Question

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = 7 cos2(x) − 14 sin(x), 0 ≤ x ≤ 2π (a) Find the interval on which f is increasing. (Enter your answer using interval notation.)

Answer 1

Answer:

$[\frac{\pi}{2},\frac{3\pi}{2}]$

Step-by-step explanation:

Let me first state that I am assuming your function is

$f(x)=7cos^2(x)-14sin(x)$

If this is incorrect, then disregard this whole answer/explanation.

In order to find where the function is increasing or decreasing, we need to first find the first derivative, set it equal to 0, and then factor to find the values that cause the derivative to equal 0.  This is where you expect to find a max or a min value in the function itself.  But this function is not going to be easily solved for 0 once we find the derivative unless we make it in terms of either sin or cos right now, before taking the first derivative.  

Let $cos^2(x)=1-sin^2(x)$

This is a Pythagorean trig identity, and I'm assuming that if you're in calculus solving for the intervals of increasing and decreasing values that you have, at one time, used trig identities.

Rewriting:

$f(x)=7(1-sin^2(x))-14sin(x)$ which simplifies to

$f(x)=7-7sin^2(x)-14sin(x)$ and in order of descending values of x:

$f(x)=-7sin^2(x)-14sin(x)+7$

Now we can find the derivative.  For the first term, let u = sin(x), therefore,

$f(u)=u^2$, u' = cos(x), and f'(u) = 2u.  The derivative is found by multiplying f'(u) by u', which comes out to 2sin(x)cos(x)

The derivative for the next 2 terms are simple, so the derivative of the function is

$f'(x)=-7[2sin(x)cos(x)]-14cos(x)$ which simplifies down to

$f'(x)=-14sin(x)cos(x)-14cos(x)$

We will set that equal to zero and solve for the values that cause that derivative to equal 0.  But first we can simplify it a bit.  You can factor out a -14cos(x):

$f'(x)=-14cos(x)(sin(x)+1)$

By the Zero Product Property, either

-14cos(x) = 0 or sin(x) + 1 = 0

Solving the first one for cos(x):

cos(x) = 0

Solving the second one for sin(x):

sin(x) = -1

We now look to the unit circle to see where, exactly the cos(x) = 0.  Those values are

$\frac{\pi}{2},\frac{3\pi}{2}$

The value where the sin is -1 is found at

$\frac{3\pi}{2}$

We set up a table (at least that's what I advise my students to do!), separating the intervals in ascending order, starting at 0 and ending at 2pi.

Those intervals are

0 < x < $\frac{\pi}{2}$, $\frac{\pi}{2}$, and $\frac{3\pi}{2}$

Now pick a value that falls within each interval and evaluate the derivative at that value and determine the sign (+ or -) that results.  You don't care what the value is, only the sign that it carries.  For the first interval I chose

$f'(\frac{\pi}{4})=-$ so the function is decreasing here (not what you wanted, so let's move on to the next interval).

For the next interval I chose:

$f'(\pi)=+$ so the function is increasing here.

For the last interval I chose:

$f'(\frac{7\pi}{4})=-$

It appears that the only place this function is increasing is on the interval

$[\frac{\pi}{2},\frac{3\pi}{2}]$