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Stationary point of inflection

I have this question that I have been looking at for hours and I'm well and truly over it.

Would like some help please.

Obviously it's linked to Derivatives and Concavity, but I'm not sure how to go about it.

Attachment 28430

Thanks for your time.

Re: Stationary point of inflection

I would let the quadratic function be:

$\displaystyle f(x)=ax^2+bx+c$

and then:

$\displaystyle g(x)=f(x)+\frac{\ln(x)}{x^4}=ax^2+bx+c+\frac{\ln(x )}{x^4}$

Now, require $\displaystyle g'(1)=0$ and $\displaystyle g''(1)=0$ and you will get two equations from which you will be able to solve for $\displaystyle a$ and $\displaystyle b$, where $\displaystyle c$ remains a parameter, as it merely shifts the function vertically and so has no bearing on the $\displaystyle x$-coordinate of critical values.