Find all global extrema and where they occur on the given interval.

f(x)= e^(-3x-x^(2)) interval [-2,0]

find the global maximum value and where it occurs

Find the global minimum value and where it occurs

I found the derivative of the function and identified the critical point as being x=-3/2 but I am lost as where to continue from here. Help me out please!

Critical point, this is where f' = 0. This is the ONLY point on this interval where the slope changes from pos to neg or vice versa.

f' = (-3 -2x)e^(-3x-x^(2)

Ok, so x = - 3/2. Is this a peak or a valley? Since there is only one change, one critical point, it is a min or a max. Therefore the extremes, x = -2 and x = 0 will give you the other.

Two methods
A. Take the second derivative. If it's positive, critical point is a min. If neg, cp is a max. You still have to solve for the other two points. So I like B. better

B. Solve for all three points. Determine the min and max

f(-3/2) = e^(-3(-3/2)-(-3/2)^(2)) = e^(9/2 - 9/4) = e^(9/4) = 9.49

f(-2) = e^(-3(-2) - (-2)^2) = e^(6 - 4) = e^2 = 7.39

f(0) = e^(-3(0) - (0)^2) = e^0 = 1

Max: 9.49 at x = -3/2 Min: 1 at x = 0

Plug and chug has its advantages.

ƒ'(x) = (-3 - 2x)e^(-3x - x²)
e^(-3x - x²) does not equal zero anywhere in [-2, 0], so if ƒ'(x) = 0, then -3 - 2x = 0 --> 2x = -3
--> x = -1.5
ƒ''(x) = (-3 - 2x)²e^(-3x - x²) - 2e^(-3x - x²)
ƒ''(-1.5) = -2e^(2.25) which is less than zero, so -1.5 is a local maximum
In fact, it is a global maximum in [-2.0]. ... {It's a global maximum, period.}
The global maximum value, then, is ƒ(-1.5) = 9.4877 (rounded)
The global minimum in [-2, 0] is ƒ(0) = 1.

I know you found the derivative and the critical point, but let me just take you through the steps for future reference. Since you found the derivative, I am guessing that it is not a challenge for you.