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 June 4th, 2015, 05:25 AM #1 Newbie   Joined: May 2015 From: sweden Posts: 12 Thanks: 0 prove function invertible Hi! i need help solving this problem. Prove that f(x)= x^7+5x^3+3 is invertible and find the derivative to the inverse function at the point 9 Im not really sure how to do this. i know it has to be injective to be invertible but i dont know how to prove it with this function, and if it it the way i should do it? please help
 June 4th, 2015, 05:34 AM #2 Senior Member     Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 f(x) is continuous. The derivative of the function $\displaystyle f'(x)=7x^6+15x^2$ is always positive. It means that f(x) is a monotonic function. Hence it is invertible.
June 4th, 2015, 05:44 AM   #3
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 Originally Posted by skaa f(x) is continuous. The derivative of the function $\displaystyle f'(x)=7x^6+15x^2$ is always positive. It means that f(x) is a monotonic function. Hence it is invertible.
oh, was it really that simple. thanks. What about the other part of the problem.
Do i need to find the inverse function? because i tried doing that and it didnt really work

 June 4th, 2015, 05:56 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 If g(x) is the inverse function to f(x) then f(g(x))= x. By the chain rule, f'(g(x))g'(x)= 1 so that g'(x)= 1/f'(g(x)). The derivative of g(x) at x= 9 is 1 over the derivative of f at the x value such that f(x)= 9. What is x there? (Hint- it's easy!)
June 4th, 2015, 06:05 AM   #5
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 Originally Posted by Country Boy If g(x) is the inverse function to f(x) then f(g(x))= x. By the chain rule, f'(g(x))g'(x)= 1 so that g'(x)= 1/f'(g(x)). The derivative of g(x) at x= 9 is 1 over the derivative of f at the x value such that f(x)= 9. What is x there? (Hint- it's easy!)
f(x)= 9 --> x=1

Derivative of g(x) is 1/ the derivative of f(1)? or did i understand wrong?
Because the derivative of that would just be zero...?

June 4th, 2015, 06:40 AM   #6
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Quote:
 Originally Posted by skaa $\displaystyle f'(x)=7x^6+15x^2$ is always positive. It means that f(x) is a monotonic function. Hence it is invertible.
The derivative is not always positive because it is zero at $x = 0$. The fact that the derivative is zero somewhere invalidates your claim that the function is always increasing. It is always increasing, but it is necessary to show this directly by examining the value of the function for small $x$ (positive and negative).
Quote:
 Originally Posted by chaarley Derivative of g(x) is 1/ the derivative of f(1)? or did i understand wrong? Because the derivative of that would just be zero...?
You have correctly found that $x=1$ when $f(x) = 9$, but to find the derivative of $g(x)$ at this point, you should apply $x=1$ to the derivative of $f(x)$ as given by skaa in the post I quoted above.

June 4th, 2015, 08:03 AM   #7
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 Originally Posted by chaarleey f(x)= 9 --> x=1 Derivative of g(x) is 1/ the derivative of f(1)? or did i understand wrong? Because the derivative of that would just be zero...?
No, take the derivative of f(x) and evaluate that at x= 1.

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