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Thread: Finals question help! (calculus)

  1. #1
    Newbie rIBBON:lacedx's Avatar
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    Finals question help! (calculus)

    Hi everyone,

    My Calculus teacher gave us some review sheets of what to expect on the final. I've narrowed it down to this question that I can't figure out no matter what. Can anyone help me? Here is the problem:

    Find the value of x where f(x)= x^2-3(x+2)^(1/2) has its absolute minimum.

    I know I'm supposed to take the derivative and set that equal to zero to find the critical points. However, I keep getting stuck and can't seem to find the critical points because I get lost in the math.

    Thank you for any help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by rIBBON:lacedx View Post
    Hi everyone,

    My Calculus teacher gave us some review sheets of what to expect on the final. I've narrowed it down to this question that I can't figure out no matter what. Can anyone help me? Here is the problem:

    Find the value of x where f(x)= x^2-3(x+2)^(1/2) has its absolute minimum.

    I know I'm supposed to take the derivative and set that equal to zero to find the critical points. However, I keep getting stuck and can't seem to find the critical points because I get lost in the math.

    Thank you for any help!
    I will assume that by absolute minimum you mean global minimum.

    $\displaystyle f(x)=x^2-3(x+2)^{1/2}$

    $\displaystyle f'(x)=2x-3(1/2)(x+2)^{-1/2}=2x-\frac{3}{2(x+2)^{1/2}}$

    setting $\displaystyle f'(x)=0$ gives us:

    $\displaystyle 2x-\frac{3}{2(x+2)^{1/2}}=0\ \ \ \ \ ..(1)$

    simplifying:

    $\displaystyle \frac{4}{3}x=\frac{1}{(x+2)^{1/2}}$

    or after squaring:

    $\displaystyle \frac{16}{9}x^2 (x+2)-1=0$

    Now this is a cubic, which has three roots, of these two are spurious (that
    is are not zeros of the derivative of $\displaystyle f(x)$, and the other is
    a genuine root. It is probably better to go for a direct numerical solution of
    $\displaystyle (1)$ rather than solve this cubic.

    A quick sketch of $\displaystyle f(x)$ shows the minimum is close to $\displaystyle x=0.5$,
    and formula itteration of:

    $\displaystyle x_{n+1}=\frac{3}{4(x_n+2)^{1/2}}$

    gives the minimum occurs at $\displaystyle x \approx 0.4766$ and is $\displaystyle \approx -4.494$

    RonL
    Last edited by CaptainBlack; Jan 7th 2007 at 11:08 PM.
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