If the derivative of a function f is given by | vedclass

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(A) Given $f'(x) = (x - a)^{2m} (x - b)^{2n + 1}$.Critical points are found by setting $f'(x) = 0$,which gives $x = a$ and $x = b$.Consider the point

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$x = a$ gives neither a maximum nor a minimum.

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(A) Given $f'(x) = (x - a)^{2m} (x - b)^{2n + 1}$.Critical points are found by setting $f'(x) = 0$,which gives $x = a$ and $x = b$.Consider the point $x = a$. For a small positive value $h$,we check the sign of $f'(x)$ around $x = a$:For $x = a - h$,$f'(a - h) = (a - h - a)^{2m} (a - h - b)^{2n + 1} = (-h)^{2m} (a - b - h)^{2n + 1}$. Since $2m$ is even,$(-h)^{2m} = h^{2m} > 0$. Since $a > b$,$(a - b - h) > 0$ for small $h$,so $f'(a - h) > 0$.For $x = a + h$,$f'(a + h) = (a + h - a)^{2m} (a + h - b)^{2n + 1} = (h)^{2m} (a - b + h)^{2n + 1}$. Since $a > b$,$(a - b + h) > 0$,so $f'(a + h) > 0$.Since the sign of $f'(x)$ does not change as $x$ passes through $a$,$x = a$ is a point of inflection and gives neither a maximum nor a minimum.

If the derivative of a function $f$ is given by $f'(x) = (x - a)^{2m} (x - b)^{2n + 1}$,where $m$ and $n$ are positive integers and $a > b$,then which of the following is true?

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