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1.) Compute the domain of the function $$f(x)=\frac{1}{\lfloor x^2-7x+13\rfloor}.$$

 

2.) Allie and Betty play a game where they take turns rolling a standard die. If a player rolls $n$, she is awarded $f(n)$ points, where \[f(n) = \left\{\begin{array}{cl} 6 & \text{ if }n\text{ is a multiple of 2 and 3}, \\2 & \text{ if }n\text{ is only a multiple of 2}, \\0 & \text{ if }n\text{ is not a multiple of 2}.\end{array}\right.\]Allie rolls the die four times and gets a 5, 4, 1, and 2. Betty rolls and gets 6, 3, 3, and 2. What is the product of Allie's total points and Betty's total points?

 Apr 27, 2019
edited by lolzforlife  Apr 27, 2019
 #1
avatar+6251 
+6

\(f(x)=\dfrac{1}{\lfloor x^2-7x+13\rfloor}\)

 

\(\text{The floor function will map }[0,1) \to 0\\ \text{which results in undefined division by 0}\\ \text{Thus the interval }[0,1) \text{ must be restricted from the floor function}\\ \text{thus }x^2-7x+13<0 \text{ OR } x^2-7x+13 \geq 1\)

 

\(x^2 - 7x+13 < 0\\ \left(x-\dfrac 7 2\right)^2 +\dfrac 3 4 < 0\\ \text{this clearly cannot occur with real numbers}\)

 

\(x^2 - 7x+13 \geq 1\\ x^2 - 7x + 12 \geq 0\\ \left(x-\dfrac 7 2\right)^2 - \dfrac 1 4 \geq 0\\ \left(x-\dfrac 7 2\right)^2 \geq \dfrac 1 4\\ \left(x-\dfrac 7 2\right) \geq \dfrac 1 2 \text{ OR }\left(x-\dfrac 7 2\right) \leq -\dfrac 1 2\\ x \geq 4 \text{ OR }x \leq 3\\ x \in (-\infty, 3]\cup [4,\infty)\)

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 Apr 27, 2019
 #2
avatar+46 
+2

Can you answer number 2 please?

lolzforlife  Apr 27, 2019
 #3
avatar+118667 
+3

How about thanking Rom for the answer you already have.

Or asking him questions if your do not fully understand?

Melody  Apr 28, 2019
 #4
avatar+46 
+2

Sorry for sounding impatient.  I figured out number 2 so it no longer has to be answered!  Thanks @Rom and @Melody.

lolzforlife  Apr 28, 2019

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