If the value of `c` prescribed inRolles theorem for the function `f(x)=2x(x-3)^n` on the interval `[0,2sqrt(3)]i s3/4` , write the value of `n` (a positive integer).

Answer

` n=3`

Explanation

ATQ,
$$ f(x)=2 x(x-3)^{n} $$
Differentiating the given function wrt `x`, we get
$$ f^{\prime}(x)=2\left[x n(x-3)^{n-1}+(x-3)^{n}\right] $$
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