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DeoxNA
Group Title
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?
 one year ago
 one year ago
DeoxNA Group Title
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?
 one year ago
 one year ago

This Question is Closed

mark_o. Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
The derivative is:\[f'(x)=\frac{ 2(x^2+6)2x \cdot 2x }{ (x^2+6)^2 }=\frac{ 2x^2+124x^2 }{ (x^2+6)^2 }=\frac{ 122x^2 }{ (x^2+6)^2 }\]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:\[f'(x)=0 \Leftrightarrow 122x^2=0 \Leftrightarrow 6x^2=0 \Leftrightarrow (\sqrt{6} x)(\sqrt{6} + x)=0\]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\[[\sqrt 6, \infty)\]
 one year ago

DeoxNA Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago
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