Question 828243
1) after x weeks, the number of people(or population) living in a community is given by the function {{{f(x)=2x^2-4x+100}}}
After that, you typed "provided that x is less than or equal to 1".
That does not make sense I think you meant
provided that {{{x>=1}}} , or
provided that x is greater than or equal to 1.
 
a.) The average rate of change of the population from the second week (x=2) to the fifth week (x=5) can be found by dividing the difference in population by the number of weeks passed.
So, in {{{2-5=3}}} weeks the population went from
{{{f(2)=2*2^2-4*2+100=2*4-8+100=8-8*100}}} to
{{{f(5)=2*5^2-4*5+100=2*25-20+100=50-20+100=130}}}
The average rate of change of the population from the second week (x=2) to the fifth week (x=5) is
{{{(f(5)-f(2))/(5-2)=(130-100)/(5-2)=30/3=highlight(10)}}}
That average rate is the slope of the line connecting the points for {{{x=2}}} and {{{x=3}}} on the graph.
{{{graph(300,300,-2,8,-20,180,10x+80,2x^2-4x+100)}}}
 
b.) find the rate of change of the population after the second week.
At the point {{{x=2}}} (after the second week, the change of the population is given by the value of the derivative of f(x) at {{{x=2}}} .
If you have been taught about derivatives, you would know that the derivative of {{{f(x)}}} is
{{{df/dx}}}{{{(x)=2*(2x)-4=4x-4}}}
and for {{{x=2}}} its value is
{{{df/dx}}}{{{(2)=4*2-4=8-4=highlight(4)}}}
That rate at the point with {{{x=2}}} is the slope of the tangent to the graph at the point with {{{x=2}}} .
{{{graph(300,300,-2,8,-20,180,4x+92,2x^2-4x+100)}}}
 
2) During a flu epidemic, medical researchers found out that the number of people infected in x days is given by the function {{{p(x)=15x^2-x^3}}} for {{{0<=x<=10}}} (for x between 0 and 10, inclusive) .
(Clumsier alternative wordings:
for x greater than or equal to 0, but lesser than or equal to 10,
or if you want a more unwieldy mouthful, for x such that 0 is less than or equal to x, and x is less than or equal to 10).
a.) find the average rate of infection for the first seven days (from day 1 to day 7).
As done above,
{{{p(1)=15*1^2-1^3=15-1=14}}} , and
{{{p(7)=15*7^2-7^3=15*49-343=735-343=392}}}
{{{(f(7)-f(1))/(7-1)=(392-14)/(7-1)=378/6=highlight(63)}}}
b) what was the rate of infection after the fourth day?
The derivative is
{{{dp/dx}}}{{{(x)=15*(2x)-3x^2=30x-3x^2}}}
and for {{{x=4}}} its value is
{{{dp/dx}}}{{{(4)=30*4-3*4^2=120-3*16=120-48=highlight(72)}}}
That rate at the point with {{{x=4}}} is the slope of the tangent to the graph at the point with {{{x=4}}} , with {{{p(4)=15*4^2-4^3=15*16-64=240-64=176}}}.
The graphs of {{{p(x)}}} with one or the other line are shown below.
{{{graph(300,300,-2,12,-75,675,63x-49,15x^2-x^3)}}}
{{{graph(300,300,-2,6,-30,270,72x-112,15x^2-x^3)}}} It's hard to see that the line at left is the tangent at {{{x=4}}} , but a close-up does not help much either. The function is almost a straight line in the vicinity of that point.