f(x) = x^3+27x^2 - 54.
To
examine for negative
values:
Solution:
f(x) =
x^3+27x^2 -54 < 0 for x = 0 obviuosly.
f(-26) =
(-26)^3+27*(-26)^2 -54 = 26^2*1 -54 > 0
f(-27) =
(-27)^3+27*(-27)^2 -54 = -54.
Therfore f(x) has a root
between -26 and -27.
Therefore f(x) < 0 for x
< -27.
Similarlt f(1) = 1^3 +27*1^2-54 28-54
< 0
f(2) = 2^3+27(2^2)-54 = 62>
0.
So there is a root between 1 and
2.
Also f(-1.5) = (-1.5)^3 +27(-1.5)^2 - 54 =
3.375>0
f(-1.4) = (-1.4)^3+27(-1.4)^2-54 = -3.824
< 0.
So f(x) has a real root between -1.4 and
-1.5
Let these roots be x1 , x2 , x3 in increasing
order.
Then f(x) = (x-x1)(x-x2)(x-x3). So when x <
x1 all 3 factors are gative. So f(x) is negative.
When x1
< x < x2 , f(x) = (+ve)(-ve)(-ve) = -ve. So when x is between lower two
roots f(x) positive.
When x2 < x < x3, f(x)
= ( f(x) = (+v1) (+ve)(-ve) = -ve.
When When x > x3
all factors are positive. So f(x) > 0.
Thesefore
f(x) negative when x< x1 or x <
-26.9...
f(x) - ve for x2 < x < x3, where x2
and x3 are the roots between (-1.5 and -1.4) and (1 and 2).
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