Download Math 111 - Solution of Test 1 Problem 1. The graph of y = f(x) is

Calculus I - Fall 2007 - Yahdi
Math 111
- Solution of Test 1
Problem 1. The graph of y = f (x) is given below. For each of the graphs 1, 2, 3, 4 and 5 explain
how it is obtained from the graph of f (x) and give its corresponding formula.
• Graph 1 is obtained from the graph the f (x) by a vertical shift of 3 units upward, therefore the
corresponding formula is f (x) + 3 .
• Graph 3 is obtained from the graph the f (x) by horizontal shift of 4 units to the right, therefore the
corresponding formula is f (x − 4) .
• Graph 4 is obtained from the graph the f (x) by a vertical compress by a factor
1
3,
therefore the
1
3 f (x)
corresponding formula is
.
• Graph 5 is obtained from the graph the f (x) by a vertical reflection about the x-axis −→ −f (x), then
horizontal shift of 4 units to the left, therefore the corresponding formula is −f (x + 4) .
• Graph 2 is obtained from the graph the f (x) by horizontal shift of 6 units to the left −→ f (x + 6),
then a vertical stretch by a factor 2, therefore the corresponding formula is 2f (x + 6) .
√
Problem 2. Find the domain of the function f (x) =
x2 − 4x + 3
.
x−4
√
x2 − 4x + 3
exists if x2 − 4x + 3 > 0 and x − 4 6= 0.
x−4
• Solve x2 − 4x + 3 = 0, then test a number in each interval to find the sign of this expression.
x2 − 4x + 3 = 0 ⇐⇒ (x − 1)(x − 3) = 0 ⇐⇒ x = 1 or x = 3.
Testing a number in each interval, we get:
x = 0 =⇒ 02 −4(0)+3 = 3 positive, x = 2 =⇒ 22 −4(2)+3 = −1 negative, and x = 4 =⇒ 42 −4(4)+3 = 3
positive. Therefore,
positive
1
negative
3
positive
.
• Solving x − 4 = 0, we get x = 4.
• So f (x) exists for x 6 1, x > 3, x 6= 4. In other terms, the domain is (−∞, 1] ∪ [3, 4) ∪ (4, +∞) .
f (x) =
Problem 3. A small-appliance manufacturer finds that it costs $9,000 to produce 1,000 toaster ovens a
week and $12,000 to produce 1,500 toaster ovens a week. Assume it is a linear relationship.
Express the cost as a function of the number x of toaster ovens produced.
1
2
Let x be the number of toaster ovens produced a week, and let y = f (x) be the cost to produce these x
toaster. Since it is a linear function then, using the point-slope form we have y = m(x − x1 ) + y1 . We have:
(
x1 = 1, 000toasters −→ y1 = $9, 000,
12, 000 − 9, 000
3, 000
y2 − y1
Therefore, m =
=
=
= 6. Thus
x2 − x1
1, 500 − 1, 000
500
x2 = 1, 500toasters −→ y2 = $12, 000.
y = 6(x − 1, 000) + 9, 000 = 6x + 3, 000
Problem 4. Find the formula of the function f(x) that best fits the data in the following table:
x
0
2
4
6
f(x) 1.21 2.7225 6.12562 13.7826
The data fits the model of an exponential function f (x) = bax because we have constant ratios in y-values
2.7225
6.12562
13.7826
for equally spaced x-values. Indeed,
=
=
= 2.25. We need to choose two points:
1.21
2.7225
6.12562
• (0, 1.21) −→ f (0) = 1.21 −→ ba0 = 1.21 −→ b = 1.21
√
2.7225
• (2, 2.7225) −→ f (2) = 2.7225 −→ ba2 = 2.7225 −→ a2 = 2.7225
a
=
2.25 = 1.5
=
=
2.25
=⇒
b
1.21
Conclusion: f (x) = 1.21(1.5)x
Problem 5. Under ideal conditions a certain bacteria population is known to triple every three hours.
Suppose that there are initially 25 bacteria.
(1) Give a formula for the size of this population after t hours.
(2) How long will it take for the population to reach 2,500?
Denotes by P (t) the bacteria population at the time t, where t is measured in hours and the given initial
population is b = P (0) = 25. The population triples every 3 hours, therefore:
(1) After 3 hours −→ only 1 period of 3 hours −→ P (3) = 3(25) =⇒ P (3) = 3(25)
After 6 hours −→ 2nd period of 3 hours −→ P (6) = 3[3(25)] = 32 (25)
We have a pattern: after t hours, we have 3t periods of 3 hours. Since the initial population 25 will
t
triple each 3 hours, then it will triple 3t times after t hours. Therefore, P (t) = 3 3 25
(2) We have to solve for t equation: P(t) = 2, 500
t
t
2, 500
P (t) = 25, 000 =⇒ 3 3 25 = 2, 500 =⇒ 3 3 =
= 100
25
Apply the logarithm “ln” to both sides of the last equation and use ln AB = B ln A :
t
ln 3 3 = ln 100 =⇒
t
t
ln 100
ln 100
ln 3 = ln 100 =⇒ =
=⇒ t = 3
≈ 12.5754 hours
3
3
ln 3
ln 3
Problem 6. Find a formula for the inverse function f −1 for each of the following functions
√
(1) f (x) = x + 1.
(2) f (x) = 4 + ln 3x.
(1) We need to solve for x,
f (x) = y. We have:
√
x + 1 = y ⇔ x + 1 = y2 ⇔ x = y2 − 1
Now, interchange y with x, we get: f −1 (x) = y 2 − 1 .
(2) We need to solve for x, f (x) = y. We have:
4 + ln 3x = y ⇔ ln 3x = y − 4 ⇔ eln 3x = e(y−4) ⇔ 3x = e(y−4) ⇔ x =
Now, interchange y with x, we get: f −1 (x) =
e(x−4)
.
3
e(y−4)
3