Yes, and when you did it for (ii), you got e^x - 1, which you deduced was >=0 for x >=0. However, note that e^x - 1 equals 0 when x = 0. So this is already the best you can do using this simple method - if you take α>1 then dxd(ex−αx)=ex−α which is < 0 when x = 0.
Yes, and when you did it for (ii), you got e^x - 1, which you deduced was >=0 for x >=0. However, note that e^x - 1 equals 0 when x = 0. So this is already the best you can do using this simple method - if you take α>1 then dxd(ex−αx)=ex−α which is < 0 when x = 0.
I see what you're saying. So how would dividing by x help?