Results 1 to 5 of 5

Math Help - finding the coordinantes for the tangent line.

  1. #1
    Member
    Joined
    Jun 2007
    Posts
    79

    finding the coordinantes for the tangent line.

    Find the coordinates of each point on the graph of the given function where the tangent line is horizontal.
    I have this problem:
    f(x)=(sqrt x)(x-3).
    I used the product rule:
    (sqrt x)(x-3)'+(sqrt x)'(x-3)
    coming out (sqrt x)+x-3. I know this cannot be right.
    What am I doing wrong with this?
    The answer is suppose to be (1,-2).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    You should get f'(x)=\frac{3(x-1)}{2\sqrt{x}}

    As you can see, what makes the numerator equal to 0?.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2007
    Posts
    79
    Quote Originally Posted by galactus View Post
    You should get f'(x)=\frac{3(x-1)}{2\sqrt{x}}

    As you can see, what makes the numerator equal to 0?.
    From what you wrote, I understand how the numerator get to zero, But I am lost on how to get it to that particular form.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    f(x)=(x-3)\sqrt{x}
    f'(x)=(x-3)'\sqrt{x}+(x-3)(\sqrt{x})'=\sqrt{x}+\frac{x-3}{2\sqrt{x}}=\frac{3x-3}{2\sqrt{x}}=\frac{3(x-1)}{2\sqrt{x}}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,714
    Thanks
    633
    Hello, driver!

    Find the coordinates of each point on the graph of the given function
    where the tangent line is horizontal: . f(x)\;=\;\sqrt{x}(x-3)

    We have: . f(x)\;=\;x^{\frac{1}{2}}(x - 3) \;=\;x^{\frac{3}{2}} - 3x^{\frac{1}{2}}

    Then: . f'(x)\;=\;\frac{3}{2}x^{\frac{1}{2}} - \frac{3}{2}x^{-\frac{1}{2}}\;=\;0\quad\Rightarrow\quad\frac{3}{2}  \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)\:=\:0\quad\Rightarrow\qu  ad \sqrt{x} - \frac{1}{\sqrt{x}} \;=\;0

    Multiply through by \sqrt{x}\!:\;\;x - 1\:=\:0\quad\Rightarrow\quad x \,=\,1


    There is a horizontal tangent at (1,\,-2)

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: January 12th 2011, 02:38 PM
  2. Finding a tangent line
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 2nd 2011, 11:01 AM
  3. finding the tangent line
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 7th 2010, 11:07 PM
  4. Finding the tangent line
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 19th 2009, 11:27 AM
  5. Finding the tangent line!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 1st 2008, 10:55 AM

Search Tags


/mathhelpforum @mathhelpforum