Recent content by fyziky

thank you hgfalling you are wise. Since f(x) approaches infinity from both directions and we do have a max/min value it must be a min. also since it approaches infinity from both sides if the min is negative it will cross the X axis twice so to have a unique solution thi min must be the zero...

thank you very much for the hints but unfortunately I am still stuck. I still am unable to solve for the zeros, and when i use maple the answer i get is dependent on the lambert function. when i try different plots varying my b i get one zero when b=1/2e. maybe that is the solution but i still...

the question is simple but i can't seem to think of a solution. for what b>0 does ln(x)=b x^{2} have exactly one solution, and not 0. I've tried playing with ln rules but can't seem to think of a solution

Hello all, My question is as follows: f:[1,\infty) is defined by f(x)=\sqrt{x}+2x (1\leqx<\infty) Given \epsilon>0 find \delta>0 such that if |x-y|<\delta then |f(x)-f(y)|<\epsilon It seems I am being asked to show continuity, and not uniform continuity, so my approach is this, but I am...