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Math Help - Given a function, find points, find limits, etc.

  1. #1
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    Given a function, find points, find limits, etc.

    Given the function f(x) = x^2ln(x), x is contained in the set (0,1)

    A) Find the coordinates of any points where the graph of f has a horizontal tangent line.

    B) Find the coordinates of any points of inflection on graph f

    C) Find lim as x --> 0^+ f(x) AND lim as x --> 0^+ f'(x)

    D) Sketch graph of f using info obtained in a,b,c, with coordinates, max, min, etc.


    The first 2 parts are based on units I did a WHILE ago ...

    Do I do this for A?

    dy/dx = x + 2xln(x)
    0 = x  + 2xln(x) (because 0 = horizontal line)

    And solve for x?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Do I do this for A?


    (because 0 = horizontal line)

    And solve for x?



    absolutely
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Lord Darkin View Post
    Given the function f(x) = x^2ln(x), x is contained in the set (0,1)

    A) Find the coordinates of any points where the graph of f has a horizontal tangent line.

    B) Find the coordinates of any points of inflection on graph f

    C) Find lim as x --> 0^+ f(x) AND lim as x --> 0^+ f'(x)

    D) Sketch graph of f using info obtained in a,b,c, with coordinates, max, min, etc.


    The first 2 parts are based on units I did a WHILE ago ...

    Do I do this for A?

    dy/dx = x + 2xln(x)
    0 = x + 2xln(x) (because 0 = horizontal line)

    And solve for x?
    FoR C)

    \lim_{x\to0^+}x^2\ln{x}=\lim_{x\to0^+}\frac{\ln{x}  }{x^{-2}}=\lim_{x\to0^+}\frac{x^{-1}}{-2x^{-3}} by L'Hopital's rule.
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  4. #4
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    Part a, my math solver on the calculator says -1.7632 (I had to use the absolute value of x to put in for the natural log).

    But I'm confused here since x is supposed to be contained in the set from 0 to 1.

    VonNemo19, thanks for the help, I'll look into that in more depth once I understand part a and b more.
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  5. #5
    MHF Contributor Calculus26's Avatar
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    you have x + xln(x) = 0

    x(1+2ln(x)) = 0

    x= 0 or e^(-1/2) = .606

    0 is not in (0,1) so e^(-1/2) is the only pt on the interval with a horizontal tangent
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  6. #6
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    y = x^2 \cdot \ln{x}

    y' = x + 2x\ln{x} = x(1 + 2\ln{x}) = 0

    reject x = 0 as a solution since x \in (0,1)

    \ln{x} = -\frac{1}{2}

    x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}}
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  7. #7
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    Ahh, I get it.

    So how does everything look?

    Part A

    x = 0.6065

    y = -0.184 (Makes sense? - plugged it back in f(x))

    Answer: (0.607, -0.184)

    Part B

    y'' = 3 + 2lnx
    0 = 3 + 2lnx
    x = e^{-\frac{3}{2}}
    x = 0.2231
    y = -0.446 (Plugged back in f ' (x) )

    Answer: (0.223, -0.446)

    Part C

    \lim_{x\to0^+}[x^2\ln{x}] = \lim_{x\to0^+}\frac{\ln{x}}{x^{-2}} = \lim_{x\to0^+}\frac{x^{-1}}{-2x^{-3}} = 0



    \lim_{x\to0^+}[x + 2xlnx] = (\lim_{x\to0^+}[x] + \lim_{x\to0^+}[2xlnx]) = \lim_{x\to0^+} -2x = 0

    Answer: Zero (For Both)

    Part D is just graphing, I should be fine with that.
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  8. #8
    No one in Particular VonNemo19's Avatar
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    looks great.

    however \lim_{x\to0^+}2x\ln{x} yields an inderminate form. rewrite as a quotient and Apply L'Hopital's rule.
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  9. #9
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    ^yes, I used L'Hopital's rule, but didn't show it there (takes a while for me to type the latex). It came out to be -2x.

    The only weird thing is, when I graph this on my calc, the point of inflection for part b doesn't make sense since the y value is -0.446 in part b but the graph on my calc is more like -0.2.
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  10. #10
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    Quote Originally Posted by Lord Darkin View Post
    ^yes, I used L'Hopital's rule, but didn't show it there (takes a while for me to type the latex). It came out to be -2x.

    The only weird thing is, when I graph this on my calc, the point of inflection for part b doesn't make sense since the y value is -0.446 in part b but the graph on my calc is more like -0.2.
    Hint:
    put x=\frac{1}{t}.
    t\to \infty ,x\to 0
    now find,
    \lim_{t\to \infty }-2\left (\frac{\ln t}{t}  \right )
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  11. #11
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Lord Darkin View Post
    ^yes, I used L'Hopital's rule, but didn't show it there (takes a while for me to type the latex). It came out to be -2x.

    The only weird thing is, when I graph this on my calc, the point of inflection for part b doesn't make sense since the y value is -0.446 in part b but the graph on my calc is more like -0.2.
     <br />
\frac{d^2y}{dx^2}=\frac{d}{dx}[x+2x\ln{x}]=1+(2x\frac{1}{x}+2\ln{x})=3+2\ln{x}=0\Rightarrow\  ln{x}=-\frac{3}{2}\Rightarrow{x}=e^{-3/2}<br />
    Attached Thumbnails Attached Thumbnails Given a function, find points, find limits, etc.-x-2lnx.bmp  
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  12. #12
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    Quote Originally Posted by Raoh View Post
    Hint:
    put x=\frac{1}{t}.
    t\to \infty ,x\to 0
    now find,
    \lim_{t\to \infty }-2\left (\frac{\ln t}{t} \right )

    What is the purpose of doing that? I'm just wondering since it seems like I still get lim f '(X) = 0.

    Also, I understand that e=^(-3/2) but I'm confused about the y value for the point of inflection. My calculator shows that the lowest point of the graph of f(x) is -0.183 so why does the point of inflection I have in part b have y=-0.446?
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  13. #13
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Lord Darkin View Post
    What is the purpose of doing that? I'm just wondering since it seems like I still get lim f '(X) = 0.

    Also, I understand that e=^(-3/2) but I'm confused about the y value for the point of inflection. My calculator shows that the lowest point of the graph of f(x) is -0.183 so why does the point of inflection I have in part b have y=-0.446?
    f(e^{(-3/2)})\approx{-.075}\neq-.446
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  14. #14
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    Quote Originally Posted by Lord Darkin View Post
    What is the purpose of doing that? I'm just wondering since it seems like I still get lim f '(X) = 0.

    Also, I understand that e=^(-3/2) but I'm confused about the y value for the point of inflection. My calculator shows that the lowest point of the graph of f(x) is -0.183 so why does the point of inflection I have in part b have y=-0.446?
    sorry about that i thought that would help (Post 8).
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  15. #15
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    Oh!!! I substituted in the x value that I got in part b into the first derivative, not the original!!!

    Need to remember that.

    Thanks everyone!

    Problem Solved!
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