Question

use the first derivative test to find the functions extreme values and to find where the extreme values exist. justify your answer

Answer 1

Below ( kinda long)

Explanation:

Use fist derivative to find where the derivative = 0

Derivative = 4 x^3 + 12 x^2

set = to zero

4x^3 + 12x^2 = 0

4x^2 ( x + 3) = 0 shows x = 0 or -3

Now take the SECOND derivative

12x^2 + 24x and sub in the values found above ( 0 and -3)

if the result is >0 this point is a local min, if < 0 it is a local max and if = 0 it is neither a min or a max....it is an INFLECTION point

12x^2 + 24x for x = -3 this equals +36 <=====local min

for x = 0 this equals 0 <======inflection point

SO at -3 there is a local min extreme....find the value at -3 :

x^4 + 4x^3 - 2 =

(-3)^4 + 4 ( -3^3) - 2 = - 29 the local min is at (-3,-29)