summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of…

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of f(x)=6xe^(-0.5x). find the location of any local extrema of f(x). select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a. there is a local maximum at x = . there is no local minimum. (type an integer or a decimal. use a comma to separate answers as needed.) b. there is a local minimum at x = . there is no local maximum. (type an integer or a decimal. use a comma to separate answers as needed.) c. there is a local maximum at x = and there is a local minimum at x = . (type integers or decimals. use a comma to separate answers as needed.) d. there are no local extrema.

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of f(x)=6xe^(-0.5x). find the location of any local extrema of f(x). select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a. there is a local maximum at x = . there is no local minimum. (type an integer or a decimal. use a comma to separate answers as needed.) b. there is a local minimum at x = . there is no local maximum. (type an integer or a decimal. use a comma to separate answers as needed.) c. there is a local maximum at x = and there is a local minimum at x = . (type integers or decimals. use a comma to separate answers as needed.) d. there are no local extrema.

Answer

Explanation:

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

Use the product - rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = 6x) and (v=e^{- 0.5x}). (u^\prime=6) and (v^\prime=-0.5e^{-0.5x}). Then (f^\prime(x)=6e^{-0.5x}+6x(-0.5e^{-0.5x})=6e^{-0.5x}(1 - 0.5x)).

Step2: Set the derivative equal to zero

Set (f^\prime(x)=0). Since (e^{-0.5x}\neq0) for all real (x), we solve (1 - 0.5x = 0). [ \begin{align*} 1-0.5x&=0\ 0.5x&=1\ x& = 2 \end{align*} ]

Step3: Determine the nature of the critical point

Take the second - derivative of (f(x)). Using the product - rule on (f^\prime(x)=6e^{-0.5x}-3xe^{-0.5x}). Let (u = 6 - 3x) and (v = e^{-0.5x}), (u^\prime=-3) and (v^\prime=-0.5e^{-0.5x}). Then (f^{\prime\prime}(x)=-3e^{-0.5x}+(6 - 3x)(-0.5e^{-0.5x})=e^{-0.5x}(-3-3 + 1.5x)=e^{-0.5x}(1.5x - 6)). Evaluate (f^{\prime\prime}(2)): (f^{\prime\prime}(2)=e^{-1}(1.5\times2 - 6)=e^{-1}(3 - 6)=- 3e^{-1}<0). So (x = 2) is a local maximum.

Answer:

A. There is a local maximum at (x = 2). There is no local minimum.