# How do you find the local max and min for y = x^3 − 12x + 9?

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)$= {x}^{3} - 12 x + 9 \Rightarrow f ' \left(x\right) = 3 {x}^{2} - 12$

Equating to zero :$3 {x}^{2} - 12 = 0$

rArr3(x^2-4)=0rArrx=±2

$f \left(- 2\right) = {\left(- 2\right)}^{3} - 12 \left(- 2\right) + 9 = 25$

and $f \left(2\right) = {\left(2\right)}^{3} - 12 \left(2\right) + 9 = - 7$

hence T.P's at (-2 ,25) and (2 ,-7)

$f ' ' \left(x\right) = 6 x$
Nature of T.P's

f''(-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]}