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

1 Answer
Jan 10, 2016

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)