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Homework Help: Curve Sketching

  1. Dec 26, 2009 #1
    I got this lobe Curve sketching question on my assignment. i know most of it but i think i got the drevetive part wrong. anyway, i will write the whol question with all the soultions i reached.

    1] Use the Funtion F(x) = x2-4 [tex]\sqrt{x}[/tex]

    a) State the domain

    b) Determine the intercepts.
    c)Is the graph Symmetric about the X-axis or Y axis ?

    d) Find the Asymptotes
    e) where does the function increase ? where does the function decrease ?

    f. Determine the Maximum and minimum values.

    Please, if someone could tell me which part is wrong, and direct me in the right way!

    Thank, you
  2. jcsd
  3. Dec 27, 2009 #2
    With the exception of a one small issue in part e ("x > 0 or x > 1" should just be x > 1) you have everything correct including your derivatives.

    You can find the max and min values with the critical points you have from part e, x=0 and x=1, in the manner that you specified above. Just a heads up, x=0 is a little trickier because it's the end of the domain but if you sketch a rough graph you should be able to figure it out.
  4. Dec 27, 2009 #3


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    The domain is "all non-negative real numbers" which is slightly different from "all positive real numbers".

    Also while your values for the x-intercepts are correct, your method is peculiar. You have [itex]x^2- 4\sqrt{x}= 0[/itex] and then, immediately, [itex]x^4- 16x= 0[/itex]. The intermediate steps should be [itex]x^2= 4\sqrt{x}[/itex] and [itex]x^4= 16x[/itex].
  5. Dec 28, 2009 #4
    Thank you so much Vaal and HallsOfIvy , really appreciate you insights

    I drew The graph of the function on my graphic calcu.

    and both 0 has y-value of 0 and 1 has y-value of -3 .
    but the Second derivative test shows that bpth are minimums (F"(x) > 0) , which makes no sense.

    F"(X) = 2+x-3/2

    F"(0) = 2 (since is't greater then 0 it's a minimum)
    F"(1) = 3 (Since it's greater then 0 it's a minimum )

    It makes sense that 1 is a minimum, but why 0 ? if someone could explain that point to me, would be really appreciated
  6. Dec 28, 2009 #5
    Minimums and maximums are determined by the points of inflection.

    Points of inflection are points in which the concavity changes. Assuming you've written down the second derivative correctly, when I graph it it seems to me that it is always concave up so there are no points of inflection.
  7. Dec 28, 2009 #6

    Now this is getting me confused :D.

    ya Inflection point is a way to know the minimums, and maximums. but the main test is to find the critical points, then do the second test.

    and ya the graph is always concave up except between interval(0,1)
  8. Dec 28, 2009 #7


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    No, inflection points have little or nothing to do with "maximum" and "minimum". Inflection points are points are points where the first derivative changes sign (and so the second derivative is 0). Functions that have NO minimum or maximum may have inflection points and, conversely function that do have inflection points may have no maximum or minimum.

    Last edited by a moderator: Dec 30, 2009
  9. Dec 29, 2009 #8
    Thanks HallsOfIvy , again, for your help.
    my last questions is, and am sorry if i asked alot, Is 0 a Minimum ? if not , then why it's F"(0) is Greater then 0 (which means that it's minimum)

    I think that 0 is a minimum, because The Second drevative test said so,

    i don't think that it's a minimu, because it is the end point, and f(-0.0001) doesn;t exist .
  10. Dec 29, 2009 #9
    That is a good question. The second derivative test actually will not work on critical values that lie on the very edge of the original function's domain(i.e. the point where x=0 in this case). From the graph of the original function you can see x=0 is the highest point point in the neighborhood of x=0 but less than the value of the function as it increases towards infinity which would mean it is a relative maximum if anything. Sense x=0 is the edge of the domain it's neighborhood is not really defined (because any values to the left of x=0 aren't in the domain) so I think it actually isn't even a local minimum. Perhaps HallsofIvy can confirm this.
  11. Dec 30, 2009 #10


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    [tex]f(x)= x^2- 4\sqrt{x}= x^2- 4x^{1/2}[/tex]
    [tex]f'(x)= 2x- 2x^{-1/2}[/tex]
    [tex]f"(x)= 2+ x^{-3/2}[/tex]

    No, the second derivative is NOT greater than 0 at x= 0, it is not even defined there.

    For x very close to 0, f'(x) is a large negative number so f is decreasing away from x= 0 and x= 0 is a local maximum.

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