Math 118 - Sample Midterm
Problem 1.
f(x)=(9-x2)2.
Find the functions’ critical values; determine where the function is increasing and where
it’s decreasing; find its inflection points; determine where it is concave up and where it is
concave down; sketch the graph.
f’(x)=3(9-x2)*-2x=-6x(9-x2)
f’(x)=0 => x=-3,0,3 are the critical pts.
f’(-4)<0 => x<-3 func is decreasing.
f’(-2)>0 => -3<x<0 func increasing.
f’(2)<0 => 0<x<3 func decreasing.
f’(4)>0 => 3<x
func increasing
f’’(x)=6[1*(9-x2)+x*-2x] =
=6[9-x2-2x2]=6*(9-3x2)
f’’(x)=0 => x=±3 – inflection points
f’’(-3) >0 => -3 is min & x<-3 concave down
f’’(0) <0 => 0 is max & -3<x<3 concave up
f’’(3) >0 => 3 is min & 3<x concave down
90
80
70
60
50
40
30
20
10
0
-4
-3
-2
-1
0
1
2
3
4
Problem 2.
A coffee bean farming family in Brazil spends about 1$ for producing 1 lb. of premium
coffee. They sell it to the distributors for 4$. For this price the distributor buys 10000 lb.
of their coffee. The distributor is willing to buy their coffee for more, but threatens to buy
200 lb. less for every 10 cent increase in the price. What price would leave the farming
family with the greatest profit?
x - increase in price in dollars.
P(x)=(4+x-1)(10000-2000x)=1000(10-2x)(3+x)
P’(x)=1000[(10-2x)*1+(3+x)*-2]=1000(10-2x-6-2x)=1000(4-4x)
P’(x)=0 => x=1. So optimal selling price would be 4+1=5 dollars a pound.
P(1)=1000*8*4=32000$ is the optimal profit.
Problem 3.
Nestle Bank gives it’s customers a 6% interest rate compounded quarterly. Wonka bank
wants to give the same annual effective interest rate, but in Wonka bank everything is
compounded continuously. What interest rate should Wonka bank offer its customers?
(1+(0.06/4))4*1=1.061363550625.
ex=(1+(0.06/4))4*1=1.061363550625
x=ln(1.061363550625)=0.5955…
Problem 4.
A craftswoman wants to make lava lamps and sell them at a local fair. She can make up
to 20 lamps till the fair begins. The soothsayer says that if she sells x lamps, her profit
will be exactly p(x)=x2-18ln(x). Believing the soothsayer, how many lamps should she
produce in order to get a maximal profit? What will her profit be then?
p’(x) = 2x-18/x. p’(x)=0 => 2x=18/x => x2=18/2=9 => x=±3 => x=3.
p’’(x) = 2+18/x2. p’’(3)=2+18/9=2+2=4 > 0 so min point.
p(1)=1-18ln(1)=1-18*0=1; p(20)=400-18ln(20)≈400-3=397.
So max profit is 397$ achieved on production of 20 lava lamps.
Problem 5.
Find the integrals of the following functions:
1.  [ex-ln(x2)+((x2-2)/x½)] dx
2.  [(12y-16)/(3y2-8y-5)] dy
3.  [ue-u/2] du
1.  [ex-ln(x2)+((x2-2)/x½)] dx =  [ex-ln(x2)+(x2/x½)-(2/x½)] dx =
=  ex dx -  2ln(x) dx + (x1.5) dx - (2*x-½)] dx =
= ex – 2 ln(x) dx + x2.5/2.5 – 2*x0.5/0.5 + C = ex – 2 ln(x) dx + x2.5/2.5 – 4x0.5 + C
Now do  ln(x) dx by parts:
Set U=ln(x), dU=(1/x)dx and then dV is equal to the rest i.e. dV=dx. V= dx=x
 ln(x) dx = xln(x) -  x(1/x)dx = xln(x) -  1dx = xln(x) – x
And the whole integral becomes: ex – 2(xln(x)-x) + x2.5/2.5 – 4x0.5 + C.
2.  [(12y-16)/(3y2-8y-5)] dy
Set x=(3y2-8y-5). dx=(6y-8)dy. dy=dx/(6y-8)
So  [(12y-16)/(3y2-8y-5)] dy = [(12y-16)/x]*[dx/(6y-8)] = [(12y-16)/(6y-8)]*1/x* dx
= (2/x)dx = 2(1/x)dx = 2ln(x) + C = 2ln(3y2-8y-5) + C.
3.  [ue-u/2] du
Set U=u, dV= e-u/2du; V=e-½ udu= -2e-u/2du ;
 [ue-u/2] du = UdV = UV-VdU = -2ue-u/2 - -2e-u/2du = -2ue-u/2 +2(-2e-u/2) + C =
= -2ue-u/2 - 4e-u/2 + C = -2e-u/2(u-2) + C