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ASHA 777 [7]
2 years ago
9

The rabbit population on a small island is observed to be given by the function P(t) = 130t − 0.3t^4 + 1100 where t is the time

since observations of the island began.When is the maximum population attained What is the maximum population? When does the rabbit population disappear from the island?
Mathematics
1 answer:
Montano1993 [528]2 years ago
7 0
The maximum occurs when the derivative of the function is equal to zero.
P(t)=-0.3t^{4}+130t+1100 \\ P'(t)=-1.2t^{3}+130 \\ 0=-1.2t^{3}+130 \\ 1.2t^{3}=130 \\ t^{3}= \frac{325}{3}  \\ t=4.76702
Then evaluate the function for that time to find the maximum population.
P(t)=-0.3t^{4}+130t \\ P(4.76702)=-0.3*4.76702^{4}+130*4.76702+1100 \\ P(4.76702)=1564.79201
Depending on the teacher, the "correct" answer will either be the exact decimal answer or the greatest integer of that value since you cannot have part of a rabbit.
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If a(x) = 3x + 1 and b (x) = StartRoot x minus 4 EndRoot, what is the domain of (b circle a) (x)?
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Answer:

[1,\infty)

Step-by-step explanation:

b(x)=\sqrt{x-4}

a(x)=3x+1

Since we want to know the domain of (b \circ a)(x), let's first consider the domain of the inside function, that is, that of a(x)=3x+1. Every polynomial function has domain all real numbers.

So we can plug anything for function a and get a number back.

Now the other function is going to be worrisome because it has a square root. You cannot take square root of negative numbers if you are only considering real numbers which that is the case with most texts.

Let's find (b \circ a)(x) and simplify now.

(b \circ a)(x)

b(a(x))

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\sqrt{(3x+1)-4}

\sqrt{3x+1-4}

\sqrt{3x-3}

Now again we can only square root positive or zero numbers so we want 3x-3 \ge 0.

Let's solve this to find the domain of (b \circ a)(x).

3x-3 \ge 0

Add 3 on both sides:

3x \ge 3

Divide both sides by 3:

x \ge 1

So we want x to be a number greater than or equal to 1.

The option that says this is [1,\infty)

-------------------------------

Give an example why option A fails:

A number in the given set is -2.

a(x)=3x+1

b(x)=\sqrt{x-4}

So a(-2)=3(-2)+1=-6+1=-5 and b(-5)=\sqrt{-5-4}=\sqrt{-9} \text{ which is not real}.

Give an example why option B fails:

A number in the given set is 0.

a(x)=3x+1

b(x)=\sqrt{x-4}

So a(0)=3(0)+1=0+1=1 and b(1)=\sqrt{1-4}=\sqrt{-3} \text{ which is not real}.

Give an example why option D fails:

While all the numbers in set D work, there are more numbers outside that range of numbers that also work.

A number not in the given set that works is 3.

a(x)=3x+1

b(x)=\sqrt{x-4}

So a(3)=3(3)+1=9+1=10 and b(1)=\sqrt{10-4}=\sqrt{6} \text{ which is real}.

4 0
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12.2 - 2.44 = 9.76

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