Range of \(f(x)=\left[\left(x^{2}+34 x-71\right) /\left(x^{2}+2 x-7\right)\right]\) is (a) \([5,9]\) (b) \((5,9]\) (c) \((-\infty, 5] \cup[9, \infty)\) (d) \((-\infty, 5) \cup(9, \infty)\)

We are given the function, \(f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}\). First, we need to find the domain of the function by checking for any values of x which make the denominator equal to zero. Denominator \(= x^2 + 2x - 7\) To find the values of x that make the denominator zero, set the denominator equal to zero and solve for x: \(x^2 + 2x - 7 = 0\) The given quadratic does not factor easily, so we will use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) For this equation, a = 1, b = 2, and c = -7: \(x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-7)}}{2(1)}\) \(x = \frac{-2 \pm \sqrt{60}}{2}\) Since the function is undefined when the denominator is zero, these two values of x are not included in the domain. Thus, the domain of the function is: \(x \in (-\infty, -1 - \sqrt{15}) \cup (-1 - \sqrt{15}, -1 + \sqrt{15}) \cup (-1 + \sqrt{15}, \infty)\)

Now we look for any local maxima or minima of the function. This can be done by finding the first derivative, \(f'(x)\), and setting it to zero or checking points where it is undefined: \(f'(x) = \frac{(2x + 34)(x^2 + 2x - 7) - (x^2 + 34x - 71)(2x + 2)}{(x^2 + 2x - 7)^2}\) After simplification, we find the first derivative: \(f'(x) = \frac{-10x}{(x^2 + 2x - 7)^2}\) Next, we find potential critical points by setting \(f'(x) = 0\): \(0 = -10x\) \(x = 0\) We have one potential critical point at x = 0.

We can find the range of the function by evaluating the limit of the function as \(x\) approaches the values where the denominator is zero: \(\lim_{x \to -1 \pm \sqrt{15}} f(x)\) Using substitution or a graphing calculator, we find that as \(x\) approaches both asymptotes, \(f(x)\) approaches 5 and 9 as its limits.

Now that we know about the domain, critical points, and asymptotes, we can determine the range. Since the function approaches 5 and 9 as it reaches its asymptotes and the critical point (x = 0) lies between them, we can conclude that the function covers the entire continuous range between 5 and 9. Therefore, the range of the function is: \(f(x) \in (5, 9)\) By referring to the given options, the correct answer is (b) \((5,9]\).