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

Question

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

Emilia Aguilar · Accepted Answer

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]}

William Abdalla · Answer

undefined To find the extrema of the function $ f(x) = \frac{x^2}{x^2 - 3x} + 8 $ on the interval $[4,9]$, we first find the critical points by taking the derivative of the function and setting it equal to zero. Then, we evaluate the function at the critical points as well as at the endpoints of the interval to determine the extrema.

First, find the derivative of $ f(x) $ with respect to $ x $:
$$ f'(x) = \frac{d}{dx}\left(\frac{x^2}{x^2 - 3x} + 8\right) $$

To find the critical points, set $ f'(x) $ equal to zero and solve for $ x $.

Then, evaluate $ f(x) $ at each critical point as well as at the endpoints of the interval $[4,9]$ to determine the maximum and minimum values.