monotonicity

prove that for +ve values of x
x2 > (1+x)[log(1+x)]2

5 Answers

62
Lokesh Verma ·

u need to prove that the derivative is greater for the LHS than the RHS for all values of x!

This is slightly tricky.

Here first derivative will nto give the result.

So u need to take the 2nd derivative (derivative of the derivative)

Try these. U will get the result :)

62
Lokesh Verma ·

at one stage u will get . x>log(x+1)

which is true (so u will need to use this :)

1
big looser ......... ·

thanks brother, i got it

1
Grandmaster ·

but nishant how can you prove that graphs aren't intersecting thgats they don't have a common root.
and only for some values of x its greater

e.g. f(x) = -(x+2)2
g(x)=x2-3

62
Lokesh Verma ·

in the above question the values are equal at x=0.

And everywhere, one is increasing mroe than the other.

so ...

for ur question the derivative of f(x) is not greater than that of g(x) for all x. hence that does not hold !

Your Answer

Close [X]