Answer:
f(x)
= x3 ? 5x2 ? 3x in [1, 3]
(i)
Being a polynomial function, f(x) is continuous in
[1,
3]
(ii)
f?(x) exists uniquely in (1, 3),
is
derivable in (1, 3).
Since
f(x) satisfies both conditions of L.M.V. theorem.
Therefore
there exists at least one such
that
3c2
? 10c + 7 = 0
3c2
? 3c + 7c + 7 = 0
3c(c ? 1) ?
7 (c ? 1) = 0
(c ? 1) (3c
?1) = 0
Here
Hence
the theorem verified.
Now
f?(c) = 0
Hence
there is no that f?(c)
= 0
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