Question

question 6 (1 point)
the function ( f(x)=e^{x}(x - 3) ) has a critical point at ( x=)
. (enter a number for your answer.)
view hint for question 6
question 7 (1 point)
the function ( |2x - 10| ) has a singular point at ( x=)
. (enter a number for your answer.)
view hint for question 7
question 8 (2 points)
let ( f(x)=x-\frac{64x}{x + 4} ). the absolute maximum on the interval ( 0,13 ) occurs at ( x=)
. the absolute minimum on this interval
occurs at ( x=)
. note that here we are
looking for the ( x ) value (not the range value). be sure to enter a number for each
answer.
view hint for question 8

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 \(\lim_{x
ightarrow5^{-}}\frac{\vert2x - 10\vert-(2\times5 - 10)}{x - 5}=- 2\) and the right - hand derivative \(\lim_{x
ightarrow5^{+}}\frac{\vert2x - 10\vert-(2\times5 - 10)}{x - 5}=2\). 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 \(x=-20
otin[0,13]\))

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 \(x = 12\)