Question

Find the​ x-value of all points where the function below has any relative extrema. Find the​ value(s) of any relative extrema. ​G(x)equalsx cubed minus 3 x squared minus 24 x plus 2

Answer 1

`x_(1)=-2,x_(2)=4`

value of g(x) at these points are as follows

g(-2)=30

g(4)=-78

Step-by-step explanation:

Given

g(x)=
`x^(3)-3x^(2)-24x+2`

Differentiating with respect to x we get

`g'(x)=3x^(2)-6x-24`

to obtain point of extrema we equate g'(x) to zero

`g'(x)=3x^(2)-6x-24\\\\\therefore 3x^(2)-6x-24=0\\\\\Rightarrow x^(2)-2x-8=0\\\\x^(2)-4x+2x-8=0\\x(x-4)+2(x-4)=0\\(x+2)(x-4)=0`

Thus the critical points are obtained as
`x_(1)=-2,x_(2)=4`

The values at these points are as

`g(-2)=(-2)^(3)-3(-2)^(2)-24(-2)+2=30\\\\g(4)=(4)^(3)-3(4)^(2)-24(4)+2=-78`