# Question Video: Finding the Intervals of Increasing and Decreasing of a Function Involving a Trigonometric Function Mathematics • Higher Education

Determine the intervals on which the function π(π₯) = 2π₯ β sin π₯, where 0 β€ π₯ β€ 4π , is increasing and where it is decreasing.

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### Video Transcript

Determine the intervals on which the function π 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 is increasing and where it is decreasing.

Here, we are given the function π 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, with zero and four π included. Within the set of π₯-values from zero to four π, we are asked to find the intervals on which the function π is increasing or decreasing.

Recall that a function π is increasing on an interval πΌ if its first derivative, π prime of π₯, is positive for all values of π₯ in πΌ. A function π is decreasing on an interval πΌ if itβs first derivative is negative for all values of π₯ in πΌ. In other words, a function is increasing on an interval πΌ if its gradient function, or slope, is positive on πΌ and decreasing on an interval πΌ if itβs slope is negative on πΌ.

Letβ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π₯ to the power of π, where we multiply the coefficient π by the exponent π 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 π, we need to find the intervals of π₯-values within zero and four π for which it is positive, i.e. π is increasing and, for which it is negative, i.e. π 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 π in the question is not defined at zero and four π, assuming the domain of π₯ starts at zero and ends at four π. Therefore, we cannot compare the derivative of π to the value zero at the π₯-values zero and four π. And so, we cannot include zero and four π in the intervals of π₯-values for which π may be increasing or decreasing. We find that the derivative of π is positive for those π₯-values between zero and four π for which two is greater that cos π₯ and negative for those π₯-values between zero and four π for which two is smaller than cos π₯.

Now, recall that the function π¦ equals cos π₯ has a minimum π¦-value of minus one and a maximum π¦-value of one for any π₯-value and, therefore, in particular, for those π₯-values between zero and four π. 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. Therefore, the function π is increasing on the open interval of π₯-values from zero to four π.

So, we have deduced that π 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. 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 π is increasing on the open interval of π₯-values from zero to four π.