Question

How do you find the local max and min for `f(x) = (3x) / (x² - 1)`?

Answer 1

`f(x) = (3x)/(x^2-1)` has no local extrema

Explanation:

Find the critical points by equating the first derivative to zero:

`f'(x) = frac (3(x^2-1)-6x^2) ((x^2-1)^2) = -3(x^2+1)/((x^2-1)^2)`

As the derivative is negative in all the domain of the function, the function is strictly decreasing and has no local extrema.

graph{3x/(x^2-1) [-10, 10, -5, 5]}