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$$f(x) = x - (1/x) ; [3,1]$$ What is the first and second derivative of this function?

I think I found the first derivative $f'(x) = (x-1)(x+1)/x$

Cookie
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Kay
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  • Consider your function as $ \ f(x) \ = \ x^1 \ - \ x^{-1} \ $ and check your differentiation again... Also, are you being asked to find the absolute extremes on the interval $ \ [ 1 \ , \ 3 ] \ $ ? – colormegone May 20 '14 at 02:06

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note that $f'(x)=1+\frac{1}{x^2}>0$ for all $x\in[1,3]$ then $f$ is increasing in $[1,3]$, and how $f'(x)=0$ not have solutions then $f$ reach minimun in $f(1)=0$ and maximun in $f(3)=8/3$

AsdrubalBeltran
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