Wednesday, December 25, 2013

find the values of x where the function: y= x^3 -3x^2 + 2 reaches a max and min, also calculate the max & min values, sketch the function

y= x^3-3x^2+2


To find maximum
and mimimum and trace the
graph.


Solution:


We  find the
critical points x1 and x2 by setting f'(x) = 0 and solve for x. We find the the sign of 
f"(x) at these critical points. If  f " (x) is negative (or positive )at the critical
point , then  f(x) should be maximum (or minimum)
respectively.


f'(x) = (x^3-3x^2+2)' = 3x^2-3*2x
=3x^2-6x


f'(x) = 0  gives  3x^2-6x = 0. 3x(x-2) =
0


So x= 0 Or x =2 are the critical
values.


Now find the 2nd
derivative.


f''(x) = (3x^2-6x)' =
6x-6.


Find the sign of f"(x) at the critical points x = 0
and x = 2.


f '' (0) =6*0 -6  = -6. S0 f(0) is a local
maximum at x= 0. f(0) = 0^3-3*0^2+2 = 2


So f(0) = 2 is  a
maximum.


At x= 2, f " (2) = 6*2-6 = 12-6 = +6. So f(2) is
the local minimum.


f(2) = 2^3-3*2^2+2 = 8-12+2 = -2 is the
local minimum.


Tracing the
graph:


Roots:


Obviously f(1)
=  0 . So x=1 is a real root of x^3-3x^2+2 = 0. The graph crosses x axis at
x=1.


Also (x^3-3x^2+2)/(x-1) = x^2-2x-1  becomes 0 for
x^2-2x+1 =2. Or


(x-1)^2 =2.
Therefore,


x-1 = +or- sqrt2
.Or


x1 = 1+sqrt2 and x2 =
1-sqrt2. 


So f(x) has  three real roots : A pair of surds
and one rational . x1 = 1-sqrt2, x=1 and x2 = 1+sqrt2.


So
f(x) crosses x axis 3 times.


y = 2 is the y intercept wnen
x = 0. Also it is the local maximum.


 Graph  and its
nature:


At  -infinity, f(x) = -infinity as the eading term
is  x^3  has a positive coefficient 1.


The graph  of  f(x)
is strictly increasing for  all x < 0  as f'(x) =3x(x-2)  is positive for
x<2.. So f(x )  is continuously  increasing from - infinity to a maximum in the
interval (-infinity to 0). It crosses x axis at x 1 =
1-sqrt2.


It reaches a local maximim at  of f(0) = 2 at x
=0.


In the interval (0 , 2),  f(x) continuously decreasing
from maximum local maximum 2 to -2. The graoh crosses x axis at x=1 as f(1) =
0.


At x=2 , f(x) reaches a local minimum off(2) =
-2.


For   x>2 as f'(x) = 3x(x-2) > 0. So  
f(x) is strictly increasing again. The graph crosses x axis  at x2 = 1+sqrt2 and f(x)
goes infinity as x--> infinity.


Since f'(x) = 0 at
x=0 and x =2,  the graph has  tangent at x = 0 and at x= 2 || to x
axis.


The  shape of graph is  like that of a curved letter
"N "  in the interval (1-sqrt2  , 1+sqrt2), further,  with left extremity approching -
infinity  as x-->-infinity and right extremity going to +infinity as x-->
+infinity.

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