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Math Help - Need help.

  1. #1
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    Need help.

    I'm having a hard to solve this two problems. I would really appreciate your help. Look at the attachement.Thanks.
    Attached Thumbnails Attached Thumbnails Need help.-20140416_230735-1-.jpg  
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  2. #2
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    Re: Need help.

    Problem #1: The area of a triangle is $\dfrac{1}{2}bh$. In this case, the base and height are both determined by the $x$- and $y$-intercepts of the tangent line. In general, points on the parabola given are of the form $(u,1-u^2)$. The slope of a tangent line at the point $(u,1-u^2)$ is given by plugging $u$ into the derivative. The derivative of the parabola is $y' = -2x$, so the slope of the tangent line is $-2u$. Then, in point slope form, $y-y_0 = m(x-x_0)$. Plugging in:

    $y-(1-u^2) = -2u(x-u)$

    Converting to slope-intercept form:

    $y = -2ux + u^2+1 = mx+b$

    Since $-2ux = mx$, obviously, $u^2+1 = b$, so the $y$-intercept is $u^2+1$.

    Then, what is the $x$-intercept? That is when $y=0$. Plug in for $y=0$ and solve for $x$:

    $2ux = u^2+1 \Rightarrow x = \dfrac{u^2+1}{2u}$

    This gives you your base and height. So, the area is:

    $A = \dfrac{1}{2}bh = \dfrac{1}{2}\dfrac{u^2+1}{2u}(u^2+1) = \dfrac{1}{4}(u+u^{-1})(u^2+1) = \dfrac{1}{4}(u^3+2u+u^{-1})$

    Now, to minimize that, take the derivative with respect to $u$ and set it equal to zero to find critical numbers:

    $A'(u) = \dfrac{3u^2+2-u^{-2}}{4} = 0$

    Multiplying out by $4u^2$ gives:

    $3u^4 + 2u^2-1 = 0$

    $(u^2+1)\left(u^2-\dfrac{1}{3}\right) = 0$, so $u^2 = -1$ or $u^2 = \dfrac{1}{3}$. Since $\sqrt{-1}$ is not a real number, it must be that $u = \pm \sqrt{\dfrac{1}{3}}$. Since the negative answer is not an $x$-value in the first quadrant, we reject it. Hence, $u = \sqrt{\dfrac{1}{3}}$. Next, check to see if it is truly a minimum. You can use the second derivative test (if the second derivative is positive at $u = \sqrt{\dfrac{1}{3}}$, you found a minimum).

    For question 2: By the logarithm rules, $\ln|2x| = \ln 2 + \ln |x|$. Since $\ln 2 \approx 0.693$, it is a constant. Hence, just as the corollary said, $\ln|x| \text{ and } \ln|2x|$ differ by just a constant.
    Last edited by SlipEternal; April 16th 2014 at 07:39 PM.
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  3. #3
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    Re: Need help.

    thanks
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