Video: Finding the Intervals of Increasing and Decreasing of a Function Involving a Trigonometric Function

Determine the intervals on which the function 𝑓(π‘₯) = 2π‘₯ βˆ’ sin π‘₯, where 0 ≀ π‘₯ ≀ 4πœ‹ , is increasing and where it is decreasing.

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Video Transcript

Determine the intervals on which the function 𝑓 of π‘₯ equals two π‘₯ minus sin π‘₯ where π‘₯ is greater than or equal to zero and less than or equal to four πœ‹ is increasing and where it is decreasing.

Here, we are given the function 𝑓 of π‘₯ equals two π‘₯ minus sin π‘₯ and told to consider it only on the set of π‘₯-values between zero and four πœ‹, with zero and four πœ‹ included. Within the set of π‘₯-values from zero to four πœ‹, we are asked to find the intervals on which the function 𝑓 is increasing or decreasing.

Recall that a function 𝑓 is increasing on an interval 𝐼 if its first derivative, 𝑓 prime of π‘₯, is positive for all values of π‘₯ in 𝐼. A function 𝑓 is decreasing on an interval 𝐼 if it’s first derivative is negative for all values of π‘₯ in 𝐼. In other words, a function is increasing on an interval 𝐼 if its gradient function, or slope, is positive on 𝐼 and decreasing on an interval 𝐼 if it’s slope is negative on 𝐼.

Let’s determine the gradient function, or first derivative, of the function given to us in the question. In order to do this, we will differentiate the functions two π‘₯ and sin π‘₯ and minus their derivatives, in that order. Their derivative of two π‘₯ is two. Here, we have just used the standard formula for differentiating terms of the form π‘Žπ‘₯ to the power of 𝑛, where we multiply the coefficient π‘Ž by the exponent 𝑛 and decrease the exponent by one.

In our case, the coefficient is two. And the exponent is one. The derivative of sin π‘₯ is cos π‘₯. This is just a standard derivative that we should memorize. Having found the first derivative of 𝑓, we need to find the intervals of π‘₯-values within zero and four πœ‹ for which it is positive, i.e. 𝑓 is increasing and, for which it is negative, i.e. 𝑓 is decreasing.

Note that the derivative of a function is not defined at the end points of its domain, as we cannot draw a unique tangent at those points. So, the derivative of the function 𝑓 in the question is not defined at zero and four πœ‹, assuming the domain of π‘₯ starts at zero and ends at four πœ‹. Therefore, we cannot compare the derivative of 𝑓 to the value zero at the π‘₯-values zero and four πœ‹. And so, we cannot include zero and four πœ‹ in the intervals of π‘₯-values for which 𝑓 may be increasing or decreasing. We find that the derivative of 𝑓 is positive for those π‘₯-values between zero and four πœ‹ for which two is greater that cos π‘₯ and negative for those π‘₯-values between zero and four πœ‹ for which two is smaller than cos π‘₯.

Now, recall that the function 𝑦 equals cos π‘₯ has a minimum 𝑦-value of minus one and a maximum 𝑦-value of one for any π‘₯-value and, therefore, in particular, for those π‘₯-values between zero and four πœ‹. Since two is always greater than one, we have that two is always greater than cos π‘₯ for any π‘₯-value and, in particular, for all π‘₯-values between zero and four πœ‹. Therefore, the function 𝑓 is increasing on the open interval of π‘₯-values from zero to four πœ‹.

So, we have deduced that 𝑓 is increasing on all the π‘₯-values for which its derivative is defined. And so, it must be decreasing for no values of π‘₯ for which its derivative is defined. In other words, it must be decreasing for no values of π‘₯ from zero to four πœ‹. This is indeed true, as the fact that cos π‘₯ is always less than or equal to one implies that cos π‘₯ is never greater than two. So, the answer to the question is that 𝑓 is increasing on the open interval of π‘₯-values from zero to four πœ‹.

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