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The maximum and minimum value of y=x+1/x in interval (1/3,4/3) 1)2,-2. 2)10/3,2. 3) none of these. and the correct answer is 2)?
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The maximum and minimum value of y=x+1/x in interval (1/3,4/3) 1)2,-2....
Explanation:
Given, y = x 1/x in interval (1/3,4/3)
To find the maximum and minimum value of y, we differentiate y with respect to x and equate it to zero.
dy/dx = (1/x) - (1/x^2) = 0
x = 1
To confirm whether x = 1 is a point of maximum or minimum, we differentiate y^2 with respect to x and check its sign.
d(y^2)/dx = (2/x) - (2/x^3)
d(y^2)/dx at x = 1 is 0. Hence, we check the sign of d(y^2)/dx at x = 1 -
d(y^2)/dx at x = 0.9 is positive, and d(y^2)/dx at x = 1.1 is negative. Hence, x = 1 is a point of maximum.
Now, we substitute x = 1/3, 4/3 in y = x 1/x to find the minimum and maximum values of y.
y(1/3) = (1/3)^(3) = 1/27
y(4/3) = (4/3)^(3/4) = 2
Therefore, the maximum value of y is 2 and the minimum value of y is 1/27.
Answer:
The correct answer is option 2) 10/3, 2.
Given, y = x 1/x in interval (1/3,4/3)
To find the maximum and minimum value of y, we differentiate y with respect to x and equate it to zero.
dy/dx = (1/x) - (1/x^2) = 0
x = 1
To confirm whether x = 1 is a point of maximum or minimum, we differentiate y^2 with respect to x and check its sign.
d(y^2)/dx = (2/x) - (2/x^3)
d(y^2)/dx at x = 1 is 0. Hence, we check the sign of d(y^2)/dx at x = 1 -
d(y^2)/dx at x = 0.9 is positive, and d(y^2)/dx at x = 1.1 is negative. Hence, x = 1 is a point of maximum.
Now, we substitute x = 1/3, 4/3 in y = x 1/x to find the minimum and maximum values of y.
y(1/3) = (1/3)^(3) = 1/27
y(4/3) = (4/3)^(3/4) = 2
Therefore, the maximum value of y is 2 and the minimum value of y is 1/27.
Answer:
The correct answer is option 2) 10/3, 2.
Community Answer
The maximum and minimum value of y=x+1/x in interval (1/3,4/3) 1)2,-2....
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