4. in a video game, the amount of damage a player’s character can do is calculated based function
$d(l) = \frac{10l^2}{l^2 + 6}$
(a) calculate $d(2)$. interpret your answer in the context of the problem using units and
(b) calculate (and interpret in context) the aroc of $d(l)$ on the interval $0, 2$.
(c) is $d(l)$ one-to-one? explain your answer.
(d) find $l$ such that $d(l) = 6$.
(e) state the end behavior of $d(l)$. explain what the end behavior means in the context

Question

Accepted Answer

### Part (a)
# Explanation:
## Step 1: Substitute $ L = 2 $ into $ D(L) $
We have the function $ D(L)=\frac{10L^{2}}{L^{2}+6} $. Substitute $ L = 2 $ into the function:
$ D(2)=\frac{10\times(2)^{2}}{(2)^{2}+6} $
## Step 2: Simplify the expression
First, calculate the numerator and the denominator:
- Numerator: $ 10\times(2)^{2}=10\times4 = 40 $
- Denominator: $ (2)^{2}+6=4 + 6=10 $
Then, $ D(2)=\frac{40}{10}=4 $

Interpretation: When the player's character has a level $ L = 2 $, the amount of damage the character can do is 4 (the unit of damage is not specified in the problem, but we can say that at level 2, the damage output is 4 units of damage).

# Answer: $ D(2)=4 $

### Part (b)
# Explanation:
## Step 1: Recall the formula for AROC
The Average Rate of Change (AROC) of a function $ f(x) $ on the interval $[a,b]$ is given by $ \frac{f(b)-f(a)}{b - a} $. For our function $ D(L) $ on the interval $[0,2]$, $ a = 0 $ and $ b=2 $.
First, find $ D(0) $ and $ D(2) $ (we already know $ D(2) = 4 $ from part (a)).
Calculate $ D(0) $: $ D(0)=\frac{10\times(0)^{2}}{(0)^{2}+6}=\frac{0}{6}=0 $
## Step 2: Calculate AROC
Using the AROC formula: $ \text{AROC}=\frac{D(2)-D(0)}{2-0}=\frac{4 - 0}{2}=2 $

Interpretation: On the interval from level $ L = 0 $ to $ L=2 $, the average rate of change of the damage function is 2. This means that, on average, for each unit increase in the level of the character (from level 0 to level 2), the damage output of the character increases by 2 units of damage per level.

# Answer: The AROC of $ D(L) $ on $[0,2]$ is $ 2 $

### Part (c)
# Explanation:
## Step 1: Recall the definition of a one - to - one function
A function $ y = f(x) $ is one - to - one if for every $ y $ value in the range, there is exactly one $ x $ value in the domain such that $ f(x)=y $. A function is one - to - one if and only if it passes the horizontal line test, or if $ f(a)=f(b) $ implies $ a = b $ for all $ a,b $ in the domain.
Let's assume $ D(a)=D(b) $, so $ \frac{10a^{2}}{a^{2}+6}=\frac{10b^{2}}{b^{2}+6} $
## Step 2: Cross - multiply and simplify
Cross - multiply: $ 10a^{2}(b^{2}+6)=10b^{2}(a^{2}+6) $
Expand both sides: $ 10a^{2}b^{2}+60a^{2}=10a^{2}b^{2}+60b^{2} $
Subtract $ 10a^{2}b^{2} $ from both sides: $ 60a^{2}=60b^{2} $
Divide both sides by 60: $ a^{2}=b^{2} $
Take square roots: $ a=\pm b $
Since $ L $ represents the level of a character, $ L\geq0 $ (level cannot be negative). So if $ a^{2}=b^{2} $ and $ a,b\geq0 $, then $ a = b $. Wait, but let's test with some values. Let's take $ L = 1 $ and $ L=- 1 $ (but $ L=-1 $ is not in the domain as level can't be negative). Wait, actually, for non - negative $ L $, let's see if we can find two different non - negative $ L $ values with the same $ D(L) $.
Suppose $ L_1 = \sqrt{3} $ and $ L_2=\sqrt{3} $ (same value). Wait, no, let's check the derivative (or analyze the function). The function $ D(L)=\frac{10L^{2}}{L^{2}+6}=10-\frac{60}{L^{2}+6} $. As $ L $ increases, $ D(L) $ increases (since the denominator $ L^{2}+6 $ increases, so $ \frac{60}{L^{2}+6} $ decreases, and $ 10-\frac{60}{L^{2}+6} $ increases). So the function is increasing for $ L\geq0 $ (since the domain of $ L $ for a character's level is non - negative). For an increasing function, if $ D(a)=D(b) $, then $ a = b $. So the function is one - to - one on its domain ($ L\geq0 $).

