Question Video: Continuity of Trigonometric Functions | Nagwa Question Video: Continuity of Trigonometric Functions | Nagwa

# Question Video: Continuity of Trigonometric Functions

Find the set on which π(π₯) = (5π₯ β 5) cos 5π₯ is continuous. [A] π(π₯) is continuous on β because π₯ β¦ (5π₯ β 5) is a polynomial, and π₯ β¦ (cos 5π₯) is continuous on β. [B] π(π₯) is continuous on β β {5}. [C] π(π₯) is continuous on β because π₯ β¦ (cos 5π₯) is continuous on β. [D] π(π₯) is continuous on β because π₯ β¦ (5π₯ β 5) is a polynomial.

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### Video Transcript

Find the set on which the function π of π₯ is equal to five π₯ minus five multiplied by the cos of five π₯ is continuous. A) π of π₯ is continuous on the set of real numbers because the map from π₯ to five π₯ minus five is a polynomial and the map from π₯ to the cos of five π₯ is continuous on the set of real numbers. Option B) π of π₯ is continuous on the set of real numbers excluding five. Option C) π of π₯ is continuous on the set of real numbers because the map of π₯ to the cos of five π₯ is continuous on the set of real numbers. And option D) π of π₯ is continuous on the set of real numbers because the map from π₯ to five π₯ minus five is a polynomial.

We can start by recalling if two functions π of π₯ and β of π₯ are continuous on their domains, then the function π of π₯ multiplied by β of π₯ is continuous on the domain of π multiplied by β. We can see that the function π of π₯ given to us in the question is equal to the product of two functions. Itβs the product of five π₯ minus five and the cos of five π₯. In particular, we can see that our function π of π₯ is the product of a polynomial and a trigonometric function.

Using this rule, we can use the continuity of our functions π of π₯ and β of π₯ individually to deduce the continuity of their product. First, we can recall that every polynomial is continuous on the entire set of real numbers. In particular, five π₯ minus five is a polynomial. So, we can conclude that five π₯ minus five is continuous on the set of real numbers because it is a polynomial.

Next, we recall that every trigonometric function is continuous on its domain. In particular, this means that we know that the cos of five π₯ is continuous on its domain. So, we need to find the domain of the cos of five π₯. Well, we know that for any real number π₯, the cos of π₯ is defined. And we can use this to argue that the cos of five π₯ is defined for any real number π₯.

Therefore, the set of reals is the domain of the function cos of five π₯. And we can conclude that the cos of five π₯ is continuous on the set of real numbers. Therefore, what we have shown is that the function π of π₯ is the product of two functions which are both continuous on the real numbers.

Therefore, we can use this rule for continuity to conclude that π of π₯ must be continuous on the set of real numbers. Therefore, what we have shown is that the function π of π₯ is continuous on the real numbers because the map from π₯ to five π₯ minus five is a polynomial and the map from π₯ to cos of five π₯ is continuous on the real numbers.

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