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There are two positive integers $m$ and $n$. Let $$f(x)=\int_1^x {(t-a)^{2n}(t-b)^{2m+1}}dt, a\neq b$$ I need to find whether $f(x)$ is maximum/minimum at $x=a,b$.

My Approach: I differentiated $f(x)$ with respect to $x$. $$f'(x)=\frac{d}{dx}{(\int_1^x {(t-a)^{2n}(t-b)^{2m+1}}dt)}$$ By Newton-Leibniz formula, $$f'(x)=\big[(x-1)^{2n}(x-b)^{2m+1}\big]\frac{dx}{dx}-\big[ (1-a)^{2n}(1-b)^{2m+1} \big]\frac{d}{dx}{(1)}$$ Thus, $f'(x)=(x-a)^{2n}(x-b)^{2m+1}$. For critical points, $f'(x)=0$, $$0=(x-a)^{2n}(x-b)^{2m+1}$$ So, $x=a$ and $x=b$ are critical points. $$f''(x)=(x-b)^{2m+1}\frac{d(x-a)^{2n}}{dx}+(x-a)^{2n}\frac{d(x-b)^{2m+1}}{dx}$$ $$f''(x)=\big[ (x-b)^{2m+1}(2n)(x-a)^{2n-1} \big]+\big[ (x-a)^{2n}(2m+1)(x-b)^{2m} \big]$$ From here one can easily deduce that $f''(x)\geq0$. Now, if I put the value of $x=a$ or $x=b$, I'll get $f''(x)=0$ which gives that the function $f(x)$ has inflexion at $x=a$ and $x=b$. So where am I going wrong because the answer given in my book is that $x=b$ is a point of local minimum.

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    $\begingroup$ $f''(x)=0$ doesn't imply that its a zone of inflection. E.g. $f(x)=x^4$ at $x=0$ is a minimum, but has this property. $\endgroup$ – John Doe Apr 13 '17 at 15:34
  • $\begingroup$ @JohnDoe but clas.sa.ucsb.edu/staff/lee/Inflection%20Points.htm says that at inflection point $f''(x)=0$. $\endgroup$ – Shuvam Shah Apr 13 '17 at 15:40
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    $\begingroup$ Your solution is fine up to where you find the critical values. To classify them, use the First Derivative Test and make use of the fact that $2n$ is an even number while $2m+1$ is odd. $\endgroup$ – Ryan Gibara Apr 13 '17 at 15:43
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    $\begingroup$ Yes, inflection point $\Rightarrow$ $f''(x)=0$, but $f''(x)=0$ $\nRightarrow$ inflection point. Essentially, $f''(x)=0$ is not very useful in classifying a stationary point. $\endgroup$ – John Doe Apr 13 '17 at 15:45
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    $\begingroup$ Also, what Ryan is saying is to look at the derivative either side and see if it is positive or negative. $\endgroup$ – John Doe Apr 13 '17 at 15:49
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$$f'(x)=(x-a)^{2n}(x-b)^{2m+1}$$ So let $\epsilon>0$, then $$f'(b+\epsilon)=(b-a+\epsilon)^{2n}(\epsilon)^{2m+1}>0.$$ Instead $$f'(b-\epsilon)=(b-a-\epsilon)^{2n}(-\epsilon)^{2m+1}<0.$$ So if you picture it, you'll see $x=b$ is a minimum.

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  • $\begingroup$ Okay, I got it. I was so close. Thanks! $\endgroup$ – Shuvam Shah Apr 13 '17 at 16:11

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