Derivative increasing without bounds

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SUMMARY

If the derivative f'(x) is greater than 0 for all real values of x, it does not guarantee that the function f(x) increases without bounds. The function f(x) = 2x/sqrt(x² + 2) serves as a counterexample, where its derivative f'(x) = 4/(x² + 2)^(3/2) is always positive, yet f(x) approaches a horizontal asymptote of 2 as x approaches infinity. This demonstrates that a positive derivative does not imply unbounded growth.

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if f'(x) > 0 for all real values of x then x increases without bounds. I thought that was true but in the back of the book it says false and uses f(x)=2x/sqrt(x2+2) as an example. i worked out the derivative and got f'(x) = 4/(x2+2)3/2.

how does that show that the first sentence is false? I'm quite confused abou this
 
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You got the derivative right. And it's always positive, yes. But f(x) doesn't increase without bound. It approaches 2 as x->inf. Can you show that?
 
asymptotes :D
 

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