Answer:
(2, -1)
Step-by-step explanation:
Given the function f(x) = 8x³ − 12x² − 48x, the critical point of the function occurs at its turning point i,e at f'(x) = 0
First we have to differentiate the function as shown;
[tex]f'(x)= 3(8)x^{3-1}- 2(12)x^{2-1} - 48x^{1-1}\\ \\f'(x) = 24x^2 - 24x-48x^0\\\\f'(x) = 24x^2 - 24x-48\\\\At \ the\turning\ point\ f'(x)= 0\\24x^2 - 24x-48 = 0\\\\\\[/tex]
[tex]Dividing \ through \ by \ 24\\\\x^2-x-2 = 0\\\\On \ factorizing\\\\x^2-2x+x-2 = 0\\\\x(x-2)+1(x-2) = 0\\\\(x-2)(x+1) = 0\\\\x-2 = 0 \ and \ x+1 = 0\\\\x = 2 \ and \ -1[/tex]
Hence the critical numbers of the function are (2, -1)
