Roots (Zeros)

Show that the ...

Question

Roots (Zeros)

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

r(θ) = 2θ − cos²θ + √2, (−∞, ∞)

Accepted Answer

Determine if the function F of X equals 3X minus sin of x plus 5 has exactly one root in the interval from negative affinity to infinity. I've noted down the intermediate value theorem because we will use it in just a moment. We'll get back to that. Now, to solve this, let's first look at the end behavior of our function F of X. We have F of X equals 3 X minus sine of X plus 5. Let's check the limit as X approaches infinity and negative infinity. Now, we have 3 multiplied by infinity or just infinity. Minus sign of infinity. Plus 5 Now the sign term Oscillates Between -1 and positive 1. Which means that the 3 x portion of our function is dominating the sign portion. This limit, as exposes affinity, will eventually Go to infinity, due to the linear function always increasing. This is similar, or The limit going to negative affinity. Since We have 3 X This goes to negative infinity. It will always decrease. Sign has the same behavior. Which means our limit as X approaches negative infinity, will be negative infinity. Now we can check the intermediate value theorem. This states that there must be at least one route on the interval, from negative infinity to infinity in this case, if F of X changes signs. In this case, we noticed that we had the limit going to negative infinity. And then changing and going to positive infinity. This tells us that we do change signs, which means we can use the intermediate value theorem. Now, we can check if there's exactly one route. So we will check. Or exactly One roots. To do this, we will check if the function is strictly increasing or decreasing. We can do this by looking at the derivative. F of X. Now we can compute the derivative. To give us 3 Minus Cosine of X. Now, let's go ahead and look at our function. We know that cosine of X oscillates between -1 and 1 similar to sine. So, for our derivative, we can have 3 plus 1, which will be 4, or 3 minus 1, which will just be 2. Our derivative of X is between. 2 And We can then see that our derivative is always positive, which means our function F of X. As strictly Increasing Because our function is strictly increasing, we can then confirm that this has exactly one root. On the interval OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Roots (Zeros)

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

r(θ) = 2θ − cos²θ + √2, (−∞, ∞)

Key Concepts

Intermediate Value Theorem

Continuity of Functions

Derivative and Monotonicity

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