Question

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

Answer 1

The original question gives the fllowing function:
$f(x)=\frac{500}{1+19e^{-0.6x}}$
Let's tackle each option one by one.
For the fisrt statement, the initial condition is found by replacing x with 0.
That is:
$f(0)=\frac{500}{1+19e^{0}}=\frac{500}{20}=25$
Which makes the first statement false.
For the second one we need to take the limit to infinty to see is the functon reaches a maximum:
$\lim_{x\to \infty}f(x)=\frac{500}{1+19\times0}=\frac{500}{1}=500$
Which makes the statemnt correct.
For the thid one we replace x with 2, and thn x with 3 to see if 3 is indeed the firs year that the company will have more than 100 employees:
$f(2)\approx74 \\ f(3)\approx120$
Which makes the statement true.
The fourth statementis checked by takin the second derivative of the function, if the second derivative is negative then the rate is decreasing for all x, if it's possitive the rate is increasing for all x, otherwise the rate might increase or decrase for certain values.
 
$(f(x))'=500\frac{-0.6.19e^{-0.6x}}{(1+e^{-06x})^2}$
now using the derivative of a rationalfunction formula:
$(\frac{u}{v})'=\frac{u'v-uv'}{v^2}$
We get(after simplifying a lot):
$(f(x))''=5700\frac{e^{-1.2x}(-0.6e^{0.6x}+11.4)}{(1+19e^{-0.6x})^3}$
Which will be negative or high values of x. Thus the fourth statement is false.