Explain why the equation cos x = x has at least one solution.

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Welcome back, everyone. Consider the function F X equals sin x minus 2 X. Using the intermediate value theorem, which of the following statements correctly justifies that the equation sine x equals 2 X has at least one solution? A says F X equals sin x minus 2X is continuous for all X and their existing interval 0 to 0.5 inclusive. Where F of 0 and F of 0.5 have opposite signs ensuring a route in 0 or in parentheses 0 to 0.5 closed parenthesis. B says the function FX is differentiable and since the derivative of FFX changes sign, the equation must have a solution. C says FFX is always positive for all real numbers, so there is no solution. And D says since sin X is between -1 and 1 and 2 X can take any value, then there is no solution. Now, how can we use the intermediate value theorem to find which statement correctly justifies that the equation sine x equals 2 X has at least one solution? Well, we know that F of X equals sin x minus 2 X. So a solution to sin x equals 2 X is equivalent to finding a root of FF X. In other words, A solution we can find a solution by setting FFX to 0, OK? Now in terms of continuity, the function F of X is composed of two continuous functions sine X, which is continuous everywhere, and 2 X, which is also continuous everywhere. Since the difference of two continuous functions is continuous, then. That means FX is continuous for all values of X, OK? Because both of the parts of FFX are continuous. So now, we need to find an interval where FFX changes sign. So let's do that by evaluating FF 0. Now evaluating above 0, OK. Then F of 0 equals the sin of 0 minus 2 multiplied by 0, which equals 0. I know if this suggests a solution since F of 0 equals 0, but to apply IVT we need to find values A and B, such that FFA and F of B have opposite sides. So we already have one of those values, F of A. We could say that A is 0, OK? So we need to find another value. F of B, OK, to confirm the existence of a root in an interval. We're checking another point. So let's evaluate F of 0.5. Now, evaluating F of 0.5 is gonna be equal to the sign of 0.5 minus 2 multiplied by 0.5, and using approximate values, it's going to be approximately 0.4794. minus 2 multiplied by 0.5, we know that value is 1. And that is gonna be equal to -0.5 to 06. So no, since. F of 0 equals 0 and F of 0.5 is less than 0, OK? There is a sign change. There is a sign change. Between 0 and 0.5, OK? In other words, if we were to put that on a graph, that sign change helps us to know that it is a root or there is a root. So since F of X is continuous and F of 0 equals 0, which is already a root, or there exists an interval 0 to 0.5, where FFX changes sign, IVT guarantees that there's at least one solution to FFX equals 0 in this interval. Therefore, by the intermediate value theorem, the equation sine x equals 2 X has at least one real solution in the interval, 0 to 0.5 inclusive. Thus, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Explain why the equation cos x = x has at least one solution.

Key Concepts

Intermediate Value Theorem

Continuity of Functions

Behavior of the Functions

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