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

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Answer 1 · 2018-06-10T07:29:32

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

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]}

What are the extrema of f(x)=(x^2)/(x^2-3x)+8 f(x)=x2x2−3x+8f(x)=(x^2)/(x^2-3x)+8 on x in[4,9]x∈[4,9]x in[4,9]?