Find the absolute (global) Maximum and absolute (global) Minimum values of the function f(x)= (x^2)/(x^2-9) on the interval [-2,1].

Question

anonymous · Answer

check $f(-2), f(1)$  and then find the critical point by finding the derivative, set it equal to zero and solve 
check the function evaluated at  that number as well

anonymous · Answer

f(-2)= -4/5   f(1)= -1/8.. the derivative is (2x(x^2-9)-x^2(2x))/(x^2-9)^2

anonymous · Answer

-18x/(x^2-9)^2=0

anonymous · Answer

x=0

campbell_st · Answer

an easy solution to this problem is to use some graphing software and graph it...