Question

@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?

Answer 1

Nicely done @pre-algebra ;)

Answer 2

@Zarkon @JamesJ

Answer 3

Consider the function
f(t) = t - ln(1+t)

Now see what to do?

Answer 4

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?

Answer 5

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

Answer 6

your inequality is not true

Answer 7

What's the counterexample?

Answer 8

t=-1/2

Answer 9

http://www.wolframalpha.com/input/?i=t-%28t^2%29%2F2%3C%3Dln%281%2Bt%29