How do you find the relative extrema for #f(x)=(9x^(2)+1)/x#?

1 Answer
Jan 10, 2016

Answer:

Relative maxima at x= #-1/3# and minima at x= #1/3)

Explanation:

#(df)/dx= 9-1/x^2#. Equating it to 0 , #x=+-1/3#. f'(x) does not exist at x=0 hence the critical points are #-1/3, 0. and +1/3#. For using first derivative test, test the increasing/decreasing behaviour in intervals #(-oo,-1/3), and (1/3, oo)#. If f '(x) is positive, then f(x) is increasing and if it is negative, then f(x) is decreasing. Take up any test value say -1 in #(-oo,-1/3), #-1/6 in #(-1/3, 0) , #1/6 in (0, 1/3) and +1 in # (1/3, oo)#

f ' (x) is positive in #(-oo, -1/3)#,

negative in #(-1/3,0)#,
negative in #(0, 1/3) #and

positive in #(1/3, oo)#

Conclusion is there is a relative maxima at x=# -1/3 # ( function changes from increasing to decreasing) and relative minima at x= #1/3# (function changes from decreasing to increasing)