consider the function ( f(x)=x e^{-2 x}, quad 0 leq x leq 2 ).\nthis function has an absolute minimum value…

Explanation:

Step1: Find the derivative of (f(x))

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x), (u^\prime=1), (v = e^{-2x}), (v^\prime=- 2e^{-2x}). (f^\prime(x)=e^{-2x}-2xe^{-2x}=e^{-2x}(1 - 2x))

Step2: Find the critical points

Set (f^\prime(x)=0). Since (e^{-2x}>0) for all (x), then (1 - 2x = 0), so (x=\frac{1}{2})

Step3: Evaluate (f(x)) at critical point and endpoints

• When (x = 0): (f(0)=0\times e^{-2\times0}=0)
• When (x=\frac{1}{2}): (f(\frac{1}{2})=\frac{1}{2}e^{-2\times\frac{1}{2}}=\frac{1}{2e}\approx0.184)
• When (x = 2): (f(2)=2e^{-2\times2}=2e^{-4}\approx0.037)

Answer:

The absolute minimum value is (0) which is attained at (x = 0), and the absolute maximum value is (\frac{1}{2e}\approx0.184) which is attained at (x=\frac{1}{2})