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Analysis of Functions I: increase, decrease, and concavity

  1. Oct 22, 2011 #1
    Hello I need help with these problems. The direction said

    a. find the intervals on which f is increasing, b. the intervals on which f is decreasing, c. the open intervals on which f is concave up, d. the open intervals on which f is concave down and e. the x-cordincates of all inflection points
    1. f(x)=x^4-8x^2+16
    I find a,b c and d for this function. However I have trouble finding the inflection points for this function. When you find the inflection point you suppose to set up the second derivative of the function to equal 0 but I have no idea how to solve it afterward. Is there an inflection point for this equation?
    f"(x)=12x^2-16
    2. f(x)=x/(x^2+2)
    so f'(x)=-x2+2/(x2+2)2
    to find a,b I have to set this equal to 0 and solve it. How do I solve it? and what would be the second derivative and how do I solve it by setting it up to equal to 0 to know if it's concave up or down?
    3. f(x)=x2lnx
    The first derivative is 2xlnx+x. How do I solve it or know what is a, b is?and the second derivative is 2lnx+3. How do I use the second derivative to solve for c,d and e?

    Please help! Thank you!
     
  2. jcsd
  3. Oct 22, 2011 #2

    ehild

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    So you need to solve the equation 12x^2-16 = 0. What is your problem???

    Use parentheses. The formula is wrong without them.
    You will have a fraction. It can be zero if the nominator is zero.

    To find x where the derivative is zero, factor out x.

    ehild
     
  4. Oct 22, 2011 #3

    HallsofIvy

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    What? You don't know how to solve [itex]12x^2- 16= 0[/itex]? Add 16 to both sides, divide both sides by 12, then take the square root of both sides.

    you are taking Calculus and do not know how to solve equations like this? Multiply both sides of the equation by that denominator to get [itex]2- x^2= 0[/itex]

    Differentiate it, using the quotient rule again.

    Oh, c'mon! 2xln(x)+ x= x(2ln(x)+ 1)= 0. One thing you surely learned long ago is that a product is equal to 0 only if one or more of the factors is 0. So here, either x= 0 or 2ln(x)+ 1= 0: x= 0 or ln(x)= -1/2.

    2ln(x)+ 3= 0 leads to ln(x)=-3/2. Again, you should have learned in algebra or precalculus that if ln(x)= a then [itex]x= e^a[/itex]. Once you know where it zero, you can determine the intervals in which the second derivative is positive or negative.

     
  5. Oct 22, 2011 #4
    Actually I finished this before checking back haha. Please delete this post
     
    Last edited: Oct 23, 2011
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