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Thread: functions involving radical

  1. #1
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    functions involving radical

    what if the radical is in the denominator?
    $\displaystyle y=\frac{2}{x^2+8x+10}$

    the domain is $\displaystyle \{x|x^2+8x+10\ge 0\}$

    so the domain is $\displaystyle x^2+8x+10\ge0$

    $\displaystyle \frac{4ac-b^2}{4a}$

    $\displaystyle \frac{40-64}{4}$

    $\displaystyle -4$

    $\displaystyle x\ge-4$??

    range: ??

    is this correct?
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  2. #2
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    First check where $\displaystyle x^2+8x+10=0$
    Solution is $\displaystyle x= -6.5 $ and $\displaystyle -1.6$

    This means as domain is $\displaystyle x\ge0$ so the equation $\displaystyle x^2+8x+10$ is always positive.

    This means if $\displaystyle x=0$,y= $\displaystyle \frac{2}{0^2+8(0)+10}$ $\displaystyle = \frac{1}{5}$

    As $\displaystyle x^2+8x+10$ always increases when increases, So the value of this function $\displaystyle \frac{2}{x^2+8x+10}$ decreases.
    Hence max. value this function can have is $\displaystyle \frac{1}{5}$ and the graph of this function tends to zero, when x increases but can never be equal to zero.

    Range :$\displaystyle \frac{1}{5}\ge y>0$
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  3. #3
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    how did you get and ?
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  4. #4
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    Talking

    Quote Originally Posted by princess_21 View Post
    how did you get x = -6.5 and x = -1.6?
    The poster solved the quadratic equation.

    You can, for instance, use the Quadratic Formula for this....
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  5. #5
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    Quote Originally Posted by princess_21 View Post
    how did you get and ?
    I have used this formula to solve $\displaystyle x^2+8x+10=0$ :

    $\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

    Where standard equation is this one: $\displaystyle ax^2+bx+c$
    a=1, b=8, c=10

    Plug in the values to solve the equation.
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