f ′ ( x ) = 0 ⟺ Extremum ... right?
f
(
x
)
=
(
x
2
−
1
)
n
+
1
(
x
2
+
x
+
1
)
,
n
∈
N
If this function has an extremum at
x
=
1
, what is/are the possible value(s) of
n
?
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2 solutions
@Cccc Dddd I think you've missed out an n + 1 in the first term.
How is that f'(x)=0 when x=0 ??
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f'(x) only needs to be equal to 0 when x=1, right? The function has an extremum at x=1, not x=0.
I think it's true for all values of n, since ( x 2 − 1 ) will be zero . zero doesn't care to the powers
The given function can be written in the form: ( x + 1 ) n + 1 ( x − 1 ) n + 1 ( ( x + 2 1 ) 2 + 4 3 ) . The first and the third factor are always positive on the interval ( − 1 , ∞ ) and f ( 1 ) = 0 . If the n + 1 were odd, f ( x ) would change its sign at 1, and then f ( x ) could not have a local extremum at x = 1 . If the n + 1 were even, then f ( x ) ≥ 0 for all all values of x close to 1, and therefore f ( x ) would have a local minimum at 1. So the answer is: n must be odd.
f ′ ( x ) = 2 x ( x 2 − 1 ) n ( x 2 + x + 1 ) + ( x 2 − 1 ) n + 1 ( 2 x + 1 ) = ( x 2 − 1 ) n ( 4 x 3 + 3 x 2 − 1 ) x=0 => f'(x)=0 , if n is odd then f'(x) changes its sign through x=0 x = 2 n + 1