▶️ Answer/Explanation
A. i.
To find \( h(4) \), we use the definition \( h(x) = g(f(x)) \).
Substitute \( x = 4 \) into the equation: \( h(4) = g(f(4)) \).
From the table, locate \( x = 4 \) to find the value of \( f(4) \).
\( f(4) = 3 \).
Now substitute this value into \( g \): \( h(4) = g(3) \).
Using the function definition \( g(x) = 2 \ln x \), substitute \( 3 \) for \( x \).
\( g(3) = 2 \ln(3) \).
Calculate the decimal value: \( 2 \times 1.0986… \approx 2.1972… \)
Rounding to three decimal places, \( h(4) \approx 2.197 \).
A. ii.
We must identify the input values \( x \) where the output \( f(x) \) equals 3.
Looking at the row for \( f(x) \) in the table:
The value 3 appears in the column where \( x = 2 \).
The value 3 appears in the column where \( x = 4 \).
Therefore, the values are \( x = 2 \) and \( x = 4 \).
B. i.
Set the expression for \( g(x) \) equal to 3: \( 2 \ln x = 3 \).
Divide both sides by 2: \( \ln x = 1.5 \).
Convert to exponential form to solve for \( x \): \( x = e^{1.5} \).
Calculate the decimal value: \( e^{1.5} \approx 4.48168… \)
Rounding to three decimal places, \( x \approx 4.482 \).
B. ii.
As \( x \) increases without bound (\( x \rightarrow \infty \)), the natural logarithm function \( \ln x \) increases continuously.
Consequently, \( 2 \ln x \) also grows infinitely large.
The end behavior using limit notation is: \( \lim_{x \to \infty} g(x) = \infty \).
C. i.
No, the function \( f \) does not have an inverse function.
C. ii.
For a function to have an inverse, it must be one-to-one (injective), meaning every unique input \( x \) must map to a unique output \( f(x) \).
Referring to the table, we can see that \( f(2) = 3 \) and \( f(4) = 3 \).
Since two different input values (2 and 4) produce the same output value (3), the function fails the horizontal line test.
Therefore, \( f \) is not one-to-one and cannot have an inverse.
▶️ Answer/Explanation
Part A
i.
We are given the model \(P(t) = ab^t\).
At \(t=2\) (year 2015), the number of plants \(P(2) = 59\). Substituting this into the equation gives:
\(59 = ab^2\)
At \(t=8\) (year 2021), the number of plants \(P(8) = 118\). Substituting this into the equation gives:
\(118 = ab^8\)
ii.
To find \(b\), divide the second equation by the first equation:
\(\frac{118}{59} = \frac{ab^8}{ab^2}\)
\(2 = b^{8-2}\)
\(2 = b^6\)
\(b = \sqrt[6]{2} \approx 1.1225\)
Now, substitute the value of \(b\) back into the first equation to solve for \(a\):
\(59 = a(1.1225)^2\)
\(59 = a(2^{2/6}) = a(2^{1/3})\)
\(a = \frac{59}{2^{1/3}} \approx \frac{59}{1.2599}\)
\(a \approx 46.828\)
Values: \(a \approx 46.83\) and \(b \approx 1.12\)
Part B
i.
The average rate of change from \(t=2\) to \(t=8\) is given by the slope formula \(\frac{P(8) – P(2)}{8 – 2}\).
Average Rate = \(\frac{118 – 59}{8 – 2}\)
Average Rate = \(\frac{59}{6}\)
Average Rate \(\approx 9.833\)
The average rate of change is approximately \(9.83\) plants per year.
ii.
To estimate the number of plants at \(t=10\), we use the rate of change calculated above to project forward from \(t=8\).
Time difference \(\Delta t = 10 – 8 = 2\) years.
Estimated \(P(10) = P(8) + (\text{Average Rate} \times \Delta t)\)
Estimated \(P(10) = 118 + (9.833 \times 2)\)
Estimated \(P(10) = 118 + 19.666\)
Estimated \(P(10) \approx 137.67\)
The estimated number of plants is approximately \(137.67\).
iii.
The estimates will be less than the number of plants predicted by the model \(P\).
Reasoning: The model \(P(t) = ab^t\) is an exponential growth function with \(b > 1\) and \(a > 0\). The graph of such a function is concave up, meaning the rate of growth is constantly increasing. The average rate of change calculated in Part B(i) represents the slope of the secant line connecting \(t=2\) and \(t=8\). When we use this linear rate to estimate values for \(t > 8\), we are following the straight line path. Because the actual graph curves upward (becoming steeper), the straight line lies below the curve for \(t > 8\). Therefore, the linear estimate underestimates the actual exponential growth.
