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

Jun 10, 2018

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

#### Explanation:

The derivative of the function is

$y ' = \frac{2 x \left({x}^{2} - 3 x\right) - {x}^{2} \left(2 x - 3\right)}{{x}^{2} - 3 x} ^ 2 =$

$= \frac{\cancel{2 {x}^{3}} - 6 {x}^{2} \cancel{- 2 {x}^{3}} + 3 {x}^{2}}{{x}^{2} - 3 x} ^ 2 = \frac{- 3 {x}^{2}}{{x}^{2} - 3 x} ^ 2$

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]}