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Solutions for Practice Midterm 1 - QL Pre-calculus | MATH 1080, Exams of Pre-Calculus

Material Type: Exam; Class: QL Pre-calculus; Subject: Math; University: Weber State University; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 07/23/2009

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MATH 1080 PRACTICE MIDTERM 1
1. Problem 1
Solve
(1) |4x+ 3| โ‰ค 5.
Solution 1.1.We compute:
|4x+ 3| โ‰ค 5
โˆ’5โ‰ค4x+ 3 โ‰ค5
โˆ’2โ‰ค4xโ‰ค8
โˆ’1
2โ‰คxโ‰ค2.
(2)
The solution set is hโˆ’1
2,2i.
2. Problem 2
If f(x) = 3x+ 7 and g(x) = 1
x2write down formulae for f+g,fg,
fโ—ฆg, and gโ—ฆf.
Solution 2.1.From the given information, we get:
(f+g)(x) = 3x+7+ 1
x2
(fg)(x) = 3x+ 7
x2
(gโ—ฆf)(x) = 1
(3x+ 7)2
(fโ—ฆg)(x) = 3
x2+ 7.
(3)
3. Problem 3
Let f(x) = x2. Compute (and simplify as much as the possible) the
difference quotient
(4) f(2 + h)โˆ’f(2)
h.
When his small, what is the difference quotient approximately equal
to?
1
pf3
pf4
pf5
pf8

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  1. Problem 1 Solve

(1) | 4 x + 3| โ‰ค 5.

Solution 1.1. We compute:

| 4 x + 3| โ‰ค 5 โˆ’ 5 โ‰ค 4 x + 3 โ‰ค 5 โˆ’ 2 โ‰ค 4 x โ‰ค 8

โˆ’

โ‰ค x โ‰ค 2.

The solution set is

[

]

  1. Problem 2 If f (x) = 3x + 7 and g(x) = (^) x^12 write down formulae for f + g, f g, f โ—ฆ g, and g โ—ฆ f.

Solution 2.1. From the given information, we get:

(f + g)(x) = 3x + 7 +

x^2 (f g)(x) =

3 x + 7 x^2 (g โ—ฆ f )(x) =

(3x + 7)^2

(f โ—ฆ g)(x) =

x^2

  1. Problem 3 Let f (x) = x^2. Compute (and simplify as much as the possible) the difference quotient

(4)

f (2 + h) โˆ’ f (2) h

When h is small, what is the difference quotient approximately equal to? 1

P

x

y

y=g(x) y=h(x)

y=k(x)

70

y=f(x)

6

4

Figure 1. Illustration for problem 4- not exactly to scale

Solution 3.1. Let ฯ†(h) = f^ (2+h h)โˆ’ f^ (2). Then

ฯ†(h) =

f (2 + h) โˆ’ f (2) h

=

(2 + h)^2 โˆ’ 22 h

=

4 + 4h + h^2 โˆ’ 4 h

=

4 h + h^2 h = 4 + h.

For small h, ฯ†(h) โ‰ˆ 4.

Hence, A = xy = x(200 โˆ’ 2 x) = 2(100x โˆ’ x^2 ) = โˆ’2(x^2 โˆ’ 100 x) = โˆ’2((x โˆ’ 50)^2 โˆ’ 2500) (completing the square) = โˆ’2(x โˆ’ 50)^2 + 5000 โ‰ค 5000

with equality when x = 50. The maximum area is 5000 square meters.

  1. Problem 6 Solve the inequality

(9) 4 x^2 โˆ’ 3 x + 5 < 0.

Solution 6.1. Notice that

4 x^2 โˆ’ 3 x + 5 = 4

x^2 โˆ’

3 x 4

[(

x โˆ’

]

  • 5 (completing the square)

= 4

x โˆ’

x โˆ’

for every real number x. Hence, inequality 9 has no real-valued solu- tions.

  1. Problem 7 Find the roots of

(11) f (x) = x^5 โˆ’ x^4 + x^3 โˆ’ x^2

and state their multiplicity.

7/

y=f(x)

y=L(x)

b a

2

4

x

y

Figure 3. Illustration for problem 8

Solution 7.1. Observe that

f (x) = x^2 (x^3 โˆ’ x^2 + x โˆ’ 1) = x^2 (x^2 (x โˆ’ 1) + (x โˆ’ 1)) = x^2 (x^2 + 1)(x โˆ’ 1).

Hence, the real roots of f are 0 (with multiplicity 2) and 1 (with mul- tiplicity 1). If we allow for non-real roots, then the roots ยฑi each have multiplicity

  1. Problem 8 Sketch the graph of

(13) f (x) = 3x + 4 +

x โˆ’ 2

Solution 8.1. By inspection, f has a vertical asymptote at x = 2 and an oblique asymptote of y = L(x) = 3x + 4. Notice that

(14) f (0) = 4 โˆ’

For x > 2, f (x) > L(x) while for x < 2, f (x) < L(x).

2

x

y

y=f(x)

y=g(x)

y=h(x)

y=k(x)

1

โˆ’

Figure 4. Illustration for problem 10

together with the graphs of g, h, and k to illustrate how the graph of f is obtained.

  1. Problem 11 Which of these functions is one-to-one? Justify your answers. (a) f (x) = (^1) x (b) f (x) = 3x^2 โˆ’ 4 (c) f (x) = |x โˆ’ 2 |

Solution 11.1. (a) Since y = (^) x^1 if and only if x = (^1) y it follows that f is one-to-one. (b) Since f (1) = f (โˆ’1) = โˆ’1, f is not one-to-one. (c) Since f (4) = f (0) = 2, f is not one-to-one.

  1. Problem 12 State the domain of the following functions: (a) f (x) = (^) e^1 x (b) f (x) = 34 x (^3) โˆ’ 2 x 21 x (c) f (x) = x

(^2) โˆ’ 1 x+ (d) The composition f โ—ฆ g where f (x) = x 3 x+2 with^ g(x) =^

1 xโˆ’ 4.

Solution 12.1. (a) Since ex^ is defined and non-zero for all real numbers, the domain of f is R. (b) Here, f is a rational function. The denominator is 0 at x = 0 only and hence the domain of f is (โˆ’โˆž, 0) โˆช (0, โˆž). (c) By analogous reasoning to part (b) the domain of f is

(19) (โˆ’โˆž, โˆ’1) โˆช (โˆ’ 1 , โˆž).

(d) Note that

(f โ—ฆ g)(x) =

1 (xโˆ’4)^3 1 xโˆ’ 4 + 2 =

(x โˆ’ 4)^2 + 2(x โˆ’ 4)^3

=

(x โˆ’ 4)^2 (1 + 2(x โˆ’ 4))

=

(x โˆ’ 4)^2 (2x โˆ’ 4)

Thus, the domain of f โ—ฆ g is {x | x 6 = 2, 4 }.