How do u find the Maximum and the period of y=3sin1/3x?       And what is the maximum and minimum of y=tan2x?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

First you need to find the first derivative (y')of the function.

Then the zeros of y' will be the extreme points of y. 

First let us determine y':

y= 3sin(1/3)x

y'= 3(1/3)cos(1/3)x= cos(1/3)x

Now let us find the zeros,

cos(1/3)x=0

==> (1/3)x = n*pi/2 

==> x= 3n*pi/2    (n=1,2,3...)

Then the function has unlimited extreme values at x=3n*pi/2

Now to determine if these values are maximum or minimum:

y''= -(1/3)sin(1/3)x 

Let us analyze the values of y'':

y''=-(1/3) sin (1/3)3n*pi/2)

  = -(1/3)sin(n*pi/2)  

Let us substitute with n values:

n=1 ==> y''= -(1/3)sin(pi/2)= -1/3 <0 ==> Maximum

n=2 ==> y''= -(1/3)sin(pi) = 0

n=3 ==> y''= -(1/3)sin(3pi/2)= 1/3 >0  ==> minimum

n=4 ==> y''= -(1/3)sin(2pi)= 0

n=5 ==> y''=-(1.3)sin(5pi/2)= -1/3 <0 ==> maximum

Then we conclude that :

y has unlimited maximum values at x= 3(n)*pi/2 (n= 1,5,9,13...)

y has unlimited minimum values at x=3n*pi/2 (n=3,7,11,14...)

 

 

 

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

y = 3sin(1/3)x is aperiodic function

y = 3sin(x/3) = 3sin(x+T)/3. The smallest value for T is 6pi after which y =3sin (1/3)x repeats. So T is the period. And y = 3sin(1/3)x is a periodic function

There is no maximum or minimum period.

However there is maximum and minimum value of y.

y = 3sin(1/3)x is maximum , when sin(1/3)x = sin pi/2. Or (1/3)x = pi/2. Or x = 3pi/2 = 1.5 radian or x = 270 degree.So the maximum value = y = 3sin(1/3(270 deg) = 3.

y = 3sin(1/3)x minimum when sin(1/3)x = sin (3pi/2) . Or x = 3*3pi/2 = 4.5Pi = 810 degree.(Two fullrotation and then a 90 degree.). So min y = -3.

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