Methods 3/4 Questions
Educational · TutoringQuestions
1. Consider two functions, 8! p(x) = x^3 + 2x !8 and 8! q(x) = x^2 - 1 !8, defined for all real numbers 8! x !8. Determine which of the following functions are odd and which are even: 8! p(x) !8, 8! q(x) !8, 8! p(x) + q(x) !8, and 8! p(x)q(x) !8.
2. Find the equation of the parabola that results when the graph of 8! y = (x + 2)^2 !8 is reflected across:
a) the line 8! y = 0 !8
b) the line 8! x = 0 !8
3. A buoy in the ocean experiences a periodic change in depth due to tidal forces. The depth of the buoy, 8!d(t)!8 meters, can be modeled by the equation 8!d(t) = A \sin(nt + \epsilon) + c!8, where 8!t!8 is the number of hours after midnight. If the average depth of the buoy is 3 meters, the maximum depth is 6 meters, and the minimum depth occurs 6 hours after midnight, find the values of 8!A!8, 8!n!8, 8!c!8, and 8!\epsilon!8.
4. A particle moves along a straight line according to the equation 8! x = 5 + 3 \cos(2t) !8, where 8! x !8 is the position in metres and 8! t !8 is the time in seconds. Find the maximum distance of the particle from the point where 8! x = 5 !8.
5. A carnival ride has a height, 8!h!8 m, above the ground at time 8!t!8 seconds given by the equation 8!h = 8\sin(5t - 30)^\circ + 10!8. If the ride stops after 50 seconds, how far above the ground are the riders stranded?
6. Find the absolute maximum and minimum values of the function g(x) = 5 - 2x^2 on the interval [-2, 2].
Answers and Explanations
Answer 1:
- 8! p(x) !8 is odd
- 8! q(x) !8 is even
- 8! p(x) + q(x) !8 is neither odd nor even
- 8! p(x)q(x) !8 is odd
To determine if a function is odd or even, we check if 8! f(-x) = f(x) !8 (even) or 8! f(-x) = -f(x) !8 (odd). Applying this to each function, we can determine their parity.
Answer 2:
a) 8! y = -(x + 2)^2 !8
b) 8! y = (-x + 2)^2 !8
a) Reflecting across the line 8! y = 0 !8 (x-axis) inverts the sign of the function.
b) Reflecting across the line 8! x = 0 !8 (y-axis) inverts the sign of the input variable 8! x !8.
Answer 3:
8!A = 3!8, 8!n = \frac{\pi}{6}!8, 8!c = 3!8, 8!\epsilon = 0!8
Since the average depth is 3 meters and the maximum depth is 6 meters, the amplitude 8!A!8 is half the difference between the maximum and average depths, so 8!A = 3!8. The average depth is also the vertical shift 8!c!8, so 8!c = 3!8. Since the minimum depth occurs 6 hours after midnight and the period is 12 hours, the angular frequency 8!n!8 is 8!\frac{\pi}{6}!8. The phase shift 8!\epsilon!8 is 0 since the maximum depth occurs at midnight.
Answer 4:
The maximum distance is 3 metres.
The cosine function oscillates between -1 and 1. Therefore, the maximum value of 8! 3 \cos(2t) !8 is 3, which occurs when 8! \cos(2t) = 1 !8. This gives a maximum distance of 3 metres from the point where 8! x = 5 !8.
Answer 5:
To find the height after 50 seconds, we substitute 8!t = 50!8 into the equation: 9!h = 8\sin(5(50) - 30)^\circ + 10!9 9!h = 8\sin(250 - 30)^\circ + 10!9 9!h = 8\sin(220)^\circ + 10!9 Using a calculator to find 8!\sin(220^\circ)!8, we get: 9!h \approx 8(-0.6428) + 10!9 9!h \approx -5.1424 + 10!9 9!h \approx 4.8576!9 So, the riders are stranded approximately 4.86 m above the ground.
We used the given equation to find the height at 8!t = 50!8 seconds by substituting the value of 8!t!8 and evaluating the trigonometric function.
Answer 6:
The absolute maximum value is 5 at x = 0, and the absolute minimum value is -3 at x = ±2.
To find the absolute maximum and minimum values, we need to evaluate the function at its critical points and endpoints. The critical point is where g’(x) = 0, which is at x = 0. The endpoints are x = -2 and x = 2. Evaluating g(x) at these points, we get g(0) = 5, g(-2) = -3, and g(2) = -3. Therefore, the absolute maximum value is 5 at x = 0, and the absolute minimum value is -3 at x = ±2.