Question 638252
If i have this function : 
f( x ) =1+ 12x – x^3 
which is the minimun relative?
---
Take the derivative.
f'(x) = -3x^2+12
---
Solve:
-3x^2 + 12 = 0
x^2 = 4
x = +2 or x = -2
------
At this point you might use the 2nd devivative test.
f''(x) = -6x
---
f(-2) = 12, so relative min at x = -2
f(2) = -12, so relative max at x = 2
============
{{{graph(400,400,-10,10,-25,25,1+12x-x^3)}}}
============
Cheers,
Stan H.
============