Video: Continuity of Trigonometric Functions

Find the set on which 𝑓(π‘₯) = (5π‘₯ βˆ’ 5) cos 5π‘₯ is continuous. [A] 𝑓(π‘₯) is continuous on ℝ because π‘₯ ↦ (5π‘₯ βˆ’ 5) is a polynomial, and π‘₯ ↦ (cos 5π‘₯) is continuous on ℝ. [B] 𝑓(π‘₯) is continuous on ℝ βˆ’ {5}. [C] 𝑓(π‘₯) is continuous on ℝ because π‘₯ ↦ (cos 5π‘₯) is continuous on ℝ. [D] 𝑓(π‘₯) is continuous on ℝ because π‘₯ ↦ (5π‘₯ βˆ’ 5) is a polynomial.

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Video Transcript

Find the set on which the function 𝑓 of π‘₯ is equal to five π‘₯ minus five multiplied by the cos of five π‘₯ is continuous. A) 𝑓 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) 𝑓 of π‘₯ is continuous on the set of real numbers excluding five. Option C) 𝑓 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) 𝑓 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 𝑔 of π‘₯ and β„Ž of π‘₯ are continuous on their domains, then the function 𝑔 of π‘₯ multiplied by β„Ž of π‘₯ is continuous on the domain of 𝑔 multiplied by β„Ž. We can see that the function 𝑓 of π‘₯ given to us in the question is equal to the product of two functions. It’s the product of five π‘₯ minus five and the cos of five π‘₯. In particular, we can see that our function 𝑓 of π‘₯ is the product of a polynomial and a trigonometric function.

Using this rule, we can use the continuity of our functions 𝑔 of π‘₯ and β„Ž 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 𝑓 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 𝑓 of π‘₯ must be continuous on the set of real numbers. Therefore, what we have shown is that the function 𝑓 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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