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Q.
If $f '(x) = (x - a)^{2n} (x - b)^{2p + 1}$ where n and p are positive integers, then :
Application of Derivatives
Solution:
Given, $f '(x) = (x - a)^{2n} \, (x - b)^{2p +1}$
$\therefore $ For maxima or minima
f '(x) = 0
$ \Rightarrow \left(x - a\right)^{2n} = 0 $ or $\left(x - b\right)^{2p+1} = 0 $
$\Rightarrow x = a $ and $x = b$
Now, $f '\left(a - h\right) = \left[\left(a - h - a\right)^{2n}.\left(a - h - b\right)^{2p+1}\right]$
$= h^{2n}\left(a - b - h\right)^{2p+1}$
and $f '\left(a + h\right) = \left[\left(a + h - a\right)2n.\left(a + h - b\right)^{2p+1}\right]$
$= h^{2n}\left(a - b + h\right)^{2p+1}$
Thus, f '(x) does not change sign as x passes through 'a', so x is not a point of maximum or minimum.