classify the critical points of function using second derivitive test function is x(x-2)^2+1?

Answer 1

f(x)=x(x-2)^2+1
=x(x^2-4x+4)+1
=x^3-4x^2+4x+1
f'(x)=3x^2-8x+4
setting thisto zero
3x^2-6x-2x+4=0
=(3x-2)(x-2)=0
or x=2/3 or 2
f''(x)=6x-8
at x=2/3 f''(x)=-ve so f(x) has a local max
and at x=2 f''(x) is +ve so f(x) has a local min.

Answer 2

d( x(x-2)²+1 )/dx =
d( x(x²-4x+4)+1 )/dx =
d( x³-4x²+4x+1 )/dx = 3x²-8x+4

When 3x²-8x+4=0 the critical points are:
3x²-8x+4=0

x=[8+-(8²-4*3*4)^¹/2] / (2*3)
x=[8+-(64-48)^¹/2] / 6
x=[8+-16^¹/2] / 6
x=[8+-4] / 6

x1 = 2/3 ... or ... x2 = 2

=>
y1 = (2/3)³-4(2/3)²+4(2/3)+1
y1 = 8/27 - 4(4/9) + 8/3+ 27/27
y1 = 8/27 - 16/9 + 8/3+ 27/27
y1 = 8/27 - 48/27 + 72/27+ 27/27 = 59/27

and
y2 = (2)³-4(2)²+4(2)+1
y2 = 8 - 16 + 8 + 1 = 1

=>
(2/3, 59/27) and (2, 1)
.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.
Second derivitive:
d( 3x²-8x+4 )/dx = 6x-8

when x=2/3 in the second derivitive:
y''= 6(2/3)-8 = 4 - 8 = -4 < 0
The point (2/3, 59/27) is minimum point.

when x=2 in the second derivitive:
y''= 6(2)-8 = 12 - 8 = 4 > 0
The point (2/3, 59/27) is maximum point.

₢

Answer 3

f(x)=x(x-2)^2+1
f(x)=x^3-4x^2+4x+1
Df=3x^2-8x+4
D2f=6x-8
Df=0, 3x^2-8x+4=0 ,x1=(8+4)/6=2 ,x2=(8-4)/6=1/3
the critical points is 2 ; 1/3.
then (2,f(2))=(2,1) is the minim point and (1/3,f(1/3))is the maximum point
the minim of function is 1 and the maximum is f(1/3)=25/27+1=52/27
OR: D2f(2)=12-8=4>0
D2f(1/3)=2-8=-6<0 and 2 is the minim point ,1/3 is the maximum point