View an example | All parts showing X Find the absolute extrema if they exist, as well as all values of x where they occur, for the function f(x) = 2x5 - 36x2 + 162x - 4 on the domain [0, 13]. .. Let f be a function defined on some interval. Let c be a number in the interval. Then f(c) is the absolute maximum of f on the interval if f(x) f(c) for every x in the interval. A function has an absolute extremum at c if it has either an absolute maximum or an absolute minimum there. To find the absolute extrema of the function f(x) = 2x - 36x- + 162x - 4 on the domain [0,13], evaluate f for all critical numbers for fin (0,13) and for the endpoints 0 and 13. The largest value of f found in this way is the absolute maximum, and the smallest value of f found in this way is the absolute minimum. To find the critical numbers for f(x) = 2x - 36x + 162x - 4 in (0, 13), find those numbers c in the domain of f for which f'(c) = 0 or f'(c) does not exist. Begin by finding the derivative of f(x) = 2x3 - 36x2 + 162x - 4. f(x ) = 2x - 36x2 + 162x - 4 "'(x) = 6x2 -72x + 162 Set f'(x) = 0 and solve for x on the interval (0, 13). 0 = 6x2 - 72x + 162 0 = 6(x - 3)(x - 9) x = 9,3 Identify any values of x on the interval (0, 13) for which f'(x) = 6x2 - 72x + 162 is undefined. There are no values of x on the interval (0, 13) for which f'(x) is undefined. Therefore, the critical numbers for fin (0,13) are 3 and 9. Evaluate the function at these values and at the endpoints 0 and 13. x-value Value of Function 0 3 212 13 412 Recall that the smallest of the values of the function found above is the absolute minimum. Therefore, the absolute minimum of the function is - 4, which occurs at x = 0 and x = 9. The table showing the value of the function at each critical value and endpoint is redisplayed below. x-value Value of Function - 4 212 13 412 Recall that the largest of the values of the function found above is the absolute maximum. Therefore, the absolute maximum of the function is 412, which occurs at X = 13