Bo Lin
5th Quiz - Math 32 Section 107& 108
Solution
1.
Evaluate
√
5
(1) (2 pts)2 2√
(Hint: you can include ” 2” in your answer).
4
(2) (3 pts)(3 3)− 3 .
(3) (3 pts)log25 125.
√
1
5
2+ 12
= 2√2 · 2 2 = 4 √2.
Solution. (1) √
2 2 = 2√
5
Or, 2 2√= 25 = 32 = 16 · 2 = 4 2.
4
1
1
(2) (3 3)− 3 = √1 4 = √
= √
= 19 .
√
3
3
81·9
(3 3)4
(3 3) 3
√
√
3
4
3
4
Or, notice that 3 3 = 3 2 . Then (3 3)− 3 = 3 2 ·(− 3 ) = 3−2 = 91 .
log5 125
(3) log25 125 = log 25 = 23 .
5
√
Partial
credits:
(1)
correct, though we usually simplify it
√ 32 is considered
√
√ 5
1
5
√
to 4 2; ( 2) and 2 , 1 pt. (2) 3 729 is considered correct because I don’t
expect you to discover that 729 = 93 ;
1
√ 4
(3 3) 3
, 1 pt.
√
Remarks: The n-th root of a number x is written as n x, while if n = 2 we
usually omit the 2. The exponents always appear at the top-right corner, not
√
√ 3
top-left. It is reasonable to say you need the observation 3 3 = 3 to do (2),
while I tried to help you by putting (1) as a hint.
√
2. (4 pts) Estimate√ the value of 6. That is, find
√ an interval with length
≤ 12 that contains 6. (for example to estimate 2, an eligible answer is
√
2 ∈ (1, 1.5))
Solution. First we
by the property, 2
length is 1. Then
2
and
√ 2.5 , which is
6 ∈ (2, 2.5).
notice
√ that
√ 4 <√6 < 9 and 4, 9 are perfect squares. Then
= 4 < 6 < 9 =√
3. So we get an interval (2, 3) whose
we need to compare
6 and 2.5. It suffices to compare 6
√
6.25 > 6. So 6 < 2.5. Then we get an eligible interval:
Partial credits and bonus: (2, 3), 2 pts; (1, 5) or intervals with similar length,
1 pt. (2.4, 2.5) or other interval with length less than 12 , 1 bonus pt.
√
Remark: Actually 6 ≈ 2.44948974278. From this example you may discover that computing square roots of integers by hand is not easy. In practice
mathematicians developed other algorithms to compute these numbers more
efficiently.
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Bo Lin
Let the exponential function f be f (x) = 3x .
(1) (2 pts)Is f an odd function? Is f a one-to-one function?
(2) (3 pts)Let function g be defined as g(x) = log3 x. Sketch the graph of g.
(Hint: g and f have some relation)
(3) (3 pts)Solve the inequality of unknown x: g(x) < 1.
3.
Solution. (1) f is neither odd or even. One counterexample could be f (1) =
31 = 3 while f (−1) = 3−1 = 31 6= ±3. f is one-to-one because it is increasing.
(2) g is the inverse function of f . The domain of g is (0, ∞). The graph of
g is:
(3) You can either see from the graph of do it algebraically. Since f is
increasing, so is g. Then log3 x < 1 is the same thing as x < 3. In addition x
belongs to the domain of g. So the solution is (0, 3).
Partial credits: (1) 1 pt for each correct answer. (2) Wrong domain, 1 pt
off; x-intercept not 1, 1 pt off. With horizontal asymptote (i.e. g is bounded),
1 pt off. (3) (−∞, 3), 2 pts.
Remark: when sketch the graph of a function, you should concern about the
following issues: domain and range, x,y-intercept, increasing or decreasing or
neither, odd or even of neither, asymptotes. When solve an inequality, you’d
better write you answer as an interval or a union of intervals.
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