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Several calculus problems

  1. Jan 2, 2007 #1
    Sorry for so many questions, I'm just trying to understand everything before my test coming up soon.

    1. Let R be the region in the first quadrant under the graph of y=1/sqrt(x) for
    4<=x<=9.

    a) If the line x=k divides the region R into two regions of equal area, what is the value of k?
    My thoughts: For this one, I know how to find the area of R, but I don't know how to start finding the line that divides the region into multiple parts of equal area.

    2. Let f be a function given by f(x)=x^3 - 6x^2 + p, where p is an arbitrary constant.

    a) Write an expression for f '(x) and use it to find the relative maximum and minimum values of f in terms of p. Show the analysis that leads to your conclusion.
    My thoughts: f '(x) = 3x^2 - 12x. Setting this equal to 0 and solving for x gives: x=0, 4. These are critical points, and I know that to find max/min points I need to look at second derivative. But here it says just use first derivative? I'm also not sure what to write for analysis part.

    b) For what values of the constant p does f have 3 distinct real roots?
    I don't have any thoughts on this one :frown:

    c) Find the value of p such that the average value of f over the closed interval [-1, 2] is 1.
     
  2. jcsd
  3. Jan 2, 2007 #2

    Hootenanny

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    From the question you know that;

    [tex]\int_{4}^{k}\frac{dx}{\sqrt{x}} = \int_{k}^{9}\frac{dx}{\sqrt{x}}[/tex]

    Do you follow?

    Consider the gradient either side of your local extrema, more specifically look at the sign of the gradient (i.e. +ve or -ve). What can this tell you about the local extrema?
    Factorise the cubic as see where it takes you (take a factor of x2 out)...
    Do you know the expression for the average of a function?
     
    Last edited: Jan 2, 2007
  4. Jan 2, 2007 #3

    cristo

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    We want two equal areas, so perform the following integral.
    [tex]
    \int_4^k\frac{dx}{\sqrt x}=\int_k^9\frac{dx}{\sqrt x} [/tex]


    You need to take the second derivative in order to classify the critical points.

    Look at the discriminant: http://en.wikipedia.org/wiki/Cubic_equation under the section "the nature of the roots"

    Any thoughts?

    edit: late
     
  5. Jan 2, 2007 #4

    Hootenanny

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    I beg to differ there. Once can classify local extrema by examining the sign of the first derivative either side of the critical point.
     
  6. Jan 2, 2007 #5

    cristo

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    Apologies, of course you can. I really meant that, in my opinion, in order to classify the critical points, it is simpler to take the second derivative.
     
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