Question

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)

Answer 1

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.

Answer 2

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.