Wait, but let's check with the equation $ D(a)=D(b) $ for non - negative $ a $ and $ b $. From $ \frac{10a^{2}}{a^{2}+6}=\frac{10b^{2}}{b^{2}+6} $, we get $ 10a^{2}(b^{2}+6)=10b^{2}(a^{2}+6) $, $ 10a^{2}b^{2}+60a^{2}=10a^{2}b^{2}+60b^{2} $, $ 60a^{2}=60b^{2} $, $ a^{2}=b^{2} $. Since $ a\geq0 $ and $ b\geq0 $, we have $ a = b $. So the function is one - to - one.

# Answer: The function $ D(L) $ is one - to - one. Explanation: For non - negative levels $ L_1 $ and $ L_2 $, if $ D(L_1)=D(L_2) $, then $ \frac{10L_1^{2}}{L_1^{2}+6}=\frac{10L_2^{2}}{L_2^{2}+6} $, which simplifies to $ L_1^{2}=L_2^{2} $. Since $ L_1,L_2\geq0 $ (level can't be negative), we have $ L_1 = L_2 $. So the function is one - to - one.

### Part (d)
# Explanation:
## Step 1: Set up the equation
We want to find $ L $ such that $ D(L)=6 $. So we set $ \frac{10L^{2}}{L^{2}+6}=6 $
## Step 2: Solve for $ L $
Multiply both sides of the equation by $ L^{2}+6 $ (assuming $ L^{2}+6
eq0 $, which is always true for real $ L $):
$ 10L^{2}=6(L^{2}+6) $
Expand the right - hand side: $ 10L^{2}=6L^{2}+36 $
Subtract $ 6L^{2} $ from both sides: $ 4L^{2}=36 $
Divide both sides by 4: $ L^{2}=9 $
Take the square root of both sides: $ L=\pm3 $
Since $ L $ represents the level of a character, $ L\geq0 $, so we take $ L = 3 $ (we discard $ L=-3 $ as level cannot be negative).

# Answer: $ L = 3 $

### Part (e)
# Explanation:
## Step 1: Analyze the end - behavior
To find the end - behavior of $ D(L)=\frac{10L^{2}}{L^{2}+6} $ as $ L
ightarrow\infty $, we can divide the numerator and the denominator by $ L^{2} $:
$ D(L)=\frac{10L^{2}/L^{2}}{L^{2}/L^{2}+6/L^{2}}=\frac{10}{1 + \frac{6}{L^{2}}} $
As $ L
ightarrow\infty $, $ \frac{6}{L^{2}}
ightarrow0 $. So $ \lim_{L
ightarrow\infty}D(L)=\frac{10}{1+0}=10 $

Interpretation: As the level $ L $ of the character becomes very large (approaches infinity), the amount of damage $ D(L) $ that the character can do approaches 10 units of damage. This means that there is an upper limit of 10 on the damage output of the character, no matter how high the level of the character gets, the damage will get closer and closer to 10 (but never exceed 10, and will approach 10 as the level becomes extremely high).

# Answer: As $ L
ightarrow\infty $, $ D(L)
ightarrow10 $. In context, as the character's level becomes very large, the damage output approaches 10 units of damage.

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step 1: Substitute \( L = 2 \) into \( D(L) \)

Step 2: Simplify the expression

Step 1: Recall the formula for AROC

Step 2: Calculate AROC

Step 1: Recall the definition of a one - to - one function

Step 2: Cross - multiply and simplify

Answer:

Part (b)