Defining Local Extrema
True or False?
The local extrema of f ( x ) are the points where f ′ ( x ) = 0 .
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3 solutions
Moderator note:
As asked by Isaac, these two sets are not even subsets of each other.
So you have given an example of a function with a point x 0 where f ′ ( x 0 ) = 0 and yet it is not a local extremum.
Is it also possible for a point x 0 to be a local extremum and yet f ′ ( x 0 ) = 0 ?
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At any extremum, f ′ ( x ) = 0 or f ′ ( x ) = undefined
Then, Yes it's possible for a point
x
0
to be a local extremum and yet
f
′
(
x
)
=
0
Ex.
:
f
(
x
)
=
⌊
x
⌋
For,
x
=
integer
,
f
′
(
x
)
=
0
and still there is an extremum.
We can also have an example of straight line as a counterexample to this...
A function has a local maximum for a given point in the domain if , in the delta neighbourhood of that point,the function attains the maximum value.Hence first derivative=0 corresponds to a point of local extremum but the converse need not be always true.
You can say that if f '(x) = 0 it's potentially a local extrema (verify inflexion point), there's a stronger criteria envolving the function being 'n' derivable, but not:
if local extrema then f '(x) = 0 (It would be true if known that f(x) is continuous). Since not every function is continuous it can be valid something like:
f(x) = 0 if x != 0 otherwise f(x) = 1
Plot the graph:
f(0) is a local extrema but there isn't a f '(0) = 0.
if f '(x) = 0 then it need not be a local extremum it can be a point of inflection either way the statement only satisfies some but not all
Counter Example :
f ( x ) = x 3 f ′ ( x ) = 3 x 2 ⇒ f ′ ( 0 ) = 0 But from the graph of x 3 (shown above), it has no local extremum at x = 0