As others have said, std::max_element() and std::min_element() return iterators, which need to be dereferenced to obtain the value.
The advantage of returning an iterator (rather than just the value) is that it allows you to determine the position of the (first) element in the container with the maximum (or minimum) value.
For example (using C++11 for brevity):
#include <vector>
#include <algorithm>
#include <iostream>
int main()
{
std::vector<double> v {1.0, 2.0, 3.0, 4.0, 5.0, 1.0, 2.0, 3.0, 4.0, 5.0};
auto biggest = std::max_element(std::begin(v), std::end(v));
std::cout << "Max element is " << *biggest
<< " at position " << std::distance(std::begin(v), biggest) << std::endl;
auto smallest = std::min_element(std::begin(v), std::end(v));
std::cout << "min element is " << *smallest
<< " at position " << std::distance(std::begin(v), smallest) << std::endl;
}
This yields:
Max element is 5 at position 4
min element is 1 at position 0
Using std::minmax_element() as suggested in the comments above may be faster for large data sets, but may give slightly different results. The values for my example above would be the same, but the position of the "max" element would be 9 since...
If several elements are equivalent to the largest element, the iterator to the last such element is returned.