Question

What are the local extrema, if any, of `f (x) =(x^3-3)/(x+6)`?

Answer 1

The one real number critical point of this function is `x approx -9.01844`. A local minimum occurs at this point.

Explanation:

By the Quotient Rule, the derivative of this function is

`f'(x)=((x+6)*3x^2-(x^3-3)*1)/((x+6)^2)=(2x^3+18x^2+3)/((x+6)^2)`

This function equals zero if and only if `2x^3+18x^2+3=0`. The roots of this cubic include on negative irrational (real) number and two complex numbers.

The real root is `x approx -9.01844`. If you plug in a number just less than this into `f'`, you'll get a negative output and if you plug a number just greater than this into `f'`, you'll get a positive output. Therefore, this critical point gives a local minimum value of `f` (and `f(-9.01844) approx 244` is the local minimum value (output).