Determine the absolute extrema of the function 𝑓(𝑥)=𝑥3−18𝑥2 96𝑥 , on the interval [1,10]

Answer: absolute minimum is at f(10) = -704 absolute maximum is at f(1) = 79 Solution: f(x) = x³ - 18x² + 96x f'(x) = 3x² - 36x + 96 then solve for x where f'(x) = 0 f'(x) = 3x² - 36x + 96 0 = 3x² - 36x + 96 0 = 3(x² - 12x + 32) 0 = x² - 12x + 32 0 = (x - 4)(x - 8) x = 4 x = 8 Since 4 and 8 is in the interval [1,10] we can solve for f(x) using 1,4,8, and 10 f(x) = x³ - 18x² + 96x f(1) = 1³ - 18(1)² + 96(1) = 79 f(4) = 4³ - 18(4)² + 96(4) = -128 f(8) = 8³ - 18(8)² + 96(8) = -544 f(10) = 10³ - 18(10)² + 96(10) = -704 making the value at f(10) the absolute minimum and the value at f(1) the absolute maximum