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lexus1427
27.06.2019 •
Mathematics
It's calculus work. the screenshot below should answer my question. 80 points
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Ответ:
a) In order for
to be continuous at
, we need to have
By definition of
, we know
. Meanwhile, the limits are
so
is indeed continuous at
.
b) We use the first derivative test (FDT) here, but when we compute the derivative of a piecewise function, we have to be careful at the points where the pieces "split off", because it's possible that the derivative does not exist at these points, yet an extreme value can still occur there. (Consider, for example,
at
.)
In this case,
We find the critical points for each piece over their respective domains:
On the first piece:
which does fall in [-2, 2]. The FDT shows
for
less than and near -1, and
for
greater than and near -1, so
is a local maximum.
On the second piece:
so it does not contribute any critical points.
Where the pieces meet:
By checking the conditions for continuity mentioned in part (a), we can determine that
does not exist, but that doesn't rule out
as a potential critical point.
We have
so
for
less than and near 1, and
so
for
greater than and near 1. So the FDT tells us that
is a local minimum.
Finally, at the endpoints of the domain we're concerned with, [-2, 2]:
We have
and
.
So, on [-2, 2],
attains an absolute minimum of
and an absolute maximum of
.
Ответ:
hard brother mods gonna ban u watch bro bro
Step-by-step explanation: