Maximum and Minimum Value of y = x - 1/x in the Interval (1/3, 4/3)

To find the maximum and minimum values of the function y = x - 1/x in the interval (1/3, 4/3), we need to analyze the behavior of the function within this interval.

1. Critical Points

A critical point occurs when the derivative of the function is equal to zero or undefined. Let's differentiate y = x - 1/x with respect to x:

dy/dx = 1 + 1/x^2

Setting dy/dx equal to zero, we have:

1 + 1/x^2 = 0

Simplifying this equation, we get:

1/x^2 = -1

Taking the reciprocal of both sides, we obtain:

x^2 = -1

This equation has no real solutions, which means there are no critical points within the interval (1/3, 4/3).

2. Endpoints

Next, let's evaluate the function at the endpoints of the interval (1/3, 4/3).

When x = 1/3, y = (1/3) - 3 = -8/3.
When x = 4/3, y = (4/3) - (3/4) = 13/4.

3. Second Derivative Test

Since there are no critical points within the interval, we can use the second derivative test to determine the nature of the function at the endpoints.

Taking the second derivative of y = x - 1/x, we have:

d^2y/dx^2 = 2/x^3

Substituting the values of x at the endpoints (1/3 and 4/3) into the second derivative, we get:

d^2y/dx^2 (1/3) = 2/(1/3)^3 = 2/1 = 2
d^2y/dx^2 (4/3) = 2/(4/3)^3 = 2/(64/27) = 27/32

Since the second derivative is positive at x = 1/3 and x = 4/3, the function is concave up at both endpoints.

4. Conclusion

From the above analysis, we can conclude that:

- The maximum value of y = x - 1/x in the interval (1/3, 4/3) occurs at x = 4/3, where y = 13/4.
- The minimum value of y = x - 1/x in the interval (1/3, 4/3) occurs at x = 1/3, where y = -8/3.

Therefore, the maximum value is 13/4 and the minimum value is -8/3 within the given interval.