suppose that $f(x)=(8 - 2x)e^{x}$. note: several parts of this problem require answers entered in interval…

Explanation:

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

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = 8 - 2x) and (v=e^{x}). (u^\prime=- 2) and (v^\prime = e^{x}). So (f^\prime(x)=-2e^{x}+(8 - 2x)e^{x}=(6 - 2x)e^{x}).

Step2: Find the critical values

Set (f^\prime(x)=0). Since (e^{x}\gt0) for all (x), we solve (6 - 2x = 0), which gives (x = 3).

Answer:

A. (x = 3)

Step3: Determine intervals of increase and decrease

Choose test - points. For (x\lt3), let (x = 0), then (f^\prime(0)=(6-0)e^{0}=6\gt0), so (f(x)) is increasing on ((-\infty,3)). For (x\gt3), let (x = 4), then (f^\prime(4)=(6 - 8)e^{4}=-2e^{4}\lt0), so (f(x)) is decreasing on ((3,\infty)).

Answer:

B. ((-\infty,3)) C. ((3,\infty))

Step4: Find local maxima and minima

Since (f(x)) changes from increasing to decreasing at (x = 3), (f(3)=(8 - 2\times3)e^{3}=2e^{3}) is a local maximum.

Answer:

D. (x = 3) E. NONE

Step5: Find the second - derivative of (f(x))

(f^\prime(x)=(6 - 2x)e^{x}), using the product rule again with (u = 6 - 2x), (u^\prime=-2), (v = e^{x}), (v^\prime = e^{x}). Then (f^{\prime\prime}(x)=-2e^{x}+(6 - 2x)e^{x}=(4 - 2x)e^{x}).

Step6: Determine concavity

Set (f^{\prime\prime}(x)=0), then ((4 - 2x)e^{x}=0). Since (e^{x}\gt0) for all (x), we solve (4 - 2x = 0) and get (x = 2). For (x\lt2), let (x = 0), (f^{\prime\prime}(0)=(4-0)e^{0}=4\gt0), so (f(x)) is concave up on ((-\infty,2)). For (x\gt2), let (x = 3), (f^{\prime\prime}(3)=(4 - 6)e^{3}=-2e^{3}\lt0), so (f(x)) is concave down on ((2,\infty)).

Answer:

F. ((-\infty,2)) G. ((2,\infty))

Step7: Find inflection points

Since the concavity changes at (x = 2), the inflection point is at (x = 2).

Answer:

H. (x = 2)