What are the extrema of #f(x)=(x^2)/(x^2-3x)+8 # on #x in[4,9]#?

1 Answer
Jun 10, 2018

Answer:

The given function is always decreasing and therefore has neither maximum nor minimum

Explanation:

The derivative of the function is

#y'=(2x(x^2-3x)-x^2(2x-3))/(x^2-3x)^2=#

#=(cancel(2x^3)-6x^2cancel(-2x^3)+3x^2)/(x^2-3x)^2=(-3x^2)/(x^2-3x)^2#

and

#y'<0 AA x in [4;9]#

The given function the function is always decreasing and therefore has neither maximum nor minimum

graph{x^2/(x^2-3x)+8 [-0.78, 17, 4.795, 13.685]}