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2 years ago

If f(x)=x^3−x, what is the average rate of change of f(x) over the interval [1, 5]? a)29.5 b)30 c)43 d)120

1 answer:

7 0

$\bf slope = m = \cfrac{rise}{run} \implies 
\cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby 
\begin{array}{llll}
average~rate\\
of~change
\end{array}\\\\
-------------------------------\\\\
f(x)= x^3-x \qquad 
\begin{cases}
x_1=1\\
x_2=5
\end{cases}\implies \cfrac{f(5)-f(1)}{5-1}
\\\\\\
\cfrac{[5^3-5]~-~[1^3-1]}{4}\implies \cfrac{125-5~~-~~0}{4}\implies \cfrac{120}{4}\implies 30$

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1. Suppose that a theater charges a school group $4.50 per student to show a special film. Suppose that the theater's operating

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The functions I(x) and E(x) of the theatre's income and expenses are illustrations of linear functions.

<h3>The equation of the theatre's income</h3>

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<h3>The equation of the theatre's expenses</h3>

The theatre expense per student is $1.25, and the operating cost on the staff is $130

The equation of the theatre's expenses would be:

E(x) = 1.25x + 130

<h3>Complete the table</h3>

Using the formulas I(x) = 4.5x and E(x) = 1.25x + 130, the complete table is:

Students, x 0 10 20 30 40 50 60 70

Income, I 0 45 90 135 180 225 270 315

Expenses, E 130 142.5 155 167.5 180 192.5 205 217.5

<h3>The graph of the theatre's income and expenses</h3>

See attachment

<h3>The pattern by which theatre's income and expenses increase</h3>

The functions I(x) and E(x) are linear functions.

So, the pattern with which the functions increase is a linear pattern.

<h3>The number of students when the theatre's income and expenses are equal</h3>

This means that:

I(x) = E(x)

So, we have:

4.5x = 1.25x + 130

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3.25x = 130

Divide both sides by 3.25

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Hence, the number of students is 40

<h3>The theatre profit</h3>

This is the difference between the theatre expenses and their income.

So, we have:

P(x) = E(x) - I(x)

This gives

P(x) = 1.25x + 130 - 4.5x

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P(x) = 130 - 3.25x

<h3>Solution to the inequalities</h3>

We have:

E(x) < 255

This gives

1.25x + 130 < 255

Subtract 130 from both sides and divide by 1.25

x < 100 students

Also, we have:

I(x) > 675

This gives

4.5x > 657

Solve for x

x > 146 students

Hence, the number of students for the inequalities are less than 100 and greater than 146

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2 years ago

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