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

Jan 10, 2016

Relative maxima at x= $- \frac{1}{3}$ and minima at x= #1/3)

Explanation:

$\frac{\mathrm{df}}{\mathrm{dx}} = 9 - \frac{1}{x} ^ 2$. Equating it to 0 , $x = \pm \frac{1}{3}$. f'(x) does not exist at x=0 hence the critical points are $- \frac{1}{3} , 0. \mathmr{and} + \frac{1}{3}$. For using first derivative test, test the increasing/decreasing behaviour in intervals $\left(- \infty , - \frac{1}{3}\right) , \mathmr{and} \left(\frac{1}{3} , \infty\right)$. 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 $\left(- \infty , - \frac{1}{3}\right) ,$-1/6 in $\left(- \frac{1}{3} , 0\right) ,$1/6 in (0, 1/3) and +1 in $\left(\frac{1}{3} , \infty\right)$

f ' (x) is positive in $\left(- \infty , - \frac{1}{3}\right)$,

negative in $\left(- \frac{1}{3} , 0\right)$,
negative in $\left(0 , \frac{1}{3}\right)$and

positive in $\left(\frac{1}{3} , \infty\right)$

Conclusion is there is a relative maxima at x=$- \frac{1}{3}$ ( function changes from increasing to decreasing) and relative minima at x= $\frac{1}{3}$ (function changes from decreasing to increasing)