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Thread: Domains and Inverse functions

  1. #1
    Feb 2011

    Post Domains and Inverse functions

    Hi, while I was working on my pre-calc textbook. I came across some questions that I need help on. Showing me how you did it would be more appreciable than giving me the answer. Thank you!

    1) A figure is in the shape of a rectangle surmounted by an equilateral triangle. The figure has perimeter 25m. among all such figures, find length ''x'' of the base of the figure with the largest area. State the domain of your function.
    Domains and Inverse functions-.png

    2)Find the expression for the inverse function f^-1

    f(x)= 1/(x-2)^2 if x>2

    5 + x if x< -5

    3) A cone of height 20m and radius 8m contains half the volume of water that it could contain if it were full. What is the height (or depth) ''h'' of the water?

    Domains and Inverse functions-b.png
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  2. #2
    Dec 2009
    1) The area of the figure is the sum of the areas of the rectangle and the equilateral triangle. The area of the rectangle (with base x and length y) is given by $\displaystyle A = xy$. The area of the equilateral triangle is given by $\displaystyle A = \dfrac{\sqrt{3}}{4}x^2$.

    The sum of these areas is given by $\displaystyle A = xy + \dfrac{\sqrt{3}}{4}x^2$.

    The perimeter of the figure is given by $\displaystyle 25 = 2y + 3x$. Therefore, $\displaystyle y = \dfrac{25 - 3x}{2}$.

    Now, substitute $\displaystyle y = \dfrac{25 - 3x}{2}$ in the area formula.

    $\displaystyle A = x\dfrac{25 - 3x}{2} + \dfrac{\sqrt{3}}{4}x^2 = \dfrac{25}{2}x - \dfrac{3}{2}x^2 + \dfrac{\sqrt{3}}{4}x^2$

    Thus, the area function is a parabola whose maximum will occur at its vertex. Its its vertex can be found either graphically or algebraically.

    2) f(x) is a piece-wise function, so tackle the problem piece by piece. Start with the first piece:

    $\displaystyle f(x) = \dfrac{1}{\left(x-2\right)^2}$

    Find its domain and range since they correspond to the range and domain of the inverse function, respectively.

    Here is the domain of f(x) (I'll let you find its range): $\displaystyle (2,\infty)$

    Then, find the inverse of $\displaystyle y = \dfrac{1}{\left(x-2\right)^2}$ by switching y and x and solving for the "new" y.

    Repeat the steps for the second piece: $\displaystyle f(x) = 5 + x$

    Here is the final answer:

    $\displaystyle f(x)=\begin{cases}-2 + \sqrt{x},\quad& (0,\infty)\\
    x - 5_,\quad& (-\infty,0)\end{cases}$

    3) The problem mentions the volume of a cone, so start with the formula for the volume of a cone:

    $\displaystyle V = \dfrac{1}{3}\pi r^2 h$

    You know the radius and height of the cone, so substitute these values into the volume formula.

    $\displaystyle V = \dfrac{1}{3}\pi \cdot 8^2 \cdot 20 = \dfrac{1280\pi}{3}$

    But the volume of the cone is only half the volume if it were full, so $\displaystyle V = \dfrac{640\pi}{3}$.

    The radius of the water is still 8 meters, and you just found its volume to be $\displaystyle \dfrac{640\pi}{3}$ cubic meters. Use the volume formula to solve for the height h.
    Last edited by NOX Andrew; Feb 27th 2011 at 06:25 AM.
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  3. #3
    MHF Contributor
    skeeter's Avatar
    Jun 2008
    North Texas
    (1) base of rectangle = $\displaystyle x$

    height of rectangle = $\displaystyle \dfrac{25-3x}{2}$

    $\displaystyle A = \dfrac{x(25-3x)}{2} + \dfrac{\sqrt{3} \cdot x^2}{4}$

    note that $\displaystyle x > 0$ and $\displaystyle 25-3x > 0$
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