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CALCULUS, EXTREME VALUES of function on interval and where they occur

  1. Nov 10, 2011 #1
    CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

    find the extreme value of function on the interval and where they occur.
    Please Help, teacher did during class but didnt get explanation.

    1. y=cos(x-(pi/4)) , at the interval of [0,7pi/4]
    My attempt: y'= -sin (x-(pi/4))
    y= pi/4, 5pi/4

    f(0)= -1/sqrt(2) f(7pi/4)= 0 f(pi/4)= 0 f(5pi/4)=-1
    Max of 1 @ x=pi/4
    Min of -1 at x=5pi/4
    teacher said f(0) was wrong i dont know why

    2. y= 4x/(x^2+1)
    My attempt: y'= -4x/(x^2+1)^2
    f(0)=0
    Max of 0 at x=0
    dont know why i got that wrong

    tnx a lot
     
    Last edited: Nov 10, 2011
  2. jcsd
  3. Nov 10, 2011 #2
    Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

    For f(0) it's -sin(-pi/4) = - (-sqrt2 / 2) = sqrt2 / 2 = 1/sqrt2
    Could you show your work for how you got f'(x) for #2?
     
  4. Nov 11, 2011 #3
    Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

    For f(0) the negitive of sin is later added onto -1/sqrt(2) right? making it positive? ok i see what i did.
    y= 4x(x^2+1)^-1
    y'=-4x(x^2+1)^-2

    oh ... shoot i was supposed to use quotient rule right?

    thanks alot i kind of get calculus a little bit more
     
    Last edited: Nov 11, 2011
  5. Nov 11, 2011 #4
    Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

    For #1, when you say f(0), you mean f(0)=cos(0-(pi/4))? Or f'(0)?

    For #2, you're supposed to use the quotient rule.
     
  6. Nov 11, 2011 #5
    Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

    yes, i mean the cos one not f'(0)
    when you are looking for critical points you plug in 0 of f(0) into the original/position equation right? i think i did that right just forgot about the pos/neg stuff
     
  7. Nov 12, 2011 #6

    HallsofIvy

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    Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

    You mean x, not y.

    f(0)= cos(-pi/4)= cos(pi/4)= 1/sqrt(2). Cosine is an even function.

    What interval was this on? The entire real line? A continuous function does not necessarily have a max or min on such an interval. Here, it should be clear that y is positive for positive x so 0 cannot be a maximum. In any case, you have the derivative wrong. Try it again.
     
  8. Nov 14, 2011 #7
    Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

    ok. i think i got #2 down.

    2. y'= 4(x^2+1) - 4x(x)
    ________________
    (x^2+1)^2

    = 4x^2+4-8x^2
    _______________
    (x^2+1)^2

    0 is only critical pt?
    f(0) = 0

    sine chart for 0
    -2 = neg and 2 = neg so..... no extreme values??
     
    Last edited: Nov 14, 2011
  9. Nov 16, 2011 #8
    Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

    Am i correct?? did i perform the steps right??? Thanks
     
  10. Nov 16, 2011 #9

    SammyS

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    Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

    How is it that x = 0 is a critical point?

    The derivative looks OK now.
     
  11. Nov 17, 2011 #10

    HallsofIvy

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    Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

    No, 0 is not a critical point. f'(0) exists and is not 0. The numerator of the derivative is [itex]4- 4x^2[/itex] which is 0 at x= 1 and x= -1. The denominator is never 0.


    You mean "sign", not "sine".
     
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