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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?
Verified Answer
Consider the function y = x2 – 6x + 9. The maximum value of y ob...
This question is part of UPSC exam. View all Mechanical Engineering courses
Most Upvoted Answer
Consider the function y = x2 – 6x + 9. The maximum value of y ob...
Y = x^2 - 6x + 9
= (x - 3) ^ 2
so x = 5 has maximum value
f(5) = 4
c is correct
= (x - 3) ^ 2
so x = 5 has maximum value
f(5) = 4
c is correct
Community Answer
Consider the function y = x2 – 6x + 9. The maximum value of y ob...
To find the maximum value of the function y = x^2 - 6x + 9 over the interval 2 to 5, we can use calculus.
1. Find the derivative of the function y = x^2 - 6x + 9 with respect to x.
Taking the derivative, we get:
dy/dx = 2x - 6
2. Set the derivative equal to zero and solve for x to find the critical points.
Setting dy/dx = 0, we have:
2x - 6 = 0
2x = 6
x = 3
3. Check the endpoints of the interval.
Evaluate the function at the endpoints of the interval:
When x = 2, y = 2^2 - 6(2) + 9 = 4 - 12 + 9 = 1
When x = 5, y = 5^2 - 6(5) + 9 = 25 - 30 + 9 = 4
4. Check the critical point.
Evaluate the function at the critical point:
When x = 3, y = 3^2 - 6(3) + 9 = 9 - 18 + 9 = 0
5. Compare the values obtained.
The maximum value of y is the highest value obtained among the endpoints and critical points. In this case, the maximum value is y = 4, which occurs when x = 5.
Therefore, the correct answer is option B) 3.
1. Find the derivative of the function y = x^2 - 6x + 9 with respect to x.
Taking the derivative, we get:
dy/dx = 2x - 6
2. Set the derivative equal to zero and solve for x to find the critical points.
Setting dy/dx = 0, we have:
2x - 6 = 0
2x = 6
x = 3
3. Check the endpoints of the interval.
Evaluate the function at the endpoints of the interval:
When x = 2, y = 2^2 - 6(2) + 9 = 4 - 12 + 9 = 1
When x = 5, y = 5^2 - 6(5) + 9 = 25 - 30 + 9 = 4
4. Check the critical point.
Evaluate the function at the critical point:
When x = 3, y = 3^2 - 6(3) + 9 = 9 - 18 + 9 = 0
5. Compare the values obtained.
The maximum value of y is the highest value obtained among the endpoints and critical points. In this case, the maximum value is y = 4, which occurs when x = 5.
Therefore, the correct answer is option B) 3.
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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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