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Math Help - Comphrensive Curve Sketching

  1. #1
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    Joined
    Sep 2009
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    6

    Comphrensive Curve Sketching

    If anyone could take some time to answer any or all of the problems below, it would be a massive help. I've gone through my homework and highlighted a few problems I've had as I've gone through. Much of it involves lack of mechanical skill on my part...if someone would like to explain exactly how the textbook is arriving at the sol'ns below, again, it would be a huge help.

    Thanks,

    Graeme

    ---------------------

    1. What's going on when the textbook/sol'n manual busts out the absolute value symbols, such as in:

    Since 12x^2 + 4 > for all x, we have
    f''(x) > 0 <-> x^2 - 1 > <-> |x| > 1

    2. Mechanics; I feel like I'm limited by my ability to manipulate functions. Could someone explain how the sol'n manual as gone from point A to point B in the following examples:

    (a) f'(x) = \frac{2x\sqrt{x+1} - x^2 * 1/(2\sqrt{x+1)}}{x+1} = \frac{x(3x+4)}{2(x+1)^{3/2}}

    (b) f'(x) = \frac{\sqrt{x^2 +1} - x*\frac{2x}{2\sqrt{x^2+1}}}{(x^2 + 1)^{1/2})^2} = \frac{x^2 + 1 - x^2}{(x^2 +1)^{3/2}}

    (c) f'(x) = x * \frac{1}{2}(5-x)^{-1/2}(-1)+(5-x)^{1/2} *1 = \frac{1}{2}(5-x)^{-1/2}[-x+2(5-x)] = \frac{10-3x}{2\sqrt{5-x}}


    3. And I'm most likely missing something but how the hell does x^2 end up as 1:

    lim \frac{x/x}{\sqrt{x^2 + 1}/x} =
    x->inf

    lim \frac{x/x}{\sqrt{x^2 + 1}/\sqrt{x^2}} =
    x->inf

    lim \frac{1}{\sqrt{1 + 1/x^2}}
    x->inf

    lim \frac{1}{\sqrt{1 + 0}} = 1
    x->inf

    4. How do you know when there are no asymptotes? Are there just no HAs if there denominational power is smaller than the numerator highest power? How about VAs?

    e.g. y = x-3x^{1/3}

    5. How do you when to make a common denominator? For example, why can't I just evaluate the intervals of increase and decrease right off the bat with the following function, as opposed to doing what the book does:

    e.g. y' = 1 - x^{-2/3} = 1 - \frac{1}{x^{2/3}} = \frac{x^{2/3} - 1}{x^{2/3}}

    6. Could someone take me through how to find intercepts for the following function (I don't get the whole x=(2n+1)pi/2 thing):

    f(x) = \frac{cosx}{2+sinx}
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  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    1. The absolute value occurs from the symmetry of the function y=x^2 across the y-axis:

    x^2-1>0\;\rightarrow\; x^2>1\;\rightarrow\; (x>1\mbox{ or } x<-1)\;\rightarrow\; |x|>1.

    2. Most of these problems appear to relocate the denominator of the numerator into the denominator, as in

    \frac{\frac{1}{2}}{2}=\frac{1}{2\cdot 2}=\frac{1}{4}.

    For example, in (a),

    \begin{aligned}<br />
\frac{2x\sqrt{x+1}-\frac{x^2}{2\sqrt{x+1}}}{x+1}&=\frac{\frac{4x(x+1)  }{2\sqrt{x+1}}-\frac{x^2}{2\sqrt{x+1}}}{x+1}\\<br />
&=\frac{\frac{4x^2+4x-x^2}{2\sqrt{x+1}}}{x+1}\\<br />
&=\frac{3x^2+4x}{2\sqrt{x+1}(x+1)}=\frac{x(3x+4)}{  2(x+1)^{\frac{3}{2}}}.<br />
\end{aligned}

    3. In the second step, we use the rule that

    \frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}

    together with the fact that

    x=\sqrt{x^2}

    for x\ge 0.

    4. A horizontal asymptote occurs when \lim_{\small x\rightarrow\pm\infty}f(x) attains a finite value.

    5. The book may have simplified to a fraction in order to determine more easily when the value of the derivative was positive or negative. In particular,

    \frac{x^{\frac{2}{3}}-1}{x^{\frac{2}{3}}}

    is positive when both the numerator and denominator have the same sign, and negative when they have differing sign.

    6. Because the denominator of \frac{\cos x}{2+\sin x} is never 0, we can multiply it out in the equation

    \frac{\cos x}{2+\sin x}=0

    to obtain an equivalent statement

    \cos x=0.

    The solutions here are just those values of x, in radians, for which the corresponding point on the unit circle lies on the y-axis: \frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\cdots
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