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.\nf(x)=5x^{3}e^{-x};-1,5\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 minimum is at x =, but there is no absolute maximum.\nb. the absolute maximum is at x = and the absolute minimum is at x =.\nc. the absolute maximum is at x =, but there is no absolute minimum.\nd. there are no absolute extreme values for f(x) on -1,5.
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}(3 - x)=0) and (e^{-x}>0) for all (x), then (x = 0) or (x=3) (because (5x^{2}=0) gives (x = 0) and (3 - x=0) gives (x = 3))
Step3: Evaluate (f(x)) at critical points and endpoints
- For (x=-1): (f(-1)=5\times(-1)^{3}\times e^{-(-1)}=-5e\approx - 13.59)
- For (x = 0): (f(0)=5\times0^{3}\times e^{-0}=0)
- For (x = 3): (f(3)=5\times3^{3}\times e^{-3}=\frac{135}{e^{3}}\approx 6.79)
- For (x = 5): (f(5)=5\times5^{3}\times e^{-5}=\frac{625}{e^{5}}\approx 4.19)
Answer:
B. The absolute maximum is (\frac{135}{e^{3}}) at (x = 3) and the absolute minimum is (-5e) at (x=-1)