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Example 7
Find the values of x for which the function f (x)
= x3 - 12x + 5 is (I) decreasing and (II) increasing.
Solution : f (x) =
x3 - 12x + 5
\
f ' (x) = 3x2 - 12
= 3 (x2
- 4)
Now ' f ' is decreasing
if f ' (x) < 0
i.e. 3 (x2
- 4) <0
i.e. (x2
- 4) < 0
i.e. x2
< 4
i.e. -2 < x <
2
\ f
(x) decreases in (-2, 2)
Similarly 'f ' increases
if f ' ( x ) > 0
i.e. 3(x2
- 4 ) > 0
i.e. x2
> 0
i.e. x < -2 and
x >2
Example 8
Show that f (x) = x3 - 6x2
+ 15x + 7 is always increasing.
Solution : f (x) = x3
- 6x2 + 15x + 7
\ f ' (x) = 3x2
- 12x + 15
= 3 (x2 - 4x
+5 )
\ f ' (x)
= 3 [ x2 - 4x + 4 + 1]
= 3 [ (x -2)2
+ 1 ]
Now 3 [ (x - 2)2 + 1] > 0
for all x
\ f ' (x)
> 0 for all x
\ f (x) is
always increasing
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