| 
||||||||||||
  | 
||||||||||||
  | ||||||||||||
| 
 Hi Elvina. The distance from the origin to any point (x,y) is simply sqrt(x2 + y2) (you can see this by drawing a right triangle to it and using the Pythagorean Theorem). Therefore the distance d to the point (x,y) where y = ex can be found by calculating 
 Note that d is always positive, so the smallest d means the smallest d2. Thus you can ignore the d2 (i.e. we are ignoring the square root from earlier) and just evaluate df/dx where f(x) = x2 + e2x. When you get this derivative, set it equal to zero and solve for x. This gives you the value of x at which y = ex is closest to the origin. If you need further assistance write back. Cheers,  | ||||||||||||
  | 
||||||||||||
|     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.  | ||||||||||||