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

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

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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...)

neela | High School Teacher | (Level 3) Valedictorian

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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.