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Math 121, Practice Questions for Final Hints and Answers
Note. In all questions b > 0 and b 6 = 1 where b is an exponential or logarithmic base.
- Determine whether the following functions are inverse functions. If they are inverse func- tions, sketch the both functions together on the same coordinate axis without using a graphing calculator.
(a) F (x) = 2x − 5, G(x) = x + 5 2
Answer. F (G(x)) = 2(G(x)) − 5 = 2 x + 5 2 − 5 = x + 5 − 5 = x
G(F (x)) = F (x) + 5 2
2 x − 5 + 5 2
2 x 2 = x.
Because both G(F (x)) = x and F (G(x)) = x, F and G are inverse functions. The graphs, which are lines, are left to the reader. Note, though, that they are reflections of each other about the line y = x.
(b) p(x) = x − 5 2 x , q(x) = 2 x x − 5
Answer. p(q(x)) = q(x) − 5 2 x
2 x x− 5 −^5 2 x
2 x − 5(x − 5) 2 x(x − 5)
− 3 x + 25 2 x^2 − 10 x
= x. Therefore, p and
q are not inverse functions.
- (a) Find the inverse function of f (x) = x^5 − 3. Verify that the function you have found is the inverse function to f.
Answer. Let x = y^5 − 3 and solve for y to obtain y = (x + 3)^15. Thus f −^1 (x) = (x + 3)^15.
Verification: f (f −^1 (x)) = (f −^1 (x))^5 − 3 = (x + 3) − 3 = x, and
f −^1 (f (x)) = (f (x) + 3)^15 = (x^5 − 3 + 3)^15 = (x^5 )^15 = x.
(b) What is the inverse function of h(x) = log 14 x? Verify that this is the inverse function, and graph h(x) and its inverse on the same coordinate axis without using a graphing calculator.
Answer. The inverse function is h−^1 (x) =
)x .
Verification follows by using the inverse property of logs and exponentials:
h−^1 (h(x)) =
4
)h(x)
4
)log 1 / 4 x = x, and
h(h−^1 (x)) = log 14 h−^1 (x) = log (^14)
)x = x
The graphs are left for the reader, however, again, they are reflections of each other about the line y = x.
- Suppose f is a function and g is its inverse function, suppose also the domain of f is [3, 10] and the range of f is [− 2 , 20] with f (3) = −2 and f (10) = 20.
(a) Is f one-to-one?
Answer. Yes — because f has an inverse function.
(b) Is g one-to-one?
Answer. Yes — because g has an inverse function.
(c) Find the domain and range of g.
Answer. The domain of g is [− 2 , 20] and the range of g is [3, 10].
(d) If possible, find: g(3), g(−2), g(10), g(20).
Answer. g(−2) = 3 and g(20) = 10, there is not enough information to find g(3) and g(10).
- Solve each equation without using a calculator:
(a) log 3 81 = x
Answer. 3 x^ = 81, or 3x^ = 3^4 and so x = 4.
(b) 2^3 x−^1 = 32
Answer. 23 x−^1 = 2^5 and so 3x − 1 = 5 and so x = 2.
(c) ln eπ^2 = x
Answer. x = ln eπ^2 = π^2 ln e = π^2 , thus x = π^2.
(d) b^3 x+2^ = (b^4 )x+
Answer. 3 x + 2 = 4(x + 1) and so 3x + 2 = 4x + 4, and so x = −2.
- Complete this question without the use of a graphing calculator or utility.
(a) Sketch the graph of f (x) = 3|x|.
(b) Write the equation y = log 3 x in exponential form.
(c) Sketch the graph of f (x) = log 3 x, then use reflections or translations to sketch the graphs of g(x) = − log 3 x, h(x) = − log 3 (x − 2), k(x) = 4 − log 3 (x − 2).
(d) Sketch the graph of f (x) = log 3 x^4.
Answer. You may check your graphs using a graphing utility. Note that for (a), if x ≥ 0, the graph looks like the graph of y = 3x^ and the function is even, so the rest of the graph is a reflection about the y-axis.
Therefore 2x + 1 = 300x − 100, or 298x = 101 and so x =
. You should plug this in and verify that it works because sometimes extraneous solutions are introduced when combining the logs.
- Suppose $18,000 is invested at an annual interest rate of 9% compounded daily.
(a) How much will it be worth after 12 years?
Answer. According to the compound interest formula
B = 18 , 000
= 18 , 000(1.00024657534)^4380 = $52, 997. 18.
(b) How long will it take until the investment is worth $500,000?
Answer. Solve 500, 000 = 18, 000(1.00024657534)^365 t. Hence
= (1.00024657534)^365 t
and so
365 t log(1.00024657534) = log
so t =
log
18000
log(1.00024657534)
≈ 36 .94 years.
- The magnitude of an earthquake of intensity I on the Richter scale is
M = log
I
I 0
where I 0 is the intensity of a zero-level earthquake.
Answer. (Scratch Work) Before solving any of these problems, notice that 10M^ =
I
I 0
, and so
I = 10M^ I 0 is an another form of the equation.
(a) Find the magnitude to the nearest 0.1 of the 1999 Joshua Tree earthquate that hand an intensity of I = 12, 589 , 254 I 0.
Answer. M = log(12, 589 , 254) ≈ 7 .1.
(b) Find the intensity of the 1999 Taiwan earthquake that measured 7.6 on the Richter scale.
Answer. From the scratch work formula, I = 10^7.^6 I 0 = 39, 810 , 717 I 0.
(c) On March 3, an earthquake in Orange County measured 2.9 on the Richter scale. Compare the intensity of the 1999 Taiwan earthquake with this earthquake. In other words, how many times more intense was the Taiwan earthquake?
Answer.
M 1
M 2
107.^6 I 0
102.^9 I 0
= 10(7.^6 −^2 .9)^ = 10^4.^7 ≈ 50 , 118. Therefore, the 7.6 earthquake was
approximately 50,118 times as intense as a 2.9 earthquake.
- The population of bacteria in a vat of potato salad at Bob’s Big Buffet is modeled by P (t) = P 0 ekt. At noon there were 1000 bacteria present and at 12:30pm there were 1300 bacteria present. How many bacteria were present at: (a) 11:30am when the Buffet opened? (b) At 1:30pm when the Buffet closed? (Assume no one ate any potato salad because of its funny smell).
Answer. Let t be measured in hours and let t 0 = 12 noon. Now P 0 = 1000 is the population when t = 0. Thus P (t) = 1000ekt^ and so we need to find k. Also, P (.5) = 1300, therefore 1300 = e.^5 k. Therefore k = 2 ln(1.3) ≈ .5247. Thus:
(a) At 11:30am, there were approximatley 1000e(−.5)(.5247)^ ≈ 769 bacteria present in the vat.
(b) At 1:30pm, there were approximately 1000e(1.5)(.5247)^ ≈ 2197 bacteria in the vat.
