My Math Forum Intermediate Value Theorem

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 April 15th, 2013, 09:41 PM #1 Newbie   Joined: Apr 2012 Posts: 21 Thanks: 0 Intermediate Value Theorem Hi all Strange question here involving the intermediate value theorem. Consider the equation x^3 + x +e^x = 0 Use the intermediate value theorem to show there is a solution on the interval (-1,0) and show that there is exactly one real solution of the equation. I've been trying at this one for a couple of hours now and I cant seem to make good headway. My equation fall apart before too long. I have an idea on solving the second half of the question, showing there is exactly on real solution, and that is by graphing it and then possibly trying to manipulate it? Any help is greatly appreciated. Thank you!
 April 15th, 2013, 10:02 PM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Intermediate Value Theorem It is not as difficult as it may seem, let $y= f(x) = x^3 + x + e^x$ $f(-1)= (-1)^3 + (-1) + e^{-1} = negative$ $f(0)= 0^3 + 0 + e^0 = positive$ So... the y value went from negative to positive... it must have hit zero at least once... that's the intermediate value theorem. To show it hit zero only once take the derivative, $f'(x) = 3x^2 + 1 + e^x$ So... f'(x) is positive for all real x (do you see why?), this means the function is always increasing... therefore it cannot cross the x axis more than once.
 April 16th, 2013, 12:40 AM #3 Newbie   Joined: Apr 2012 Posts: 21 Thanks: 0 Re: Intermediate Value Theorem Oh my. I just had the biggest facepalm moment. Thank you so much, you just saved me a sleepless night

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