Part C
The ecologist should have more confidence in the model for \(t=6\) years.
Reason: \(t=6\) falls within the interval of the collected data (between \(t=2\) and \(t=8\)), which is known as interpolation. Predictions made within the range of observed data are generally more reliable. Conversely, \(t=20\) is far outside the range of observed data (extrapolation). Over a long period (12 years past the last data point), environmental conditions could change, or the population might reach a carrying capacity, rendering the exponential growth model invalid.
▶️ Answer/Explanation
\( F = (0, 210) \)
\( G = (3, 200) \)
\( J = (6, 190) \)
\( K = (9, 200) \)
\( P = (12, 210) \)
\( a = 10 \)
\( b = \frac{\pi}{6} \)
\( c = 3 \)
\( d = 200 \)
i. (a) \( h \) is positive and increasing.
ii. The graph is concave up, and the rate of change of \( h \) is increasing.
Part A: Determine Coordinates
1. Identify Vertical Parameters: The center of the clock is 200 cm from the floor, so the midline of the function is \( y = 200 \). The hour hand is 10 cm long, which is the amplitude.
– Maximum height = \( 200 + 10 = 210 \) cm.
– Minimum height = \( 200 – 10 = 190 \) cm.
2. Identify Time Parameters: At \( t = 0 \), the hand points to 12 (up), which corresponds to the maximum height. The period is 12 hours.
– Maxima occur at \( t = 0, 12, 24, \dots \)
– Minima occur at \( t = 6, 18, \dots \) (halfway through the cycle).
– Midline crossings occur at \( t = 3, 9, 15, \dots \)
3. Map Points to Coordinates:
– Point F: This is a maximum peak. Since \( t=0 \) is the first maximum, \( F = (0, 210) \).
– Point G: This is the midline crossing on the way down, occurring \( \frac{1}{4} \) of the period after the peak. \( t = 3 \). \( G = (3, 200) \).
– Point J: This is the minimum (trough), occurring \( \frac{1}{2} \) of the period after the peak. \( t = 6 \). \( J = (6, 190) \).
– Point K: This is the midline crossing on the way up, occurring \( \frac{3}{4} \) of the period after the peak. \( t = 9 \). \( K = (9, 200) \).
– Point P: This is the next maximum, occurring one full period after \( F \). \( t = 12 \). \( P = (12, 210) \).
Part B: Find Constants for \( h(t) = a\sin(b(t+c)) + d \)
– \( d \) (Vertical Shift): This is the midline height. \( d = 200 \).
– \( a \) (Amplitude): This is the length of the hand. \( a = 10 \).
– \( b \) (Frequency Coefficient): The period is 12 hours. Using the formula \( \text{Period} = \frac{2\pi}{b} \):
\( 12 = \frac{2\pi}{b} \implies b = \frac{2\pi}{12} = \frac{\pi}{6} \).
– \( c \) (Phase Shift): A standard sine function \( \sin(bt) \) starts at the midline and goes up. We need our function to be at a maximum at \( t=0 \).
A sine wave hits its maximum when the argument is \( \frac{\pi}{2} \).
\( b(0 + c) = \frac{\pi}{2} \)
\( \frac{\pi}{6}(c) = \frac{\pi}{2} \)
\( c = 3 \).
(Alternatively, a cosine function starts at a max. \( \cos(x) = \sin(x + \frac{\pi}{2}) \). A shift of 3 hours is \( \frac{1}{4} \) of the 12-hour period, which aligns the sine start (midline) to the cosine start (max)).
Part C: Analysis on Interval \( (t_1, t_2) \)
The interval \( (t_1, t_2) \) corresponds to the section of the graph from point \( J \) (minimum at \( t=6 \)) to point \( K \) (midline at \( t=9 \)).
i. Properties of \( h \):
– Value: The function values rise from 190 to 200. These are all positive numbers. So, \( h \) is positive.
– Direction: The graph is going upwards from the minimum. So, \( h \) is increasing.
– Conclusion: Option (a) is correct.
ii. Concavity and Rate of Change:
– Concavity: On the interval from a minimum to an inflection point (the midline), the graph is shaped like part of a cup holding water. This shape is concave up (or \( h”(t) > 0 \)).
