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December 7th, 2018, 10:38 AM  #1 
Newbie Joined: Aug 2016 From: Romania Posts: 23 Thanks: 0  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: 2,312 Thanks: 707 
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  
Newbie Joined: Aug 2016 From: Romania Posts: 23 Thanks: 0  Quote:
 
December 7th, 2018, 11:43 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
f(1) = 3. Flip the domain and range.

December 7th, 2018, 01:19 PM  #5  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
$f(x) = x^3 + x + 1 \implies f'(x) = 3x^2 + 1 > 0 \text { for all x.}$ 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. $\therefore f(1) = 3 \implies g(f(1)) = g(3) \implies g(3) = 1.$  

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dou, function, hard, inverse, problem 
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