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Math Help - Derivative help

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    Derivative help

    I am having issues with this one problem,

    Let f be the function defined for \frac{\pi}{6} less than or equal to x less than or equal to \frac{5\pi}{6} by f(x)=x+sin^2x

    (a) Find all values of x for which f''(x)=1
    (b) Find the x-coordinates of all minimum points of f. Justify your answer.
    (c) Find the x-coordinates of all inflection points of f. Justify your answer.

    Thanks for all your help!
    Last edited by OnMyWayToBeAMathProffesor; January 1st 2009 at 04:49 PM.
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  2. #2
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    I am having issues with this one problem,

    Let f be the function defined for \frac{pi}{6} less than or equal to x less than or equal to \frac{5pi}{6} by f(x)=x+sin^2x

    (a) Find all values of x for which f''(x)=1
    (b) Find the x-coordinates of all minimum points of f. Justify your answer.
    (c) Find the x-coordinates of all inflection points of f. Justify your answer.

    Thanks for all your help!
    Can you differentiate f(x)=x+sin^2x to get f'(x)? Then, can you differentiate f'(x) to get f''(x)? That's your starting point .....
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  3. #3
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    I am having issues with this one problem,

    Let f be the function defined for \frac{pi}{6} less than or equal to x less than or equal to \frac{5pi}{6} by f(x)=x+sin^2x

    (a) Find all values of x for which f''(x)=1
    (b) Find the x-coordinates of all minimum points of f. Justify your answer.
    (c) Find the x-coordinates of all inflection points of f. Justify your answer.

    Thanks for all your help!
    f(x) = x + \sin^2 x\;\;for \;\;\frac{\pi}{6}\le x \le \frac{6\pi}{6}

    (A) can you find derivate of f(x) = 1+\sin^2 x

    use product rule to find second derivative


    Now finish it

    (C) for max and mim points, put f'(x) = 0 and find all x

    then plug these values in f(x) to find all values of f(x) . FINISH IT

    (B) out of the points find out the points where f(x) is minimum.

    FINISH IT
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  4. #4
    Member Last_Singularity's Avatar
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    Just a note: when you differentiate the function once and set f'(x) = 0 and solve for x, you still need to test for the value of the second derivative because you don't know whether that point is a max, min, or neither.
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  5. #5
    Member OnMyWayToBeAMathProffesor's Avatar
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    I am sorry for the late response.

    so when i first differentiate f(x)=x+sin^2x i get f'(x)=1+2\sin\cos which is the same as f'(x)=1+\sin\cos+\sin\cos then when i differentiate again i get f''(x)=2cos^2(x)-2sin^2(x) which equals 1 x=\frac{\pi}{6} and when \frac{5\pi}{6}. how do i justify that without saying i looked at my gragh?

    b)there is no minimum in the given interval of \frac{\pi}{6}\le x \le \frac{5\pi}{6} because when f'(x)=0 that point is a maximum on the original graph. correct?

    c) infection points occur when f''(x)=0 so that is when x=\frac{\pi}{4} and x=\frac{3\pi}{4}

    are these correct?
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  6. #6
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    I am sorry for the late response.

    so when i first differentiate f(x)=x+sin^2x i get f'(x)=1+2\sin\cos which is the same as f'(x)=1+\sin\cos+\sin\cos

    Mr F says: You ought to make life easy for yourself and recognise that 2 sinx cosx = sin(2x).

    then when i differentiate again i get f''(x)=2cos^2(x)-2sin^2(x)

    Mr F says: Which (no surprise given what I said above) is the same as 2 cos(2x).

    which equals 1 x=\frac{\pi}{6} and when \frac{5\pi}{6}. how do i justify that without saying i looked at my gragh?

    Mr F says: Solve 2 cos(2x) = 1 ( a simple trig equation for you I hope).

    b)there is no minimum in the given interval of \frac{\pi}{6}\le x \le \frac{5\pi}{6} because when f'(x)=0 that point is a maximum on the original graph. correct?

