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):

OpenStudy (anonymous):

What's the counterexample?

OpenStudy (zarkon):

t=-1/2

OpenStudy (zarkon):