Question #336dc

Important Note :

We can use the " Second Derivative Test " to find the " Extremas " of a function.

The Second Derivative Test states that:

If f"(C) > Zero , where "C" is a Critical Point, then "C" is the Local Minima

If f"(C) < Zero , where "C" is a Critical Point, then "C" is the Local Maxima

Step 1 :

Find the First Derivative

We have our original function #f(x) = x ^ 4 - 2x^2 + 3#

Therefore, #f'(x) = 4x^3 - 4x#

Step 2 :

Next, we will move on to find our potential "Critical Points"

To find our Critical Points, we must set the First Derivative equal to Zero

Therefore, #f'(x) = 4x^3 - 4x = 0#

#rArr 4x(x^2 - 1) = 0#

#rArr 4x(x + 1)(x -1) = 0#

#rArr 4x = 0; (x + 1) = 0; (x -1) = 0#

#rArr x = 0; x = -1 ; x = +1#

To find out whether or not our potential Critical Points represent a Local Maxima or a Local Minima, we want to plug them into our Second Derivative.

Step 3 :

Next, we will move on to find our Second Derivative

We have, #f'(x) = 4x^3 - 4x#

f"(x) #= 12x^2 - 4#

Substitute our Critical Points to to f"(x)

f"( 0 ) #= 12(0)^2 - 4 = -4#

f"( -1 ) #= 12(-1)^2 - 4 = 12-4 = 8#

f"( +1 ) #= 12(+1)^2 - 4 = 12-4 = 8#

Observe from the results above

f"( 0 ) = -4, which is < 0 #rArr# Local Maxima at x = 0

f"( -1 ) = 8, which is > 0 #rArr# Local Minima at x = -1

f"( +1 ) = 8, which is > 0 #rArr# Local Minima at x = +1