What are the coordinates of the extrema?

Question

For the function f ( x ) = 2x^3-4x^2+x+6 Find the x-coordinates of all relative extrema. 2)Find the x-coordinate of the inflection point on the curve.

Answer 1

f ( x ) = 2x^3-4x^2+x+6

1) To find extreme points, you have to take First Order derivative and solve equation f'(x)=0. That is

f'(x)=6x²-8x+1

6x^2-8x+1=0

The two extreme point are 

x₁=(1/6) (4 - √10) (=0.13962) , x₂=(1/6) (4 + √10) (=1.1937).

take second derivative:

f''(x)=12x-8

f''(x1)=-8 + 2 (4 - √10) <0  ---> x1 is maxima

f''(x2)=-8 + 2 (4 + √10)>0  ---> x2 is min

Now you have to find f(x1), f(x2) and the coordinates in the question:

(x1, f(x1)) (x2, f(x2)).

2( Inflection point where f''(x) changes sign

f''(x)=12x-8=0

x=2/3 is inflection point