Not sure if this is in calculus section as I am not sure of approach.
f(x) = (a+b+x)/(3abx) , x>0 and a ,b are positive real numbers
i)Show that the minimum value of f(x) occurs when x = (a + b)/2
Thanks. No idea how to approach. Tried differentiation but ended with
f'(x) = -(a+b)/(3abx^2) which does not have turning point.
You have differentiated this correctly: the given function does not have a turning point or a minimum value.
Originally Posted by Lukybear
The likeliest explanation is that there is a mistake in the question (or you have copied it incorrectly). If the correct version is then there is a minimum at x = (a + b)/2.
Ah yes. That would be most likely explanation as it is proving inequalities. Thanks.