question 6 (1 point)\nthe function ( f(x)=e^{x}(x - 3) ) has a critical point at ( x=)\n. (enter a number…

Explanation:

Question 6

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

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = e^{x}), (u^\prime=e^{x}), (v=x - 3), (v^\prime = 1). (f^\prime(x)=e^{x}(x - 3)+e^{x}\times1=e^{x}(x-3 + 1)=e^{x}(x - 2))

Step2: Set the derivative equal to zero

Since (e^{x}>0) for all (x\in R), set (x - 2=0).

Question 7

Step1: Analyze the function (y=\vert2x - 10\vert)

The function (y=\vert2x - 10\vert=\begin{cases}2x - 10, & x\geq5\-(2x - 10),&x<5\end{cases}) The derivative of (y = 2x-10) is (y^\prime=2) for (x > 5) and the derivative of (y=-(2x - 10)) is (y^\prime=-2) for (x < 5). At (x = 5), the left - hand derivative (\lim_{x\rightarrow5^{-}}\frac{\vert2x - 10\vert-(2\times5 - 10)}{x - 5}=- 2) and the right - hand derivative (\lim_{x\rightarrow5^{+}}\frac{\vert2x - 10\vert-(2\times5 - 10)}{x - 5}=2). The derivative does not exist at (x = 5).

Question 8

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

(f(x)=x-\frac{64x}{x + 4}=x-\frac{64x+256-256}{x + 4}=x-64+\frac{256}{x + 4}) (f^\prime(x)=1-\frac{256}{(x + 4)^{2}}) Set (f^\prime(x)=0), then (1-\frac{256}{(x + 4)^{2}}=0) (\frac{256}{(x + 4)^{2}}=1) ((x + 4)^{2}=256) (x+4=\pm16) (x = 12) or (x=-20) (but (x=-20\notin[0,13]))

Step2: Evaluate (f(x)) at critical and endpoints

(f(0)=0-\frac{64\times0}{0 + 4}=0) (f(12)=12-\frac{64\times12}{12 + 4}=12-48=-36) (f(13)=13-\frac{64\times13}{13 + 4}=13-\frac{832}{17}=\frac{221-832}{17}=-\frac{611}{17}\approx - 35.94)

Answer:

Question 6: (2) Question 7: (5) Question 8: Absolute maximum at (x = 0), absolute minimum at (x = 12)