Question

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

Answer 1

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'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]}