If $(x-a)^{2n}(x-b)^{2m+1}$,where $m$ and $n$ are positive integers,is the derivative of a function $f$ ,then show that $x=b$ gives a minimum but $x=a$ gives neither maximum nor minimum
My attempt: $$f'(x)=(x-a)^{2n}(x-b)^{2m+1}$$ $$f''(x)=(x-a)^{2n-1}(x-b)^{2m}[2n(x-b)+(2m+1)(x-a)]$$ But at $x=a$ and $x=b$,$f''(x)=0$
How do I proceed?
Here is one way to proceed (assuming $a\neq b$).
You could show that the sign of the derivative does not change as $x$ increases through $a$, but does change as $x$ increases through $b$.
This can be done without computing further derivatives.
To illustrate:
So about $x=a$ constrain $|\epsilon|\lt |a-b|$ and write $x=a+\epsilon$
$$f'(a+\epsilon)=\epsilon^{2n}(a+\epsilon-b)^{2m+1}$$
Now show that the second term is non-zero and has the same sign whatever $\epsilon$ is (within the constraint), and that the first term is positive for $\epsilon \neq 0$. So near $x=a$ the derivative does not change sign.
And then do the same thing near $b$ with $\epsilon$ similarly constrained.