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 December 7th, 2018, 10:38 AM #1 Member   Joined: Aug 2016 From: Romania Posts: 32 Thanks: 1 Hard Inverse function f(x) problem So I have this problem f(x)=x^3+x+1 and I need to find the inverse of this function.I know that you have to use differentiation but the problem is I see this solution (f^-1)(3)=1 but I don't understand the steps to reach that conclusion.There was another one (f^-1)(3)=5 and this looked the same but I couldn't reach these result and all I got is f'(x)=3x^2+1.Now what steps do I have to follow? December 7th, 2018, 10:53 AM #2 Senior Member   Joined: Aug 2012 Posts: 2,409 Thanks: 753 Heck if I know. Check this out. https://www.wolframalpha.com/input/?...x%5E3%2Bx%2B1) Are you perhaps being asked to find the value of the inverse at a particular point? Much easier. December 7th, 2018, 11:02 AM   #3
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 Originally Posted by Maschke Heck if I know. Check this out. https://www.wolframalpha.com/input/?...x%5E3%2Bx%2B1) Are you perhaps being asked to find the value of the inverse at a particular point? Much easier.
Maybe some steps could enlight me to reach that answer.Do you have to use the cubic equation here? December 7th, 2018, 11:43 AM #4 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond f(1) = 3. Flip the domain and range. December 7th, 2018, 01:19 PM   #5
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 Originally Posted by alex77 So I have this problem f(x)=x^3+x+1 and I need to find the inverse of this function.I know that you have to use differentiation but the problem is I see this solution (f^-1)(3)=1 but I don't understand the steps to reach that conclusion.There was another one (f^-1)(3)=5 and this looked the same but I couldn't reach these result and all I got is f'(x)=3x^2+1.Now what steps do I have to follow?
What differentiation tells you here is that f(x) has an inverse for all x because

$f(x) = x^3 + x + 1 \implies f'(x) = 3x^2 + 1 > 0 \text { for all x.}$

As far as I can see, you are just asked to find the value of the inverse function.

$g(x) = f^{-1}(x) \text { and } f(x) = x^3 + x + 1.$

$f(x) = 3 \implies x^3 + x + 1 = 3 \implies x^3 + x - 2) = 0 \implies$

$x^3 - x^2 + x^2 - x + 2x - 2 = 0 \implies x^2(x - 1) + x(x - 1) + 2(x - 1) = 0 \implies$

$(x - 1)(x^2 + x + 2) = 0.$

Are there any real roots of the quadratic factor? No, because

$1^2 - 4(1)(2) = 1 - 8 = -\ 7 < 0.$

So 1 is the only real root.

$\therefore f(1) = 3 \implies g(f(1)) = g(3) \implies g(3) = 1.$ Tags dou, function, hard, inverse, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post MFP Calculus 3 July 9th, 2013 10:34 AM Niko Bellic Trigonometry 1 May 29th, 2012 08:29 PM jaredbeach Algebra 1 November 17th, 2011 11:58 AM Math_Junkie Calculus 11 October 12th, 2009 10:25 AM derhaus Algebra 1 October 2nd, 2007 06:39 PM

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