Proof
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“ is differentiable on [0,3]” => is continuous on [0,3]. Therefore, the conditions are met for applying the mean value theorem, and the intermediate value theorem on each interval (0,1), (1,2) and (2,3). Assume
has two zeros on (0,1) where
. Therefore,
and, by the mean value theorem, there exists a value
This cannot be true because “
only at
”. Therefore,
such that:
has, at most,
one zero on the interval (0,1). Similarly, has, at most, one zero on the intervals (1,2) and (2,3) by the use of the above contradiction proof on each interval. Therefore, has, at most, 3 zeros on the interval (0,3). Furthermore, “
” and “
” which implies
and
the intermediate value theorem, there exists a value interval (0,1). Also, “
” and “
” which implies
intermediate value theorem, there exists a value interval (1,2). Finally, “
” and “
” which implies
intermediate value theorem, there exists a value interval (2,3). In conclusion, (2,3) (such that zeros (
. So, by
such that
and
on the
. So, by the
such that
and
on the
and, by the
such that
on the
has exactly one zero on each of the intervals (0,1), (1,2), and , respectively) which implies that on the interval (0,1).
As a side note: This makes sense, graphically.
has exactly three
has two zeros on (0,1) where
. Therefore,
and, by the mean value theorem, there exists a value
This cannot be true because “
only at
”. Therefore,
such that:
has, at most,
one zero on the interval (0,1). Similarly, has, at most, one zero on the intervals (1,2) and (2,3) by the use of the above contradiction proof on each interval. Therefore, has, at most, 3 zeros on the interval (0,3). Furthermore, “
” and “
” which implies
and
the intermediate value theorem, there exists a value interval (0,1). Also, “
” and “
” which implies
intermediate value theorem, there exists a value interval (1,2). Finally, “
” and “
” which implies
intermediate value theorem, there exists a value interval (2,3). In conclusion, (2,3) (such that zeros (
. So, by
such that
and
on the
. So, by the
such that
and
on the
and, by the
such that
on the
has exactly one zero on each of the intervals (0,1), (1,2), and , respectively) which implies that on the interval (0,1).
As a side note: This makes sense, graphically.
has exactly three
