f = e^(x^2-2x)
f' = (2x-2) e^(x^2-2x)
since e^(x^2-2x) is always positive, f'=0 at x=1
f" = (4x^2-8x+6)e^(x^2-2x)
f"(1) > 0 so f(1)=1/e is a local minimum.
Now just check f(0) and f(2) to find the absolute extrema for the interval.
Find the largest and
smallest values of the given function over the prescribed closed, bounded interval.
F(x)=e^{x^2-2x} for 0<x<2(It,s
bounded interval)
2 answers
F'= (x^2-2x)(2x-2)e^(x^2-2x)=0
2x(x-2)(x-1)e^(x^2-2x)=0
x=0, x=2, x=1 are solutions. So the only solution in the interval has to be 0+episilon, 2-epsilon, and 1
So we test:
at x=zero + epsilon
F= 1
at x=2-epsilion
F= e^0+=1
at x=1
F=e^-1=1/e <====smallest
and the other two are largest values.
2x(x-2)(x-1)e^(x^2-2x)=0
x=0, x=2, x=1 are solutions. So the only solution in the interval has to be 0+episilon, 2-epsilon, and 1
So we test:
at x=zero + epsilon
F= 1
at x=2-epsilion
F= e^0+=1
at x=1
F=e^-1=1/e <====smallest
and the other two are largest values.
4 answers