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Math Help - Find an equation for the horizontal tangent to the curve y=x-3root x

  1. #1
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    Find an equation for the horizontal tangent to the curve y=x-3root x

    Hi
    a ) find an equation for the horizontal tangent to the curve y=x-3root x
    b) What is range of values of values of curve's slope ?
    c ) What is range of values of curve ?
    number a already I solved but my queation now in b and c
    -----
    I try to solve
    b)
    to get curve's slope must derivatives
    range of values of values of curve's slope is ( - infinity , 1 ]
    becuse when we put ( 0 ) in curve's slope It will comw 1

    c ) did here put zero in y=x-3root x

    please help me .
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  2. #2
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    Quote Originally Posted by r-soy View Post
    Hi
    a ) find an equation for the horizontal tangent to the curve y=x-3root x
    b) What is range of values of values of curve's slope ?
    c ) What is range of values of curve ?
    number a already I solved but my queation now in b and c
    -----
    I try to solve
    b)
    to get curve's slope must derivatives
    range of values of values of curve's slope is ( - infinity , 1 ]
    becuse when we put ( 0 ) in curve's slope It will comw 1

    c ) did here put zero in y=x-3root x

    please help me .
    b) y' = 1 - \frac{3}{2\sqrt{x}} ... range of the derivative is -\infty < y' < 1 ... note that y' \ne 1

    c) y' = 0 at x = \frac{9}{4}

    y'' = \frac{3}{4x\sqrt{x}} > 0 for all  x in the domain of y, therefore y has an absolute minimum at x = \frac{9}{4} ...
    y\left(\frac{9}{4}\right) = -\frac{9}{4}

    so, range of  y is y \ge -\frac{9}{4}
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