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Consider the function y = x2 – 6x + 9. The maximum value of y obtained when x varies over the interval 2 to 5 is
- a)1
- b)3
- c)4
- d)9
Correct answer is option 'B'. Can you explain this answer?
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Consider the function y = x2 – 6x + 9. The maximum value of y ob...
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Consider the function y = x2 – 6x + 9. The maximum value of y ob...
Analysis of the Function:
The given function is y = x^2 - 6x + 9. This is a quadratic function, and its graph is a parabola. To find the maximum value of y within the interval [2, 5], we need to analyze the function further.
Finding the Maximum Value:
To find the maximum value of the function, we first need to determine the critical points. Critical points occur where the derivative of the function is equal to zero or undefined. Taking the derivative of the function y = x^2 - 6x + 9, we get:
dy/dx = 2x - 6
Setting this derivative equal to zero and solving for x gives us the critical point:
2x - 6 = 0
2x = 6
x = 3
Checking Endpoints:
Next, we need to check the function values at the endpoints of the interval [2, 5], which are x = 2 and x = 5. Calculate the function values at these points:
For x = 2: y = 2^2 - 6(2) + 9 = 1
For x = 5: y = 5^2 - 6(5) + 9 = 4
Comparing Values:
Now, compare the function values at the critical point and endpoints to determine the maximum value of y within the interval [2, 5]. The maximum value is the highest among these values.
The function values are:
- At x = 2, y = 1
- At x = 3 (critical point), y = 0
- At x = 5, y = 4
Therefore, the maximum value of y obtained when x varies over the interval [2, 5] is 4, which corresponds to option (c) in the question.
The given function is y = x^2 - 6x + 9. This is a quadratic function, and its graph is a parabola. To find the maximum value of y within the interval [2, 5], we need to analyze the function further.
Finding the Maximum Value:
To find the maximum value of the function, we first need to determine the critical points. Critical points occur where the derivative of the function is equal to zero or undefined. Taking the derivative of the function y = x^2 - 6x + 9, we get:
dy/dx = 2x - 6
Setting this derivative equal to zero and solving for x gives us the critical point:
2x - 6 = 0
2x = 6
x = 3
Checking Endpoints:
Next, we need to check the function values at the endpoints of the interval [2, 5], which are x = 2 and x = 5. Calculate the function values at these points:
For x = 2: y = 2^2 - 6(2) + 9 = 1
For x = 5: y = 5^2 - 6(5) + 9 = 4
Comparing Values:
Now, compare the function values at the critical point and endpoints to determine the maximum value of y within the interval [2, 5]. The maximum value is the highest among these values.
The function values are:
- At x = 2, y = 1
- At x = 3 (critical point), y = 0
- At x = 5, y = 4
Therefore, the maximum value of y obtained when x varies over the interval [2, 5] is 4, which corresponds to option (c) in the question.
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Consider the function y = x2 – 6x + 9. The maximum value of y obtained when x varies over the interval 2 to 5 is a)1 b)3 c)4 d)9Correct answer is option 'B'. Can you explain this answer?
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