Calculus - Differentiation - Applications of Calculus.
Max/min of given equations - Test Yourself 1.
The questions on this page require the determination of both the first and second derivative of given equations.
The questions of this page focus on the following issues: |
1. Finding a minimum & maximum values. |
2. Absolute maximum/minimum values. |
3. Where is a curve increasing/decreasing. |
4. The number of solutions for an equation. |
Minimum & maximum values. | 1. The function y = x3 - 3x2 - 9x + 1 is defined in the domain [-2, 5].
Answer.(i) Maximum at (-1, 6) and minimum at (3, -26). (ii) POI at (1, -10). |
2. The cost of running a tour ($C) for x people can be expressed as C = x3 - 972x + 20,000.
Answer.(i) Need 18 people for min cost. (ii) Min cost/person = $463.11. |
3. A curve is determined by the equation
y = x3 - 12x + 4. Answer.(i) Maximum at (-2, 20) and minimum at (2, -12). (ii) POI at (0, 24). (iv) Maximum value of 20 at x = -2 and at x = 4. |
4. Consider the curve y = 20 - 5x2 - x3 for [-5, 2]. Answer.(i) Maximum at (0, 20) and minimum at (-10/3, 1.48). (ii) POI at (-5/3, 10.74). (iv) Maximum value is 20 at both x = -5 and at x = 0. |
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Absolute max/mins. | 5. For the function f(x) = 3x2 - x3 + 9x - 2:
Answer.(ii) Maximum value at (3, 25) Minimum value at (-1, -7). (iv) Minimum value of the function is -7 and that minimum occurs at x = -1 and at x = 5 in the given domain. |
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7. Consider the curve f(x) = x3 - 2x2 + x + 4 for [-2, 2].
Answer.(i) Maximum at (1/3, 112/27) and minimum at (1, 4). (ii) POI at (2/3, 110/27). (iii) (0, 4) and (-1, 0) (v) Max value is f(x) = 6. |
8. The function y = 2x3 + 3x2 - 36x + 5 is defined in the domain [-5, 5].
Answer.(i) Maximum at (-3, 86) and minimum at (2, -39). (ii) POI at (-0.5, 23.5). (iv) Maximum value is 150. |
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9. Find both the relative and the absolute maximum and minimum values for the function f(x) = x3 - 6x2 + 9x - 5 = 0 in the domain [0, 4]. |
10. Find both the relative and the absolute maximum and minimum values for the function
f(x) = 3x4 + 4x3 -12x2 - 3 in the domain [-2, 2]. Answer.The local minimum and absolute minumum are -35 at x = -2.The local maximum is -8 when x = 0 and the absolute maximum in the domain is 29. | |
11. Find the absolute maximum and minimum values of the function in the domain [1, 9].
Answer.The minimum value is 1 at x = 1 and the maximum value is 5 at x = 9. We cant do calculus with this question to obtain the values!!!. |
12. Find the absolute maximum and minimum values of the function in the domain [1, 9].
Answer.The minimum value is 2/3 at x = 1 and the maximum value is 2/19 at x = 9. We can't do calculus with this question to obtain the values!!!. |
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Curve sketching. | 13. A function is defined by .
Answer.(ii) Maximum at (0, 0) and minimum at (4, -6.4). (iii) curve is decreasing 0 ≤ x ≤ 4. |
14. For the curve :
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15. Let f(x) = x3 + kx2 + 3x - 5 where k is a constant.
Find the values of k for which f(x) has NO stationary points. Answer.-3 < k < 3. |
16. (i) Find the derivative of
(ii) Find the value(s) of m for which the graph of has no stationary points. Answer.-5 < m < 3. |
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17. For the curve y = 7 + 5x - x2 - x3,
(i) find the co-ordinates of the maximum and minimum values. (ii) Find the point where the concavity changes. (iii) Sketch the curve. (iv) For what values of x is the curve concave down? |
18. For the curve (i) Find the co-ordinates of where the maximum and minimum values occur and determine their nature. (ii) Sketch the curve. (iii) For what x values is the curve concave up. |
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19. For the curve y = 3x4 - 8x3 + 6:
Answer.(i) Min at (2, -10) (ii) HPOI at (0, 6). and POI at (4/3, -3.48). |
20. Given f(x) = (x + 2)(x - 2)3
and f '(x) = 4(x - 2)2(x + 1) = 4x3 - 12x2 + 16 Answer.(i) Min at (-1, -27) (ii) HPOI at (2, 0) POI at (0, -16). |
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Number of solutions. | 21. Consider the curve y = 2x2 (x - 3)2.
Answer.(i) Max at (1.5, 10.125) Min at (0, 0) and at (3, 0). (iii) No. of solutions = 2. |
22. A function can be expressed as
f(x) = 3x4 + 4x3 - 12x2 Answer.(i) Max at (0, 0) Min at (-2, 32) and at (-1, 5). (iii) Increasing for -2 < x < 0 and for x > 1. (iv) k < -32.(v) k = 0 or k = -5. |