Show that has a maximum and minimum, but the maximum value is less than the minimum value.

F(x)=x+

Taking first derivative and equating it to zero to find extreme points.

F^’(x)=1-

X²=1

x=1,x=-1

now to determine which of these is min. And max. We use second derivative.

f^||(x)=

f^||(1)=2 and f^||(-1)=-2

since f^||(1) is +ve it is minimum point while f^||(-1) is –ve it is maximum point

max value-> f(-1)=-1+=-2

min vaue-> f(1)=1+=2

hence maximum value is less than minimum value