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At the Smith, Inc. company, the number of employees needed to work per year is modeled by the function f(x).




f(x)=5001+19e−0.6x




Which statements are true about the number of employees?




Select each correct answer.





The rate of the number of employees per year increases for all values of x.



The third year is the first year in which the company will have more than 100 employees.


The maximum number of employees for the company is 500.


The company started with 20 employees.

At the Smith Inc company the number of employees needed to work per year is modeled by the function fxfx500119e06xWhich statements are true about the number of class=

Respuesta :

Answer:

i) In the third year the company will have more than a 100 employees is TRUE

ii) The maximum number of employees for the company is 500 is TRUE

iii) The company started with 20 employees is FALSE

   

Step-by-step explanation:

At the Smith, Inc. company, the number of employees needed to work per year is modeled by the function f(x).

 

[tex]f(x) = \frac{500}{1 + 19e^{-0.6x}}[/tex]

where x is in years.

i) In the third year the company will have more than a 100 employees is TRUE

[tex]f(3) = \frac{500}{1 + 19e^{(-0.6\times 3)}} = \frac{500}{1 + 19e^{-1.8}} = \frac{500}{1 + 3.141} = \frac{500}{4.141} = 120.748[/tex]

[tex]f(2) = \frac{500}{1 + 19e^{(-0.6\times 2)}} = \frac{500}{1 + 19e^{-1.2}} = \frac{500}{1 + 5.723} = \frac{500}{6.723} = 74.4[/tex]

iI) The maximum number of employees for the company is 500 is TRUE[tex]f(x) = \frac{500}{1 + 19e^{(-0.6\timesx)}} = \frac{500}{1 + \frac{19}{e^{(0.6\times x)}}} Let x tend towards infinity\therefore f(x) = \frac{500}{1 + \frac{19}{e^{(0.6\times x)}}} = \frac{500}{1 + 0 } = 500[/tex]

iii) The company started with 20 employees is FALSE

   [tex]f(0) = \frac{500}{1 + 19e^{(-0.6\times 0)}} = \frac{500}{1 + 19e^{0}} = \frac{500}{1 + 19} = \frac{500}{20} = 25[/tex]

ashishdwivedilVT

1)The rate of the number of employees per year increases for all values of x.

2)The third year is the first year in which the company will have more than 100 employees.

3)The company started with 25 employees.

The given function is:

[tex]f(x) = \frac{500}{1+19e^{-0.6x} }[/tex]

[tex]f'(x) = \frac{5700e^{-0.6x} }{(1+19e^{-0.6x})^2 }[/tex]

What does f'(x) always positive indicates?

It indicates that function is monotonic or strictly increasing.

f'(x) is always positive so the rate of the number of employees per year increases for all values of x.

f(2) =74.37

f(3) = 120.75

So, the third year is the first year in which the company will have more than 100 employees.

f(0) = 25

So, The company started with 25 employees.

Therefore, the following conclusions have been drawn:

1)The rate of the number of employees per year increases for all values of x.

2)The third year is the first year in which the company will have more than 100 employees.

3)The company started with 25 employees.

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