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

For 0 < 𝑥 < 2𝜋/5, find the intervals on which 𝑓(𝑥) = cos² 5𝑥 + 3 cos 5𝑥 is increasing or decreasing.

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### Video Transcript

For zero is less than 𝑥 which is less than two 𝜋 by five, find the intervals on which 𝑓 of 𝑥 equals cos squared five 𝑥 plus three cos five 𝑥 is increasing or decreasing.

We recall, first of all, that a function is increasing whenever its first derivative 𝑓 prime of 𝑥 is greater than zero. And that same function is decreasing whenever its first derivative 𝑓 prime of 𝑥 is less than zero. We therefore need to find an expression for the first derivative 𝑓 prime of 𝑥 of this trigonometric function. And we recall, first of all, a standard result for differentiating cos of 𝑎𝑥, which is that its derivative with respect to 𝑥 is equal to negative 𝑎 sin 𝑎𝑥. This allows us to differentiate the second term. The derivative of three cos five 𝑥 is three multiplied by negative five sin five 𝑥. But what about the first term?

Well, we can think of it as cos of five 𝑥 all squared and then recall the general power rule. This tells us that if we have some function to a power, then its derivative is equal to that power, so that’s two, multiplied by the derivative of the function itself, so that will be negative five sin five 𝑥, multiplied by that function to one less power. So, we reduce the power from two to one.

We therefore have 𝑓 prime of 𝑥 is equal to two multiplied by negative five sin five 𝑥 cos five 𝑥 plus three multiplied by negative five sin five 𝑥. We can factor by negative five sin five 𝑥 to give 𝑓 prime of 𝑥 is equal to negative five sin five 𝑥 multiplied by two cos five 𝑥 plus three. Our Function 𝑓 will, therefore, be increasing when this first derivative is greater than zero. Now, let’s think about how we can solve this inequality. And we’ll think about that second bracket first of all.

The graph of cos five 𝑥, first of all, is just a horizontal stretch of the graph of cos 𝑥. And so, it still has negative one as its minimum value and one as its maximum value. The graph of two cos five 𝑥 is a vertical stretch of this graph by a scale factor of two. And so, this will have negative two as its minimum and two as its maximum. Adding three is a vertical translation of this graph, which means that the minimum value for two cos five 𝑥 plus three will be one, and the maximum value will be five.

What this tells us is that two cos five 𝑥 plus three itself is always greater than zero, as its minimum value is one. And therefore, one of the factors in our product is always positive. In order for the product of two factors to be positive, they must have the same sign. And therefore, it must also be the case that 𝑓 is increasing when the first factor, negative five sin five 𝑥, is itself positive. So, our problem has reduced somewhat. We’re now just looking for the region over which negative five sin five 𝑥 is greater than zero.

We can simplify by dividing both sides by negative five. And as we’re dividing by a negative, we must reverse the inequality, to give sin five 𝑥 is less than zero. Now, remember the domain we were given for this function was zero is less than 𝑥 is less than two 𝜋 by five. If we let 𝑢 equal five 𝑥, then if 𝑥 is between zero and two 𝜋 by five, 𝑢 will be between zero and two 𝜋. So, now, we’re just looking for where sin 𝑢 is less than zero for values of 𝑢 between zero and two 𝜋.

We can answer this by sketching a graph of 𝑢 against sin 𝑢 for values of 𝑢 between zero and two 𝜋. And we see that sin 𝑢 is less than zero for values of 𝑢 between 𝜋 and two 𝜋. Remember though that 𝑢 is equal to five 𝑥, so to convert this back to an inequality in 𝑥, we need to divide by five, giving 𝜋 over five is less than 𝑥 is less than two 𝜋 over five. This is the interval on which the function 𝑓 is increasing.

By applying the same logic, we see that 𝑓 will be decreasing when its first derivative is less than zero, which in turn leads to sin 𝑢 being greater than zero. That’s when 𝑢 is between zero and 𝜋, which leads to 𝑥 being between zero and 𝜋 by five. So, we’ve completed the problem. The function 𝑓 of 𝑥 is increasing on the open interval 𝜋 by five, two 𝜋 by five and decreasing on the open interval zero, 𝜋 by five.