    Mr F says: There is no maximum at the value of x such that f'(x) = 0 (that is, x = 3pi/4).

    c) infection points occur when f''(x)=0 so that is when x=\frac{\pi}{4} and x=\frac{3\pi}{4}

    are these correct? Mr F says: They are the correct solutions to f''(x) = 0. The fact that they do or do not give points of inflection requires justification. Note that x = 3pi/4 is a solution to f'(x) = 0. It can't be a point of inflection AND a maximum turning point (as you say in Q2) .... The resolution of this is left for you to think about for the moment.
    ..
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  7. #7
    Member OnMyWayToBeAMathProffesor's Avatar
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    sorry for be i meant to say that at f'(\frac{3\pi}{4})=0 and f(\frac{3\pi}{4}) is a maximum on the original graph of f(x)=x+sin^2x. and yes, a point can not be a maximum and inflection. I know i did something wrong, but i can not figure it out.

    \frac{\pi}{4} is a point of inflection, that i know for sure but why \frac{\pi}{4} is and not \frac{3\pi}{4}, i do not know. help please?
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  8. #8
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    f\left(\frac{3\pi}{4}\right) is not a maximum.
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  9. #9
    Member OnMyWayToBeAMathProffesor's Avatar
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    now i am utterly confused about b and c. could someone please explain step by step? thank you so much.
    Last edited by OnMyWayToBeAMathProffesor; January 1st 2009 at 05:46 PM.
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  10. #10
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    I am sorry for the late response.

    so when i first differentiate f(x)=x+sin^2x i get f'(x)=1+2\sin\cos which is the same as f'(x)=1+\sin\cos+\sin\cos then when i differentiate again i get f''(x)=2cos^2(x)-2sin^2(x) which equals 1 x=\frac{\pi}{6} and when \frac{5\pi}{6}. how do i justify that without saying i looked at my gragh?

    b)there is no minimum in the given interval of \frac{\pi}{6}\le x \le \frac{5\pi}{6} because when f'(x)=0 that point is a maximum on the original graph. correct?

    no ... f'(x) = 0 is not the sole indicator that an extrema exists there. you need to perform the 1st or 2nd derivative test to confirm it. in this case, you have endpoint extrema.

    many students get confused by this true statement ... if f has an extreme value at x, then f'(x) = 0 or f'(x) is undefined. they want to turn it around and believe that the converse is always true ... it's not.

    c) infection points occur when f''(x)=0 so that is when x=\frac{\pi}{4} and x=\frac{3\pi}{4}

    are these correct?

    yes, but you need to justify that claim. an inflection point occurs where f''(x) = 0 and f''(x) changes sign.
    help any?
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  11. #11
    Member OnMyWayToBeAMathProffesor's Avatar
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    thank you so much, if i could click the thank you button more than once, i would. i have been trying so many different things. so the min is \frac{\pi}{6} because that is the endpoint and in a closed interval you have to check the end points. i remember my teacher telling us this a loong time ago, i can't believe i forgot.

    but for the points of inflection i am not sure i completely understand. the f''(x) graph i am looking at, it goes from positive y values to negative y values at x=\frac{\pi}{4} and then it goes from negative y vlaues to positive y values at x=\frac{3\pi}{4}. isn't that a sign change?

    so just checking again

    b)there is a minimum at x= \frac{\pi}{6} due to the endpoint test.
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  12. #12
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    ... and there is an endpoint maximum at x = \frac{5\pi}{6}.

    there are two inflection points ... one at x = \frac{\pi}{4} and the other at x = \frac{3\pi}{4} because f''(x) = 0 at both points and f''(x) changes sign at both of those points.
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  13. #13
    Member OnMyWayToBeAMathProffesor's Avatar
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    ok, so the critical point was the inflection point. I do believe thats the end of the problem. Thank you very much for helping me along the way and thank you everyone else.
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