Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 3

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.