What are the local extrema of #f(x)= x^3-3x^2-9x+7#?

Question

What are the local extrema of #f(x)=  x^3-3x^2-9x+7#?

Isaiah Allen · Accepted Answer

Local maximum : #f(-1)=12#. Local minimum :# f(3)=-20#.

#f = x^3(1-3/x-9/x^2+7/x^3) to +-oo#, as #x to +-oo#. f'=3(x^2-2x-3)=0, at x = -1 and 3. #f''=6x-6, <9#, at #x = -1, >0#, at #x = 3 and = 0#, at #x =1.# So, #local-max f = f(-1)=12 and local-min f = f(3)=-20#. As, #f''' ne 0, ( 1, -4 )# is a POI ( point of inflexion ). graph{(x^3-3x^2-9x+7-y)((x-1)^2+(y+4)^2-.01)=0 [-34, 34, -21, 13]}

Charles Anctil · Answer

undefined To find the local extrema of $ f(x) = x^3 - 3x^2 - 9x + 7 $, we first find its critical points by taking the derivative and setting it equal to zero. Then, we can determine whether each critical point corresponds to a local maximum or minimum by examining the sign of the second derivative.

First, find the derivative of $ f(x) $:
$$ f'(x) = 3x^2 - 6x - 9 $$

Setting $ f'(x) $ equal to zero and solving for $ x $:
$$ 3x^2 - 6x - 9 = 0 $$
$$ x^2 - 2x - 3 = 0 $$
$$ (x - 3)(x + 1) = 0 $$

So, the critical points are $ x = 3 $ and $ x = -1 $.

Next, find the second derivative of $ f(x) $:
$$ f''(x) = 6x - 6 $$

Now, evaluate $ f''(x) $ at each critical point:
$$ f''(3) = 6(3) - 6 = 18 - 6 = 12 $$
$$ f''(-1) = 6(-1) - 6 = -6 - 6 = -12 $$

Since $ f''(3) > 0 $, the point $ x = 3 $ corresponds to a local minimum. Since $ f''(-1) < 0 $, the point $ x = -1 $ corresponds to a local maximum.

Therefore, the local minimum occurs at $ x = 3 $, and the local maximum occurs at $ x = -1 $.