How do you find the local max and min for #y = x^3 − 12x + 9#?
1 Answer
May 21, 2016
local max at (-2 .25)
local min at (2 ,-7)
Explanation:
To find the stationary points (S.P's) or Turning points (T.P's) we require to differentiate the function and equate to zero.
To find their nature use the 2nd derivative test.
• If f''(a) > 0 , then min T.P
• If f''(a) < 0 , then max T.P
now f(x)
#=x^3-12x+9rArrf'(x)=3x^2-12# Equating to zero :
#3x^2-12=0#
#rArr3(x^2-4)=0rArrx=±2#
#f(-2)=(-2)^3-12(-2)+9=25# and
#f(2)=(2)^3-12(2)+9=-7# hence T.P's at (-2 ,25) and (2 ,-7)
#f''(x)=6x#
Nature of T.P'sf''(-2) = 6(-2) = -12 < 0 , (-2 ,25) is a local max.
f''(2) = 6(2) = 12 > 0 , (2 ,-7) is a local min.
graph{x^3-12x+9 [-64.07, 64.1, -32.04, 32.03]}