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

1 Answer
Nov 12, 2015

Answer:

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).