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(math013)[2009](f)final~yhong^_10012.pdf
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Math013 Calculus I Final Exam Solution, Fall 09
1. (9=3+3+3pts) Considerafunctionde.nedby
c .x if x 1
f(x)=
2
x+ cx .2 if x> 1
where c is a constant.

(i) (3 points) For what constant c can f be afunction continuous onthe whole realline(.,)? Give brief reason.
Solution For x< 1, f(x)= c . x is clearly a continous function for any constant c. Similarly, f(x)=
2
x+ cx .2 is also continuous on x> 1 for any constant c.
At x =1, we have
lim f(x)= lim (c .x)= c .1= f(1)

x1. x1. 2
lim f(x)= lim (x + cx .2) =1.c +2 = c .1
x1+ x1+
i.e.,
lim f(x)= f(1)= lim f(x)

x1. x1+
and hence f is also continuous at x =1 for any constant c.
f is continuous on . <x for any constant c.

(ii)(3points) Usingthede.nitionofderivative,explainwhatitmeansby sayingthat f isdi.erentiable at x =1.
f(1+h).f(1)
Solution That f is at x =1 means that the limit lim exists. In particular,
h0 h [(c .(1+h)].(c .1) [((1+h)2 + c(1+h).2].(c .1)
lim = lim .
h0. h h0+ h
(iii) (3 points) What is the constantc if the given function is di.erentiable at x =1?
Solution By(ii),
2h+ h2 + ch

lim (.1) =lim =2+c,
h0. h0+ h
i.e., c = .3.
2. (10 =6+2 +2points) Incertainchemicalreaction,theconcentration C of the product as a function of time t (in minutes)is given by
12t
C = C(t)= (mole/L).
6t +1
(i) (6 points) Show thatC = C(t)satis.es an equation of the form
dC

= k(2.C)a
dt
for some suitable k and a. Find also these constants, i.e., k and a.

Solution By quotient rule,
dC (6t +1)12.12t(6) 12

==
dt (6t +1)2 (6t +1)2
Letting
..a ..a
12 12 22ak
= k(2.C)a = k 2. = k =
(6t +1)2 6t +1 6t +1 (6t +1)a
it is easy to .nd out that a =2, and k =3.

(ii) (2 points) What happens to the concentration ast +?
Solution
12 12
lim C(t)=lim = =2 .
t t 6+ 1 6
t
dC
(iii) (2 points) What happens to the rate ofchange of the reaction, i.e., , as t +?
dt
Solution
dC 12
lim =lim =0.
t dt t (6t +1)2
3. (9=3+6points) Supposethatthe derivative of a di.erentiable function f(x)is
f (x)=(x .2)(x +1)2 (x .4) ,where . <x< .
(a) (3 points) At which value of x will the function f achieve a local maximum? Justify your answer for full credit.
Solution It is easy to check that :
.
. > 0 if . <x< .1, hence f(x)is increasing there
.
.
> 0 if .1 <x< 2, hence f(x)is increasing there
f (x)
. < 0 if2 <x< 4, hence f(x)is decreasing there
.
.
> 0 if4 < x, hence f(x)is increasing there
f reaches a local maximum at x =2.

(b) (6 points) How manyin.ection points does the function f have? Justify your answer for full credit.
Solution
f (x)=(x .2)(x +1)2(x .4) =(x 2 .6x +8)(x +1)2 (x)
f =(2x .6)(x +1)2 +2(x 2 .6x +8)(x +1) =(x +1)(4x 2 .16x +10) (x)
f =2(x +1)(2x 2 .8x +5)

(x)6
f=0 if x = .1, or x =2 2 , and f (x)changes signs across each of these values.

6
f has three in.ection points at x = .1