Download Solutions for Practice Midterm 1 - QL Pre-calculus | MATH 1080 and more Exams Pre-Calculus in PDF only on Docsity!
- Problem 1 Solve
(1) | 4 x + 3| โค 5.
Solution 1.1. We compute:
| 4 x + 3| โค 5 โ 5 โค 4 x + 3 โค 5 โ 2 โค 4 x โค 8
โ
โค x โค 2.
The solution set is
[
]
- Problem 2 If f (x) = 3x + 7 and g(x) = (^) x^12 write down formulae for f + g, f g, f โฆ g, and g โฆ 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 โฆ f )(x) =
(3x + 7)^2
(f โฆ g)(x) =
x^2
- Problem 3 Let f (x) = x^2. Compute (and simplify as much as the possible) the difference quotient
(4)
f (2 + h) โ 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)โ f^ (2). Then
ฯ(h) =
f (2 + h) โ f (2) h
=
(2 + h)^2 โ 22 h
=
4 + 4h + h^2 โ 4 h
=
4 h + h^2 h = 4 + h.
For small h, ฯ(h) โ 4.
Hence, A = xy = x(200 โ 2 x) = 2(100x โ x^2 ) = โ2(x^2 โ 100 x) = โ2((x โ 50)^2 โ 2500) (completing the square) = โ2(x โ 50)^2 + 5000 โค 5000
with equality when x = 50. The maximum area is 5000 square meters.
- Problem 6 Solve the inequality
(9) 4 x^2 โ 3 x + 5 < 0.
Solution 6.1. Notice that
4 x^2 โ 3 x + 5 = 4
x^2 โ
3 x 4
[(
x โ
]
- 5 (completing the square)
= 4
x โ
x โ
for every real number x. Hence, inequality 9 has no real-valued solu- tions.
- Problem 7 Find the roots of
(11) f (x) = x^5 โ x^4 + x^3 โ 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 โ x^2 + x โ 1) = x^2 (x^2 (x โ 1) + (x โ 1)) = x^2 (x^2 + 1)(x โ 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
Problem 8 Sketch the graph of
(13) f (x) = 3x + 4 +
x โ 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 โ
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
โ
Figure 4. Illustration for problem 10
together with the graphs of g, h, and k to illustrate how the graph of f is obtained.
- Problem 11 Which of these functions is one-to-one? Justify your answers. (a) f (x) = (^1) x (b) f (x) = 3x^2 โ 4 (c) f (x) = |x โ 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 (โ1) = โ1, f is not one-to-one. (c) Since f (4) = f (0) = 2, f is not one-to-one.
- Problem 12 State the domain of the following functions: (a) f (x) = (^) e^1 x (b) f (x) = 34 x (^3) โ 2 x 21 x (c) f (x) = x
(^2) โ 1 x+ (d) The composition f โฆ g where f (x) = x 3 x+2 with^ g(x) =^
1 xโ 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 (โโ, 0) โช (0, โ). (c) By analogous reasoning to part (b) the domain of f is
(19) (โโ, โ1) โช (โ 1 , โ).
(d) Note that
(f โฆ g)(x) =
1 (xโ4)^3 1 xโ 4 + 2 =
(x โ 4)^2 + 2(x โ 4)^3
=
(x โ 4)^2 (1 + 2(x โ 4))
=
(x โ 4)^2 (2x โ 4)
Thus, the domain of f โฆ g is {x | x 6 = 2, 4 }.
