Find the absolute extrema of the function on a closed interval

Question

g(x) = (8x^2)/(x-2)  , [-2,1]

looking for: minimum (smaller x value) (x,y) = minimum (larger x value) (x,y) = maxiumum (x,y) =

Answer 1

Taking the derivative we find that f'(x)=8x(x-4)/(x-2)^2. This means that the function is increasing on the interval [-2,0] and decreasing on the interval [0,1]. Hence, the minimum is attained at x=-2 and x=1 and gives that value of f(-2)=f(1)=-8 and the maximum at x=0 which gives that values of f(0)=0.