Expert: Paul Klarreich - 12/29/2007

Question
let f''(x)=4x^3-2x and let f(x) have critical numbers -1,0, and 1.Use the Second derivative test to determine if any of the critical numbers gives a relative maximum.

Answer
Questioner:   Joyice Harris
Category:  Calculus
Private:  No
 
Subject:  AP calculus
Question:  let f''(x)=4x^3-2x and let f(x) have critical numbers -1,0, and 1.Use the Second derivative test to determine if any of the critical numbers gives a relative maximum.
 
The second derivative test says;

f''(x) > 0  ==> min
f''(x) < 0  ==> max
f''(x) = 0  ==> we don't know and have to investigate more.

f''(x)=4x^3-2x:

f(-1) = 4(-1)^3-2(-1) = -4 + 2 = -2 < 0 ==> min

f(0) = 4(0)^3-2(0) = 0 ==> ??

f(1) = 4(1)^3-2(1) = 4 - 2 = 2 > 0 == max

About x = 0; the the '2nd' test is unrevealing, as the doctors say, so we have to work harder.

 
f''(x)= 4x^3 - 2x

Integrate:

f'(x) = x^4 - x^2 + C.

Since f'(0) = 0 [Why?] we have C = 0.

f'(x) = x^2(x^2 - 1)
f'(x) = x^2(x - 1)(x + 1)



IF x > 0, but < 1, we have:

f'(x > 0, but < 1) = (pos)^2(neg)(pos) = neg; graph falling.


IF x > 0, but < 1, we have:

f'(x < 0, but > -1) = (neg)^2(neg)(pos) = neg; graph falling.

Since the graph is falling on both sides, x = 0 is neither a min nor a max.
A plot of f(x) = x^5/5 - x^3/3 will confirm this.