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Maximum and minimum value

  • Thread starter ritwik06
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  • #1
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Homework Statement


Find the maximum and minimum value of:
(sin a)^4 - (sin a)^2


I am new to the world of calculus. I have just learned to evaluate the derivatives of functions. (*and nothing else- I am a high school student) For this function the derivative is
-sin 2a . cos 2a
Please tell me if I am right.

While teaching my teacher told me that we can obtain the maximum and minimum values of a function with the help of its derivative. Is it true?
If yes, please spare some time to tell me how is it possible. I know that the maximum and minimum values can be calculated by coverting this into a perfect square. But I wish to learn it through calculus, if possible.

Thanks!
regards,
Ritwik
 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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Hi ritwik06! :smile:
-sin 2a . cos 2a
Yes, that's fine! :smile:
While teaching my teacher told me that we can obtain the maximum and minimum values of a function with the help of its derivative. Is it true?
If yes, please spare some time to tell me how is it possible.
The derivative is the rate at which the function increases.

At a maximum or minimum, the function has stopped increasing, so the derivative is zero. :smile:
 
  • #3
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Hi ritwik06! :smile:


Yes, that's fine! :smile:


The derivative is the rate at which the function increases.

At a maximum or minimum, the function has stopped increasing, so the derivative is zero. :smile:

ok. thanks a lot tim.

but the problem now is that if I put:

- sin 2a . cos 2a=0
the values of 'a' that are possible are:
0, 180 ,360, ......
45, 135, 225.....
at which value will I get the maximum and at which value the minimum???
 
  • #4
580
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similarly if I want to fin th maximum and minimum of sin x+cos x

dy/dx=cos x - sinx

cos x - sin x=0
cos x=sin x
x=45, 225... (till what values of x should I limit myself??)
 
  • #5
580
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Do I need to check from [0,2*pi]????
Because after that. the values of the T-ratios repeat. Am I right?
 
  • #6
HallsofIvy
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The first thing you need to do is rewrite those x values in terms of radians rather than degrees!

And you had better learn to start thinking in terms of radians. For the functions sin(x) and cos(x), the "x" has to be interpreted in radians, not degrees.
 
  • #7
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Well, you will get an infinite number of local extrema, because the sin and cosine are periodic functions. That being said, half will be local maxima, and half will be local minima.

You have to use the second derivative test to determine whether each critical point represents a local minimum (second derivative > 0 at the x), a local maximum (second derivative < 0 at the x) or neither (second derivative = 0 at the x).

But yes, you will have to give your answer like...

f(x) = sin x

Maxima: 1/2 PI, 5/2 PI, 9/2 PI, ...
Minima: 3/2 PI, 7/2 PI, 11/2 PI, ...

(or)

Maxima: (4n-3)/2 PI, for all natural numbers n
Minima: (4n-1)/2 PI, for all natural numbers n

Hopefully that clarified what was being asked.
 
  • #8
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi csprof2000! Welcome to PF! :smile:

Don't too much work … best just to give a hint and see if the OP can work out the rest themselves! :wink:
 
  • #9
580
0
Well, you will get an infinite number of local extrema, because the sin and cosine are periodic functions. That being said, half will be local maxima, and half will be local minima.

You have to use the second derivative test to determine whether each critical point represents a local minimum (second derivative > 0 at the x), a local maximum (second derivative < 0 at the x) or neither (second derivative = 0 at the x).

But yes, you will have to give your answer like...

f(x) = sin x

Maxima: 1/2 PI, 5/2 PI, 9/2 PI, ...
Minima: 3/2 PI, 7/2 PI, 11/2 PI, ...

(or)

Maxima: (4n-3)/2 PI, for all natural numbers n
Minima: (4n-1)/2 PI, for all natural numbers n

Hopefully that clarified what was being asked.
Thanks A Lot!
 

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