Results 1 to 5 of 5

Math Help - calculus

  1. #1
    Member
    Joined
    Oct 2008
    From
    Melbourne
    Posts
    166

    Red face calculus

    hey y'all,

    I got a question (obviously)

    "By expressing y = \sinh{(x)} as a quadratic in the term e^x, show \sinh^{-1}(x) = \ln{(x + \sqrt{x^2 +1})}."

    This is my worked solution for the quadratic part:

    y = sinh(x) = \frac{1}{2}(e^x - e^{-x})
    y * 2e^x = {[\frac{1}{2}(e^x - e^{-x}]}*2e^x = {(e^x)}^2 - 1
    0 = e^{2x} + 2ye^x - 1

    And now for the inverse part:

    I get stuck i know the answer is \sinh^{-1}(x) = \ln{(x + \sqrt{x^2 +1})} (from the question) and i know somewhere in the calculations there is some sort of logarithm cancellation i.e. \ln{(e^x)} = x

    Can someone help? and can you tell me if my quadratic expression is incorrect? thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,
    Quote Originally Posted by tsal15 View Post
    This is my worked solution for the quadratic part:

    y = sinh(x) = \frac{1}{2}(e^x - e^{-x})
    y * 2e^x = {[\frac{1}{2}(e^x - e^{-x}]}*2e^x = {(e^x)}^2 - 1
    0 = e^{2x} + 2ye^x - 1
    You forgot a minus sign, the last equality should be 0 = e^{2x} {\color{blue}-} 2ye^x - 1. Now you can, for example, let X=\mathrm{e}^x, solve for X and then solve for x.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2008
    From
    Melbourne
    Posts
    166
    Quote Originally Posted by flyingsquirrel View Post
    Hi,

    You forgot a minus sign, the last equality should be 0 = e^{2x} {\color{blue}-} 2ye^x - 1. Now you can, for example, let X=\mathrm{e}^x, solve for X and then solve for x.

    Sorry mate, I've adjusted as you said, but i got stuck again. my answer was y = x +1 thats obviously a wrong answer Can you explain it a little more? thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by tsal15 View Post
    Can you explain it a little more? thank you
    Let X=\exp x. The equation 0=\exp(2x)-2y\exp x-1 becomes 0=X^2-2yX-1 which is a quadratic. The discriminant is \Delta=(-2y)^2-4\times(-1)\times 1=4y^2+4=4(y^2+1)>0 hence the two solutions we're looking for are X=\frac{2y\pm\sqrt{4(y^2+1)}}{2}=\frac{2y\pm\sqrt{  4}\cdot\sqrt{y^2+1}}{2}=y\pm\sqrt{y^2+1}.

    Can you take it from here ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2008
    From
    Melbourne
    Posts
    166
    Quote Originally Posted by flyingsquirrel View Post
    Let X=\exp x. The equation 0=\exp(2x)-2y\exp x-1 becomes 0=X^2-2yX-1 which is a quadratic. The discriminant is \Delta=(-2y)^2-4\times(-1)\times 1=4y^2+4=4(y^2+1)>0 hence the two solutions we're looking for are X=\frac{2y\pm\sqrt{4(y^2+1)}}{2}=\frac{2y\pm\sqrt{  4}\cdot\sqrt{y^2+1}}{2}=y\pm\sqrt{y^2+1}.

    Can you take it from here ?

    YES YES THANK YOU Much appreciated
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 13th 2011, 10:11 PM
  2. Replies: 2
    Last Post: June 25th 2010, 11:41 PM
  3. Replies: 1
    Last Post: February 11th 2010, 08:09 AM
  4. Calculus III But doesn't require Calculus :)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 19th 2009, 05:23 PM
  5. Replies: 1
    Last Post: June 23rd 2008, 10:17 AM

Search Tags


/mathhelpforum @mathhelpforum