UW MADISON – MATH DEPARTMENT
MATH 211 – SPRING 2017
WORKSHEET 6 - FEB 6, 2017
NAME:
——————————————————————————————————————–
Review:
x
• Laws of exponents: a x · a y = a x+ y , aa y = a x− y , (a x ) y = a x y , (ab) x = a x b x ;
• Property of logarithm: ln b = c ⇔ e c = b, ln(e x ) = x, eln x = x;
• Laws of logarithms: ln(x y) = ln x + ln y, ln( xy ) = ln x − ln y, ln(x r ) = r ln x.
• Average rate of change:
∆y
∆x
=
f (x 2 )− f (x 1 )
.
x 2 −x 1
Geometrically, it is the slope of the secant line.
• Instantaneous rate of change ⇒ We need limit! And geometrically, it is the slope of the
tangent line.
• Tricks we can use to evaluate a limit:
– Use continuity ⇒ plug in number and get a number;
– Simplify (to deal with 00 );
p );
– Multiply conjugate (to deal with 00 or ∞
∞ or 0 − 0 or ∞ − ∞ involving
p
p
(Recall: the conjugate of a + b is a − b.)
– lim x→a
polynomial
polynomial

number




0 ∞


 0 , ∞ ⇒ Factor out the highest order term
c
does not exist



0



0 =0
c
• Be careful with plug in number when it’s not continuous, the limit may not exist!
• One-Sided limits: Notice that the limit at one point exists if and only if the left-hand limit
equals the right-hand limit.
———————————1. Determine the following statements are true of false:
(a). 9 t/2 = 3 t
(b).2 ln x + ln 3 = ln(3x 2 )
(c).ln(c + d) = ln c + ln d
(d).(ln x)2 = 2 ln x
Proof.
(a) True.
(b) True.
(c) False.
(d) False.
2. Solve the following equations for x:
(a) 2e x−7 = 5
Proof. 2e x−7 = 5 ⇒ e x−7 =
5
2
⇒ x − 7 = ln 25 ⇒ x = 7 + ln 52 .
(b) ln(x + 1) = 2.
Proof. ln(x + 1) = 2 ⇒ x + 1 = e2 ⇒ x = e2 − 1
(c) ln(ln(2 + log2 (x + 1))) = 0
Proof. ln(ln(2 + log2 (x + 1))) = 0 ⇒ ln(2 + log2 (x + 1)) = 1 ⇒ 2 + log2 (x + 1) = e ⇒
log2 (x + 1) = e − 2 ⇒ x + 1 = 2e−2 ⇒ x = 2e−2 − 1
(d) ln(x − 1) + ln(2x − 1) = 2 ln(x + 1).
Proof. ln(x − 1) + ln(2x − 1) = 2 ln(x + 1) ⇒ ln[(x − 1)(2x − 1)] = ln(x + 1)2 ⇒
(x −1)(2x −1) = (x +1)2 ⇒ 2x 2 −3x +1 = x 2 +2x +1 ⇒ x 2 −5x = 0 ⇒ x 1 = 0, x 2 = 5.
Check x 1 and x 2 so that x − 1 > 0, 2x − 1 > 0, x + 1 >. Then we get the only possible
x is x = 5.
(e) 4 x−2 = 3 x+4
Proof. Take natural logarithm on both sides, ⇒ ln(4 x−2 ) = ln(3 x+4 ) ⇒ (x − 2) ln 4 =
ln 3+2 ln 4
(x + 4) ln 3 ⇒ (ln 4 − ln 3)x = 4 ln 3 + 2 ln 4 ⇒ x = 4 ln
4−ln 3
3. (City population) The population in thousands of a city is given by P(t), where t is the
year, with t = 0 corresponding to 2000. In 2000, the population of the city was 45,200.
Write a formula for P that satisfes the given description.
(a) The population is increasing by 1650 people per year.
(b) The population is doubling every 20 years.
Hint: For (a) this is a linear model, for (b) looks at the function ab t .
Proof.
(a) Assume P(t) = at + b, then P(0) = b = 45, 200. Now by description we have P(t +
1) = P(t) + 1650, thus a(t + 1) + 45, 200 − at − 45, 200 = 1650, i.e a = 1650, so
P(t) = 1650t + 45, 200.
(b) Let’s assume P(t) = ab t , then P(0) = a = 45, 200. Now by description we have
P(t + 20) = 2P(t), thus for all t must have
45, 200b t+20 = 2.45, 200b t
=⇒
b20 = 2
=⇒
4. Find the average rate of change of the function over given interval.
(a) f (x) = x 2 + 5x over [1, 3]
(b) g(t) = 2t 2 − 4t + 1 over [0, 2].
p
(c) A(v) = v + 3 over [6, 13]
(d) C(x) =
4x
x+2
over [4, 8].
Proof. Using the formula
∆y
∆x
=
f (x 2 )− f (x 1 )
x 2 −x 1
(a) 9.
(b) 0.
(c) 1/7.
(d) 2/15.
5. Simplify as much as possible
(a) f (x) =
(b) g(c) =
x 2 −2x−3
x 2 −7x+12
(c+2)2 −4
c
and find lim x−→1 f (x), lim x−→3 f (x), limc−→0 g(c).
Proof.
(a)
x+1
x−4
(b) c + 4.
lim x−→1 f (x) = −2/3, lim x−→3 f (x) = −4, limc−→0 g(c) = 4.
b = log20 2
6. Determine the domain of f (x) =
x 2 −4x
x−2
and find
lim f (x)
lim f (x).
and
x−→1
x−→2
Proof. Domain {x ∈ R : x 6= 2} and
lim f (x) = 3
x−→1
and
lim f (x) = is undefined
x−→2