closed interval method: problem 3\n(1 point)\nconsider the function $f(x)=x e^{-8 x}, quad 0 leq x leq…

closed interval method: problem 3\n(1 point)\nconsider the function $f(x)=x e^{-8 x}, quad 0 leq x leq 2$.\nthis function has an absolute minimum value equal to:\nwhich is attained at $x=$\nand an absolute maximum value equal to:\nwhich is attained at $x=$\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\nemail instructor\npage generated october 15, 2025 at 2:00:04 pm cdt\nwebwork @ 1996-2024 | theme: math4 | ww_version: 2.19 | pg_version 2.19

closed interval method: problem 3\n(1 point)\nconsider the function $f(x)=x e^{-8 x}, quad 0 leq x leq 2$.\nthis function has an absolute minimum value equal to:\nwhich is attained at $x=$\nand an absolute maximum value equal to:\nwhich is attained at $x=$\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\nemail instructor\npage generated october 15, 2025 at 2:00:04 pm cdt\nwebwork @ 1996-2024 | theme: math4 | ww_version: 2.19 | pg_version 2.19

Answer

Explanation:

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

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x) and (v=e^{-8x}). (u^\prime=1), (v^\prime=- 8e^{-8x}) (f^\prime(x)=e^{-8x}-8xe^{-8x}=e^{-8x}(1 - 8x))

Step2: Find the critical points

Set (f^\prime(x) = 0), since (e^{-8x}>0) for all (x), then (1 - 8x=0), so (x=\frac{1}{8})

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

  • When (x = 0): (f(0)=0\times e^{-8\times0}=0)
  • When (x=\frac{1}{8}): (f(\frac{1}{8})=\frac{1}{8}e^{-8\times\frac{1}{8}}=\frac{1}{8e})
  • When (x = 2): (f(2)=2e^{-8\times2}=2e^{-16})

Since (2e^{-16}\approx2\times\frac{1}{e^{16}}\approx0) (because (e^{16}\gg1))

Answer:

The absolute minimum value is (2e^{-16}) which is attained at (x = 2) The absolute maximum value is (\frac{1}{8e}) which is attained at (x=\frac{1}{8})