Local max and min (including a domain)

[King_Silver]

I have this function: f(x) = √3x + 2Cos(x)
I have to find the local maxima and minima,where -π ∠ or equal to x ∠ or equal to 2π.

1: Differentiate the function given.
f'(x) = 3/(2√x) -2Sin(x)

2: Find where f'(x) is undefined or = 0 to get critical points.

3: Find f''(x) [The second differential]

4: Sub in critical values for x, if < 0 = Minima, if > 0 = Maxima.

Question 1:What does this domain mean? and how will it affect my answer(s)?
Question 2:Is there any easier way to find where f'(x) is undefined or 0 other than subbing in random values until you get ones that work? I always get stuck at this part despite knowing everything else, without the correct critical values it is pointless to continue with a question.

 

[PeroK]

I have this function: f(x) = √3x + 2Cos(x)

Is that $\sqrt{3}x$ or $\sqrt{3x}$ in your function?

 

[King_Silver]

Is that $\sqrt{3}x$ or $\sqrt{3x}$ in your function?

$\sqrt{3x}$

 

[PeroK]

$\sqrt{3x}$

Are you sure?

 

[King_Silver]

Are you sure?

A sugar I misread! its the other one.

 

[PeroK]

A sugar I misread! its the other one.

Try that one. It might be a little easier.

 

[King_Silver]

Try that one. It might be a little easier.

f'(x) = sqrt(3) - 2Sin(x)

 

[PeroK]

f'(x) = sqrt(3) - 2Sin(x)

And where is $f'(x) = 0$?

 

[King_Silver]

And where is $f'(x) = 0$?

Couldn't tell you if my life depended on it.

 

[King_Silver]

Been trying numbers such as 0, -1,-2,-3,-4 and 1,2,3,4... etc. but they are giving decimal places. Also does "undefined" just mean like it doesn't give a solution whatsoever?

 

[PeroK]

If you re-arrange it, you get:

$sin(x) = \frac{\sqrt{3}}{2}$

If that doesn't look familiar, then you could use the inverse function $sin^{-1}$

 

[King_Silver]

If you re-arrange it, you get:

$sin(x) = \frac{\sqrt{3}}{2}$

If that doesn't look familiar, then you could use the inverse function $sin^{-1}$

Ah yea so that is = 60 so one critical point is = 60. So there is a method to find where x = 0 thanks :) !
And what about where it is undefined? :) and where does the domain come into a question like this?

 

[PeroK]

Ah yea so that is = 60 so one critical point is = 60. So there is a method to find where x = 0 thanks :) !
And what about where it is undefined? :) and where does the domain come into a question like this?

Mathematically, the trig functions are defined in terms of radians, not degrees. Also, trig functions are periodic, so $sin(x)$ will take a given value at more than one point. This is where the domain comes in. You need to find all points where $sin(x) = \frac{\sqrt{3}}{2}$ in the given domain.

You've got a nice, regular function for $f'(x)$ so that's not undefined anywhere. So, in this problem, there's no complexity like that to worry about.

I'm not sure why you're interested in where $f(x) = 0$. You don't need that to find maxima and minima.

 

[King_Silver]

Mathematically, the trig functions are defined in terms of radians, not degrees. Also, trig functions are periodic, so $sin(x)$ will take a given value at more than one point. This is where the domain comes in. You need to find all points where $sin(x) = \frac{\sqrt{3}}{2}$ in the given domain.

You've got a nice, regular function for $f'(x)$ so that's not undefined anywhere. So, in this problem, there's no complexity like that to worry about.

I'm not sure why you're interested in where $f(x) = 0$. You don't need that to find maxima and minima.

*f'(x) my bad.

So are these steps correct?

1: differentiate f(x)
2: Find where f'(x) is 0 or undefined (i.e. critical value at 60 as calculated)
3: Find f''(x)
4: sub in critical point instead of x
5: If greater than 0 it is a minimum, if less it is a maximum.

But the question stated to find the "local maxima AND minima" I will only find one of these using this method right? And will the domain affect my answers?

 

[PeroK]

*f'(x) my bad.

So are these steps correct?

1: differentiate f(x)
2: Find where f'(x) is 0 or undefined (i.e. critical value at 60 as calculated)
3: Find f''(x)
4: sub in critical point instead of x
5: If greater than 0 it is a minimum, if less it is a maximum.

But the question stated to find the "local maxima AND minima" I will only find one of these using this method right? And will the domain affect my answers?

That's the correct method. You will find all maxima and minima this way. The domain does affect your answer. If you look for any $x$, your $f(x)$ here has an infinite number of maxima and minima.

You're still working in degrees, I see!

 

[King_Silver]

That's the correct method. You will find all maxima and minima this way. The domain does affect your answer. If you look for any $x$, your $f(x)$ here has an infinite number of maxima and minima.

You're still working in degrees, I see!

*1/3pi my bad! Thanks! :D

 

[I like Serena]

Doesn't $\sin x = \frac{\sqrt 3}{2}$ have another solution when $-\pi \le x \le 2\pi$?
How about $x=\frac{2\pi}{3}$ (or $x=120^\circ$)?

 

[HallsofIvy]

Ah yea so that is = 60 so one critical point is = 60. So there is a method to find where x = 0 thanks :) !
And what about where it is undefined? :) and where does the domain come into a question like this?

No, it is NOT "= 60". It is equal to 60 degrees or \pi/3 or simply \pi/3. You can drop the "radians" in such an answer but not degrees.