determine the location and value of the absolute extreme values of f on the given interval, if they…

determine the location and value of the absolute extreme values of f on the given interval, if they exist.\n$f(x)=5x^{3}e^{-x};-1,6$\nselect the correct choice below and, if necessary, fill in the answer boxes to complete your choice.\n(type exact answers. use a comma to separate answers as needed.)\na. the absolute maximum is at ( x = ) and the absolute minimum is at ( x = ).\nb. the absolute minimum is at ( x = ), but there is no absolute maximum.\nc. the absolute maximum is at ( x = ), but there is no absolute minimum.\nd. there are no absolute extreme values for ( f(x) ) on ( -1,6 ).

determine the location and value of the absolute extreme values of f on the given interval, if they exist.\n$f(x)=5x^{3}e^{-x};-1,6$\nselect the correct choice below and, if necessary, fill in the answer boxes to complete your choice.\n(type exact answers. use a comma to separate answers as needed.)\na. the absolute maximum is at ( x = ) and the absolute minimum is at ( x = ).\nb. the absolute minimum is at ( x = ), but there is no absolute maximum.\nc. the absolute maximum is at ( x = ), but there is no absolute minimum.\nd. there are no absolute extreme values for ( f(x) ) on ( -1,6 ).

Answer

Explanation:

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

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = 5x^{3}) and (v=e^{-x}). (u^\prime=15x^{2}), (v^\prime=-e^{-x}) (f^\prime(x)=15x^{2}e^{-x}-5x^{3}e^{-x}=5x^{2}e^{-x}(3 - x))

Step2: Find the critical points

Set (f^\prime(x) = 0). Since (5x^{2}e^{-x}\gt0) for (x\in[-1,6]) (except (x = 0), and (f^\prime(0)=0)), then (3 - x=0) gives (x = 3). Also, (x = 0) is a critical point (where (f^\prime(x)=0) due to (x^{2}) factor).

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

  • For (x=-1): (f(-1)=5(-1)^{3}e^{-(-1)}=-5e\approx - 13.59)
  • For (x = 0): (f(0)=5(0)^{3}e^{-0}=0)
  • For (x = 3): (f(3)=5(3)^{3}e^{-3}=\frac{135}{e^{3}}\approx5.49)
  • For (x = 6): (f(6)=5(6)^{3}e^{-6}=\frac{1080}{e^{6}}\approx2.19)

Answer:

A. The absolute maximum is (\frac{135}{e^{3}}) at (x = 3) and the absolute minimum is (-5e) at (x=-1)