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

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Chapter 4, Review - Problem 3 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the local absolute extrema of the function, hence, you need to find the zeroes of the equation f'(x) = 0.

You need to evaluate the derivative using the quotient rule:

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

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

`f'(x) = (3x^2 + 3 - 6x^2 + 8x)/((x^2+1)^2)`

`f'(x) = (-3x^2 + 3 + 8x)/((x^2+1)^2)`

You need to solve for x the equation f'(x) =0:

`(-3x^2 + 3 + 8x)/((x^2+1)^2) = 0 => -3x^2 + 3 + 8x = 0`

`3x^2 - 8x - 3 = 0 => x_(1,2) = (8+-sqrt(64 + 36))/6`

`x_(1,2) = (8+-sqrt(100))/6 => x_(1,2) = (8+-10)/6`

`x_1 = 3 ; x_2 = -1/3`

You need to evaluate the function at critical points:

`f(3) = (3*3 - 4)/(3^2+1) => f(3) = (5)/(10) => f(3) = 1/2`

`f(-1/3) = (3*(-1/3) - 4)/((-1/3)^2+1) => f(-1/3) = (-5)/(10/9) => f(-1/3) = -9/2`

You need to evaluate the function at the end points of interval:

`f(-2) = (3*(-2) - 4)/((-2)^2+1) =>f(-2) = (-6 - 4)/(4+1)= > f(-2) = -2`

`f(2) = (3*2 - 4)/(2^2+1) => f(2) = 2/5`

Hence, the absolute maximum of the function, on the interval [-2,2], is 1/2 and it occurs at x = 3 and the absolute minimum of the function is `-9/2` and it occurs at `x = -1/3.`

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