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

Answer:

$x_{1}=-2,x_{2}=4$

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

g(-2)=30

g(4)=-78

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

$g(-2)=(-2)^{3}-3(-2)^{2}-24(-2)+2=30\\\\g(4)=(4)^{3}-3(4)^{2}-24(4)+2=-78$