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Andreyy89
3 years ago
6

Winter temperatures tend to be cold in the city of Johnstown. The table of values represents the temperature of Johnstown during

one winter week.
t | f(x)
(days)| (°F)
____|_____________
1 | 6
2 | 5
3 | 1
4 | -2
5 | -1
6 | 0
7 | -3

Part A
Use the table to approximate the key features of the function. Find the extrema, zeros, end behavior, increasing and decreasing intervals, and positive and negative intervals.

Part B
Interpret the key features from part A in the context of the problem.

Part C
Interpret the domain and the range of the function in the context of the problem.
Mathematics
1 answer:
trapecia [35]3 years ago
6 0

Answer:

Part A Part B Part C explained

Step-by-step explanation:

PART A: Extrema: relative minimums (4,-2) and maybe (7,-3) (uncertain because it’s an endpoint), relative maximums (6,0) and maybe (1,6) (uncertain because it’s an endpoint)

Zeros: (6,0), somewhere between t = 3 and t = 4 since the graph changes from positive to negative, and somewhere possibly between t = 6 and t = 7 unless the graph hits (6,0) and stays negative.

End behavior: We can guess that as x approaches infinity, the functions approaches negative infinity, and as x approaches negative infinity, the function approaches infinity.

Intervals of increase and decrease: Increasing on (4, 6), decreasing on (1, 4) and (6, 7)

PART B: The relative minimums indicate that the two lowest temperatures occurred on day 4 at -2°F and day 7 at -3°F. The relative maximums indicate that the weekly highs were day 1 at 6°F and day 6 at 0°F.

The zeros of the function represent when the temperature in Johnstown was 0°F. This happened sometime between days 3 and 4 and on day 6.

In the context of the problem, it doesn’t make sense to go an infinite number of degrees below zero. And, the end behavior is ignored because of the restricted range.

The intervals of increase indicate when the temperature is rising, and the intervals of decrease indicate when the temperature is dropping. The intervals where the values are positive indicate when the temperature is above 0°F. The intervals where the values are negative indicate when the temperature is below 0°F.

PART C: The domain is restricted to the number of days the town recorded the temperature. So, the domain is [1, 7].

The range represents the range of temperatures of Johnstown over the course of one week. So, the range is [-3, 6].

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We are given the equation:

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Where <em>P</em> is the number of eggs laid, <em>x</em> is the number of workers, and <em>y</em> is the daily operating budget (assuming in US dollars $).

A)

We want to find dy/dx.

So, let’s find our equation in terms of <em>x</em>. We can raise both sides to 10/7. Hence:

\displaystyle P^\frac{10}{7}=\Big(x^\frac{3}{10}y^\frac{7}{10}\Big)^\frac{10}{7}

Simplify:

\displaystyle P^\frac{10}{7}=x^\frac{3}{7}y

Divide both sides by<em> </em>the <em>x</em> term to acquire:

\displaystyle y=P^\frac{10}{7}x^{-\frac{3}{7}}

Take the derivative of both sides with respect to <em>x: </em>

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\Big[P^\frac{10}{7}x^{-\frac{3}{7}}\Big]

Apply power rule. Note that P is simply a constant. Hence:

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Simplify. Hence, our derivative is:

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We want to evaluate the derivative when <em>x</em> is 30 and when <em>y</em> is $10,000.

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Hence, at <em>x</em> = 30 and at <em>y</em> = 10,000, <em>P </em>is:

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\displaystyle \frac{dy}{dx}=-\frac{3}{7}\Big(30^{0.3}(10000^{0.7})\Big)^\frac{10}{7}\Big(30^{-\frac{10}{7}}\Big)

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