Question 8.4.2: Find the maximum and minimum values, for x ∈ [−1, 3], of f (......

The function is differentiable everywhere, and

Solving the quadratic equation x^{2} − \frac{5}{2}x + 1 = 0, we get the roots x = 1/2 and x = 2. Thus f^{\prime}(x) = 0 for x = 0, x = 1/2, and x = 2. These three points, together with the two end points x = −1 and x = 3 of the interval, constitute the five candidate extreme points.We find that f (−1) = 7/12, f (0) = −1, f (1/2) = −185/192, f (2) = −5/3  and  f (3) = 5/4. Thus, the maximum value of f is 5/4, at x = 3; the minimum value is −5/3, at x = 2.