global max and global min

[Ryan Rigdon]

my problem

Find, if possible, the (global) man and min values of the given function on the indicated interval.

f(x) = 2 sin^2 (x) on [0,2]


My Work:

f'(x) = 4*sinx * cosx

4(sinx)(cosx) = 0

When x = (kpie)/2 or ((2k-1)pie)/4 where k is an integer

Critical Points 0, pie/4, pie/2, 2

To find max and min plug critical points into f(x)

f(0) = 0 f(pie/4) = 1 f(pie/2) = 2 f(2) = 1.6536

Min Value f(0) = 0 Max Value f(pie/2) = 2


Does my work look right?

 

[galactus]

When x = (kpie)/2 or ((2k-1)pie)/4 where k is an integer

Critical Points 0, pie/4, pie/2, 2

Uh oh!!!. You were doing so well, then you had to go and do this. :roll:

 

[Ryan Rigdon]

then how do i fix it?

 

[Deleted member 4993]

then how do i fix it?

You DO understand what Galactus is complainining about - right??!!

He is saying "pi" is a number almost equal to 3.14159 and "pie"s are round (not square) and are to be eaten!!!

You can type ? - as <ALT>227 on your number key-pad (for conventional MS-DOS keyboards)

 

[Deleted member 4993]

my problem

Find, if possible, the (global) man and min values of the given function on the indicated interval.

f(x) = 2 sin^2 (x) on [0,2]


My Work:

f'(x) = 4*sinx * cosx

4(sinx)(cosx) = 0

When x = (k?)/2 or ((2k-1)?)/4 where k is an integer

Critical Points 0, ?/4, ?/2, 2

To find max and min plug critical points into f(x)

f(0) = 0 f(?/4) = 1 f(?/2) = 2 f(2) = 1.6536

Min Value f(0) = 0 Max Value f(?/2) = 2


Does my work look right?

Check your work by plotting it in your graphing calculator.

 

[Ryan Rigdon]

when i look at the the lowest point starts at 0,0 and the highest point is roughly 1.6,2

so i believe that Min Value f(0) = 0 Max Value f(?/2) = 2 is a good answer


i didnt know about the ,<alt>227 thing for pie. thank you.

 

[galactus]

i didnt know about the ,<alt>227 thing for pie. thank you.

You done it again!.

Bad Ryan. I know you're smarter than this.

Just for that you have to differentiate this:

\(\displaystyle y=\frac{sin(x)cos(x)tan^{3}(x)}{\sqrt{x}}\)

 

[Ryan Rigdon]

i didnt know about the ,<alt>227 thing for pie. thank you.

You done it again!.

Bad Ryan. I know you're smarter than this.

Just for that you have to differentiate this:

\(\displaystyle y=\frac{sin(x)cos(x)tan^{3}(x)}{\sqrt{x}}\)

Will do this problem galactus as soon as I can. I am preparing for a Calculus midterm that I have on Friday. I will try to get it up tmw. By the way i dont have a problem anymore with ?.

 

[galactus]

I was just ribbing you. You still do not know why?. You spelled \(\displaystyle \pi\) as 'pie'.

It's spelled 'Pi'. It's not a pastry, it's the 16th letter of the Greek alphabet used to represent various things in mathematics.