10 in the interval negative to 1. Thanks a lot for watching everyone. I hope this video helped.

Roots (Zeros)

Show that the functions in Exercises 19–26 have exactly one zero in the given interval.

g(t) = 1/(1 − t) + √(1 + t) − 3.1, (−1, 1)

Key Concepts

Intermediate Value Theorem

Continuity of Functions

Behavior of Rational and Radical Functions

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