6) We have that in order to see [tex] \lim_{x \to \infty} f(x) [/tex] we need to see the value of the function as the x-value increases to infinity. Checking the graph, we see that as x increases, the y of the graph stays around 3 (tends to it). Hence the limit in this case is C=3. In general cases, the limit as x approaches infinity could not exist (as in the sine function which is periodic), it could be plus or minus infinity or it could be areal number; in the last case we talk of a horizontal asymptote.
7) As x goes to 1, we have that the function itself is getting closer and closer to the value 4. It is important for this calculation to notice that we do not care about f(4) but rather the value of the function as x is around 4. This function could have been discontinuous at 4 (have a hole there or not be defined) but we would still have that [tex] \lim_{x \to 1} f(x)=4 [/tex] .
8) The value of the investment after 1 year will be equal to 400*(1.051); after 2 years the value will be equal to (400*1.051)*1.051 (the quantity after one year times (1+interest)) and after n years: [tex]400*1.051^n[/tex]. Substituting n=5 we get: 512.95$.
Now for the other part: [tex]400*1.051^n=1000[/tex] (We need to solve for n)
[tex]1.051^n=2.5[/tex] Taking logs each side:
[tex]n*log1.051=log2.5[/tex]
This yields n=18.42 years.
9) Let us take the cost: it has a fixed component of 350$ and a variable component of 20c=0.2$ per serving. Hence C(x)=350+0.2*x
The revenue function has only a variable component, 70 cents per serving. Similarly, R=0.7*x where x is the number of servings.
One has that P(x)=R(x)-C(x)=0.7*x-(350+0.2*x)=0.5x-350
Finally, if the Society breaks even, we have that P(x)=0. Solving that equation, we have: 0=0.5x-350 hence 0.5x=350 hence x=700; thus, 700 servings are needed in order for the society to break even.
10) Let us assume that q=a*p+b (general linear relationship). We have that a and b satisfy:
5000=a*60+b and 3500=a*75+b. We need to solve that system of equations to get the coefficient of our function. Subtracting both equations, we get that 1500=-15*a hence a=-100. Substituting back in the first equation, 5000=-100*60+b, yielding that b=11000. Hence the linear model that we were seeking is q=-100*p+11000.
11) If there is no surplus or shortage, we have that the price is chosen so that supply and demand are equal. Hence -3p+400=7p-500. 10p=900, hence p=90$. The required price is at 90$.
