# Local max and min (including a domain)

1. Dec 4, 2015

### 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.

2. Dec 4, 2015

### PeroK

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

3. Dec 4, 2015

### King_Silver

$\sqrt{3x}$

4. Dec 4, 2015

### PeroK

Are you sure?

5. Dec 4, 2015

### King_Silver

A sugar I misread! its the other one.

6. Dec 4, 2015

### PeroK

Try that one. It might be a little easier.

7. Dec 4, 2015

### King_Silver

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

8. Dec 4, 2015

### PeroK

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

9. Dec 4, 2015

### King_Silver

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

10. Dec 4, 2015

### 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?

11. Dec 4, 2015

### 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 $sin^{-1}$

12. Dec 4, 2015

### King_Silver

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?

13. Dec 4, 2015

### PeroK

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.

14. Dec 4, 2015

### King_Silver

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?

15. Dec 4, 2015

### PeroK

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!

16. Dec 4, 2015

### King_Silver

17. Dec 5, 2015

### 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 $x=120^\circ$)?

18. Dec 6, 2015

### HallsofIvy

Staff Emeritus
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.