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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