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Is that

When f(x) is increasing, x is smaller than zero and x is greater than 2?

When f(x) is decreasing, x is between 0 and 2 but not equal to 1?

The interval where f(x) is concave down is x is smaller than 1?

The interval where f(x) is concave up is x is greater than 1?

The locam maxima and local minima are -1 and 3 respectively?

And there is no inflection point?

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  • $\begingroup$ add how you reached your conclusions and add a tag like proof-verification. $\endgroup$
    – drhab
    Feb 9, 2014 at 10:58
  • $\begingroup$ I don't understand why in your first two statements you write in that odd (in the best of the cases) way, not to say in that wrong way. And in the first line what you write after the first $\;x\;$ describes the empty set. If at all (I haven't checked) it should be "whenever $\;x\;$ is smaller than zero or $\;x\;$ is greater than two, $\;f\;$ is increasing". $\endgroup$
    – DonAntonio
    Feb 9, 2014 at 11:03

1 Answer 1

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Hint: Notice that $f(x)=x+\frac{1}{x-1}$, so $f'(x)=1-\frac{1}{(x-1)^2}$ and $f''(x)=\frac{2}{(x-1)^3}$, which simplifies the calculations a lot.

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