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f(x)=2x/(x^2+6) for x>=b (b is a real number) 1. Find f'(x) 2. Hence find the smallest exact value of b for which the inverse function f^(-1) exists. My question is: isn't b arbitrary? Couldn't I just pick any real number and then the domain of the inverse would just be x>f(b) since the domain and range switch between inverses. Except I couldn't pick an imaginary number, which curiously is also the domain restriction of f(x) and f'(x) i.e. sqrt(-6). Also, what is the purpose of finding the derivative?

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integral of 2x/(x^2+6) let u=x^2+6 du=2x dx therefore 1ntegral of du/u=ln u +C integral of 2x/(x^2+6)dx=ln( ? ) +C
b is not arbitrary. The question is a little unclear, imo. See attached Geogebra graph. As you can see, if b becomes too small, f doesn't have an inverse anymore. The minimum value for b is where f has its max. That's the reason why you have to differentiate f: to calculate where it has its max.
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The derivative is:\[f'(x)=\frac{ 2(x^2+6)-2x \cdot 2x }{ (x^2+6)^2 }=\frac{ 2x^2+12-4x^2 }{ (x^2+6)^2 }=\frac{ 12-2x^2 }{ (x^2+6)^2 }\]To find the extremes of f we have to solve f'(x)=0. Looking at the denominator of f', we see that it is greater than 0 for every x, so we don't bother with it:\[f'(x)=0 \Leftrightarrow 12-2x^2=0 \Leftrightarrow 6-x^2=0 \Leftrightarrow (\sqrt{6} -x)(\sqrt{6} + x)=0\]Because x > b and b >0 (see graph) we only have one solution: x = √6. For this value, f has its maximum. So taking b = √6 ensures we have an inverse of f on the interval\[[\sqrt 6, \infty)\]

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Thanks! What was confusing me was that (and I know that it sound obvious) f^(-1) HAD to be a function and hence pass the vertical line test. I was just thinking about drawing the inverse of the function and not realizing that the inverse had to be a function and not a relation.

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