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Mathematics
OpenStudy (anonymous):

@amistre64 Show that\[t-\frac{t^2}2\leq \ln(1+t)\leq t\]for \(t>-1\). Do you know how I can do this?

OpenStudy (anonymous):

Nicely done @pre-algebra ;)

OpenStudy (anonymous):

@Zarkon @JamesJ

OpenStudy (jamesj):

Consider the function f(t) = t - ln(1+t) Now see what to do?

OpenStudy (anonymous):

I differentiated it and observed that it attains its global minimum at \(t=0\). Similarly, I took\[f(t)=t-\frac{t^2}{2}-\ln(1+t)\]and observed that it attains its global maximum at \(t=0\). Is this thus sufficient to imply that the inequality holds?

OpenStudy (jamesj):

Yes, because the functions are continuous on the interval of interest.

OpenStudy (zarkon):

your inequality is not true

OpenStudy (anonymous):

What's the counterexample?

OpenStudy (zarkon):

t=-1/2

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