# Find the local and absolute extreme values of the function on the given interval. f(x)= (3x-4)/(x^2+1), [-2,2]

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### 1 Answer

We have to determine the extreme values of f(x)= (3x-4)/(x^2+1), [-2,2]

The extreme values of a function f(x) lie at f'(x) = 0

f(x) = (3x-4)/(x^2+1)

f'(x) = [3*(x^2 + 1) - (3x - 4)2x]/(x^2 + 1)^2

f'(x) = 0

=> 3*(x^2 + 1) - (3x - 4)2x = 0

=> 3x^2 + 3 - 6x^2 + 8x = 0

=> -3x^2 + 8x + 3 = 0

=> -3x^2 + 9x - x + 3 = 0

=> -3x(x - 3) - 1(x - 3) = 0

=> (-3x - 1)(x -3) = 0

=> x = -1/3 and x = 3

As only x = -1/3 lies in [-2, 2], we consider only this.

At x = -1/3, f(x) = (3*(-1/3)-4)/((-1/3)^2+1)

=> `(-1-4)/(1/9+1)`

=> `-5/((10/9))`

=> -45/10

=> -9/2

**The extreme value of the function in the given range is -9/2 at x = -1/3**