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Critical Points, intervals, local max/min help Calculus.

  1. May 9, 2012 #1
    Critical Points, intervals, local max/min help!! Calculus.

    1. I need help with a homework problem that I just cannot get right. It asks: Answer the following questions about the functions whos derivative is given below.
    f'(x) = (sinx +1)(2cosx +[itex]\sqrt{3}[/itex] ), 0[itex]\leq[/itex]x[itex]\leq[/itex]2∏

    a. what are the critical points of f.
    b. on what intervals is f increasing or decreasing?
    c. What points, if any, does f assume local max/min values.



    I'm extremely confused with how to find critical points and max/mins. Help is greatly appreciated!
     
  2. jcsd
  3. May 9, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Critical Points, intervals, local max/min help!! Calculus.

    You should at least attempt to solve the problem. f'(x) has be zero at a critical point. Can you think of some values of x that will do that?
     
  4. May 10, 2012 #3
    Re: Critical Points, intervals, local max/min help!! Calculus.

    Hey,

    Okay, so If I set x to equal zero, then, f'(x) equals 0.

    Basically i'm having problems with this whole set of problems. I'm having a hard time wrapping my head around critical points. So what you're saying is:

    To find a critical point:
    1. find the derivative of a function
    2. figure out what x should be set to, for f'(x) to equal 0.


    Thank you for your help. I really appreciate it.
     
  5. May 10, 2012 #4

    Mark44

    Staff: Mentor

    Re: Critical Points, intervals, local max/min help!! Calculus.

    That's not true.
    f'(0) = (sin(0) + 1)(2cos(0) +√3) = 1 *(2 + √3) = 2 + √3 ≠ 0.
    Generally, yes, but in this problem you're given the derivative already.
    Yes, solve the equation f'(x) = 0 for x.
     
  6. May 10, 2012 #5
    Re: Critical Points, intervals, local max/min help!! Calculus.

    Critical point occurs when first derivative is zero or the derivative doesn't exist but original function exists when evaluated at that point. Its easy to see why this is true for when first derivative is zero it means that instantaneous slope is a horizontal tanget line which could be a local minimum maximum or inflection point since this is place where slope makes a transition from either decreasing to increasing(local minimum or absolute minimum) or from increasing to decreasing(local maximum or absolute maximum).

    One hint for you is in order for this function to be zero,then either
    sinx + 1 = 0; or 2cosx + (3)^1/2 = 0;

    Here is nice site that was good when I was learning about this.
    http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints.aspx
     
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