# Finding Critical Points Part III

#### Hckyplayer8

##### Full Member
a) Find the critical points for f (x) = x + cos x where 0 GTE x LTE 2pi
b) Determine the intervals where f is increasing and decreasing
c) Classify as a local max, min or neither
d) Determine the intervals where f is concave up or down
e) Determine points of inflection

a) f ' (x) = 1 - sin x
Setting that to zero and quick double check on the unit circle shows sin x = 0 at x = pi over 2

This is probably obvious to the math savvy, but is there a technique for selecting the x's to test f ' at for these unit circle questions? For example, sin pi/2 = 90 degrees. But Calculus likes to take those very, very, very small intervals and the next angle in the unit circle is +/- 30 degrees...

#### Dr.Peterson

##### Elite Member
You don't really need to use specific numbers to test; you can just recognize that sin(x) is never greater than 1, so f'(x) will be greater than 0 on both sides of the critical point.

On the other hand, it will never change sign except at a zero (since it's continuous), so you can pick any number you like, such as 60 and 120 degrees. There's no need for "deltas"; this part of the work is really just algebra, not calculus.

Or you could use your calculator and test 89 and 91 degrees. Trig functions exist anywhere on the unit circle, not just at the nice, memorable points!

#### Hckyplayer8

##### Full Member
You don't really need to use specific numbers to test; you can just recognize that sin(x) is never greater than 1, so f'(x) will be greater than 0 on both sides of the critical point.

On the other hand, it will never change sign except at a zero (since it's continuous), so you can pick any number you like, such as 60 and 120 degrees. There's no need for "deltas"; this part of the work is really just algebra, not calculus.

Or you could use your calculator and test 89 and 91 degrees. Trig functions exist anywhere on the unit circle, not just at the nice, memorable points!
The calculator way isn't an option because I won't be allowed to use it on the test. Your first point makes sense, so thank you!

Moving on.

Because x always sits between 0 and 1, the slope will always be positive. Thus the function will never be decreasing. f is increasing [-infinity,pi/2) and (pi/2,pos infinity]

Because the function never decreases, both local max and mins are ruled out.

Now for concavity. Here is where I perform the second derivative test. So f '' (x) = - cos x

Now what? Just pick x values an plug them in?

#### Dr.Peterson

##### Elite Member
Because x always sits between 0 and 1, the slope will always be positive. Thus the function will never be decreasing. f is increasing [-infinity,pi/2) and (pi/2,pos infinity]

Because the function never decreases, both local max and mins are ruled out.

Now for concavity. Here is where I perform the second derivative test. So f '' (x) = - cos x

Now what? Just pick x values an plug them in?
I don't think you're saying what you mean: x is not between 0 and 1, but between 0 and 2 pi; and f'(x) is between 0 and 2. But, yes, f is never decreasing, and there is no local max or min.

But f is not increasing everywhere but at pi/2; it is increasing everywhere in the given domain except pi/2.

For concavity, can you solve the equation -cos(x) > 0? You can just think of the graph of the cosine, or of the unit-circle definition.

• Hckyplayer8

#### Hckyplayer8

##### Full Member
I don't think you're saying what you mean: x is not between 0 and 1, but between 0 and 2 pi; and f'(x) is between 0 and 2. But, yes, f is never decreasing, and there is no local max or min.

But f is not increasing everywhere but at pi/2; it is increasing everywhere in the given domain except pi/2.

For concavity, can you solve the equation -cos(x) > 0? You can just think of the graph of the cosine, or of the unit-circle definition.
Indeed. That is what I meant. Thank you for correcting that.

#### Hckyplayer8

##### Full Member
To finish this problem off...

I tested f '' (0), f ''(pi/2) and f '' (2pi).

f '' (0) = -1
f '' (pi/2) = 0
f '' (2pi)= -1

Thus because f '' is less than 0 the graph is concave downward from [0,pi/2) and (pi/2,2pi]. Because I have positive gradients on either side of the zero gradient, the zero gradient of pi/2 is the inflection point.

Is that correct?

#### Dr.Peterson

##### Elite Member
Not quite, but close. Actually, f"(x) is positive from pi/2 to 3pi/2, and zero at pi/2 and 3pi/2. So it's concave down, then up, then down.

But, yes, there is a horizontal inflection point at pi/2. There is another inflection point at 3pi/2.