Question

What are the extrema of `f(x) = (3x) / (x² - 1)`?

Answer 1

The function contains no extrema.

Explanation:

Find `f'(x)` through the quotient rule.

`f'(x)=((x^2-1)d/dx(3x)-3xd/dx(x^2-1))/(x^2-1)^2`

`=>(3(x^2-1)-3x(2x))/(x^2-1)^2`

`=>(-3(x^2+1))/(x^2-1)^2`

Find the turning points of the function. These occur when the derivative of the function equals `0`.

`f'(x)=0` when the numerator equals `0`.

`-3(x^2+1)=0`
`x^2+1=0`
`x^2=-1`

`f'(x)` is never equal to `0`.

Thus, the function has no extrema.

graph{(3x)/(x^2-1) [-25.66, 25.66, -12.83, 12.83]}