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3 years ago

8

At a local fitness center members pay a a $ 14 membership fee and $ 2 for each aerobics class. Nonmembers pay $ 4 for each aerob

ics class. For what number of aerobic classes will the cost for members and nonmembers be the same?

1 answer:

mestny [16]3 years ago

8 0

14+2x=4x: divide 2x on each side

14=2x: divide by 2

7=x

7

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Step-by-step explanation

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2 years ago

Consider the following function. f(x) = 16 − x2/3 Find f(−64) and f(64). f(−64) = f(64) = Find all values c in (−64, 64) such th

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Answer:

This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64).

Step-by-step explanation:

The given function is

$f(x)=16-\frac{x^2}{3}$

To find $f(-64)$ , we substitute $x=-64$ into the function.

$f(-64)=16-\frac{(-64)^2}{3}$

$f(-64)=16-\frac{4096}{3}$

$f(-64)=-\frac{4048}{3}$

To find $f(64)$ , we substitute $x=64$ into the function.

$f(64)=16-\frac{(64)^2}{3}$

$f(64)=16-\frac{4096}{3}$

$f(64)=-\frac{4048}{3}$

To find $f'(c)$ , we must first find $f'(x)$ .

$f'(x)=-\frac{2x}{3}$

This implies that;

$f'(c)=-\frac{2c}{3}$

$f'(c)=0$

$\Rightarrow -\frac{2c}{3}=0$

$\Rightarrow -\frac{2c}{3}\times -\frac{3}{2}=0\times -\frac{3}{2}$

$c=0$

For this function to satisfy the Rolle's Theorem;

It must be continuous on $[-64,64]$ .

It must be differentiable on $(-64,64)$ .

and

$f(-64)=f(64)$ .

All the hypotheses are met, hence this does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64) is the correct choice.

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3 years ago

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To know more about imagery in sports visit:

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