– Rate of Change: The rate of change is the slope of the tangent line. At \( J \) (the minimum), the slope is 0. As it moves toward \( K \), the slope becomes more positive (steeper). Since the slope is increasing from 0 to a positive value, the rate of change of \( h \) is increasing.
▶️ Answer/Explanation
A. i. Solve \(g(x) = 3\)
We are given the equation:
\(\log_4(2x) = 3\)
To solve for \(x\), convert the logarithmic equation to its exponential form using the definition \(y = \log_b(a) \iff b^y = a\):
\(2x = 4^3\)
Evaluate the power of 4:
\(2x = 64\)
Divide both sides by 2:
\(x = 32\)
A. ii. Solve \(h(x) = e^{1/2}\)
We are given the equation:
\(\frac{(e^x)^5}{e^{1/4}} = e^{1/2}\)
First, simplify the numerator using the power of a power rule \((a^m)^n = a^{mn}\):
\(\frac{e^{5x}}{e^{1/4}} = e^{1/2}\)
Simplify the left side using the quotient rule for exponents \(\frac{a^m}{a^n} = a^{m-n}\):
\(e^{5x – 1/4} = e^{1/2}\)
Since the bases are the same, we can equate the exponents:
\(5x – \frac{1}{4} = \frac{1}{2}\)
Add \(\frac{1}{4}\) to both sides (note that \(\frac{1}{2} = \frac{2}{4}\)):
\(5x = \frac{2}{4} + \frac{1}{4}\)
\(5x = \frac{3}{4}\)
Divide by 5:
\(x = \frac{3}{20}\)
B. i. Rewrite \(j(x)\)
We are given:
\(j(x) = \log_{10}(x+1) – 5\log_{10}(2-x) + \log_{10} 3\)
Use the power rule for logarithms \(n\log_b a = \log_b(a^n)\) to move the coefficient 5:
\(j(x) = \log_{10}(x+1) – \log_{10}((2-x)^5) + \log_{10} 3\)
Rearrange to group positive terms:
\(j(x) = \log_{10}(x+1) + \log_{10} 3 – \log_{10}((2-x)^5)\)
Use the product rule \(\log_b a + \log_b c = \log_b (ac)\) for the positive terms:
\(j(x) = \log_{10}(3(x+1)) – \log_{10}((2-x)^5)\)
Use the quotient rule \(\log_b a – \log_b c = \log_b (\frac{a}{c})\):
\(j(x) = \log_{10}\left( \frac{3(x+1)}{(2-x)^5} \right)\)
B. ii. Rewrite \(k(x)\)
We are given:
\(k(x) = \sec x – \cos x\)
Rewrite \(\sec x\) in terms of cosine:
\(k(x) = \frac{1}{\cos x} – \cos x\)
Find a common denominator:
\(k(x) = \frac{1 – \cos^2 x}{\cos x}\)
Use the Pythagorean identity \(1 – \cos^2 x = \sin^2 x\):
\(k(x) = \frac{\sin^2 x}{\cos x}\)
Separate the expression into a product:
\(k(x) = \sin x \cdot \frac{\sin x}{\cos x}\)
Substitute \(\tan x\) for \(\frac{\sin x}{\cos x}\):
\(k(x) = \sin x \tan x\)
C. Solve for \(m(x) = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
We are given the equation:
\(2 \tan^{-1}(\sqrt{3}\pi x) = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
First, evaluate the right side. The principal value of \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\) is \(\frac{\pi}{3}\):
\(2 \tan^{-1}(\sqrt{3}\pi x) = \frac{\pi}{3}\)
Divide both sides by 2:
\(\tan^{-1}(\sqrt{3}\pi x) = \frac{\pi}{6}\)
Take the tangent of both sides to remove the inverse tangent:
\(\sqrt{3}\pi x = \tan\left(\frac{\pi}{6}\right)\)
Evaluate \(\tan\left(\frac{\pi}{6}\right)\), which is \(\frac{1}{\sqrt{3}}\):
\(\sqrt{3}\pi x = \frac{1}{\sqrt{3}}\)
Isolate \(x\) by dividing both sides by \(\sqrt{3}\pi\):
\(x = \frac{1}{\sqrt{3} \cdot \sqrt{3} \pi}\)
Simplify the denominator (\(\sqrt{3} \cdot \sqrt{3} = 3\)):
\(x = \frac{1}{3\pi}\)