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)

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.

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.

To find the largest and smallest values of the function F(x) over the closed, bounded interval (0, 2), we need to first find the critical points and the endpoints of the interval.

1. Finding the critical points:
To find the critical points, we need to take the derivative of the function F(x) and set it equal to zero.

F'(x) = d/dx(e^(x^2-2x))
= (2x - 2)e^(x^2-2x)

Setting F'(x) = 0, we have:
(2x - 2)e^(x^2-2x) = 0

Since e^(x^2-2x) is always positive, the critical point occurs when (2x - 2) = 0, which is x = 1.

2. Determining the values at the endpoints:
Now, we evaluate the function F(x) at the endpoints of the interval, which are x = 0 and x = 2.

F(0) = e^(0^2-2*0) = e^0 = 1
F(2) = e^(2^2-2*2) = e^0 = 1

3. Determining the maximum and minimum values:
To determine whether the critical point and endpoints correspond to a maximum or minimum, we can use either the first or second derivative test. In this case, let's use the second derivative test.

F''(x) = d^2/dx^2(e^(x^2-2x))
= (2 - 4x + 2x^2)e^(x^2-2x)

Now, evaluate F''(x) at x = 1:
F''(1) = (2 - 4*1 + 2*1^2)e^(1^2-2*1) = 0e^(-1) = 0

Since F''(1) = 0, the second derivative test is inconclusive. However, since the function F(x) is continuous over the closed, bounded interval (0, 2), we can conclude that the maximum and minimum values occur either at the critical point or at the endpoints.

4. Evaluating F(x) at the critical point and the endpoints:
F(1) = e^(1^2-2*1) = e^(-1) ≈ 0.3679
F(0) = 1
F(2) = 1

From the above calculations, we can see that the largest value of F(x) over the interval (0, 2) is 1, and the smallest value is approximately 0.3679.

Therefore, the largest value of the function F(x) over the prescribed interval is 1, and the smallest value is approximately 0.3679.