# Finding parameter values

Problem:

What value of $a$ makes $$f(x)=x^2 + \frac {a}{x}$$ have (a) a minimum at $x=2$, (b) an inflection point at $x=1$?

(a) and (b) are separate and not dependent upon each other.

What I have done:

Not much. I wrote the derivative to try and find where $x=0$, but that is where I got stuck.

$$0=2x-ax^{-2} .$$

I'm not really sure what to do.

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(a) Take first derivative of $f(x)$, set it equal to 0, replace $x$ by 2 , and solve for $a$.

(b) Take second derivative of $f(x)$, set it equal to 0, replace $x$ by 1, and solve for $a$.

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• An extrema (minimum/ maximum) is where the first derivative of the function is equal to zero:

$$f'(x)=0$$ $$2x-\frac{a}{x^2}=0$$

We have $x=2$, so:$$2(2^3)-a=0$$

Therefore, $a$ would be 16.

As pointed out in comments, you should check whether the answer is correct:

$min\{f(x)=x^2+\frac{16}{x}\}=12$ and local minimum happens at $x=2$.

Looking at the plot might make it clearer:

• An inflection point happens where the second derivative is equal to zero as well as all other higher order derivatives , meaning if the second derivative is zero but the fourth derivative is non-zero, for example, then the point is not an inflection point. What we have:

$$f''(x)=2+\frac{2a}{x^3}=0$$ Thus, at $x=1$, $a$ would be -1.

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If the second derivative and all higher derivatives are all zero, then the function is either exactly linear or non-analytic. The former is not an inflection point; the latter is rare in practice. For an analytic inflection point, there must be an $n\ge 1$ such that the $(2n+1)$th derivative is nonzero _and_ all derivatives from $2$ to $2n$ inclusive are zero. – Henning Makholm Dec 27 '11 at 13:58
@HenningMakholm: I meant examples like $y = x^4 – x$ has a 2nd derivative of zero at point $(0,0)$, but it is not an inflection point. – Gigili Dec 27 '11 at 17:18
that fails to have an inflection point not because there is a higher-order term, but because the higher-order term has even degree. As long as the next nonzero derivative after $f''(x)=0$ has odd order, there is an inflection point, such as in $x^4+x^3-x$ or $x^5-x$. – Henning Makholm Dec 27 '11 at 17:21
@HenningMakholm: I didn't say because existence of a higher order derivative is the reason, I think $(0,0)$ is not an inflection point because the fourth derivative is the first higher order non-zero derivative while the third derivative is zero as well. – Gigili Dec 27 '11 at 17:45

$F(x)=x^2+a/x$

$F'(x)=2x-a/x^2$

let $f'(x)=0$ and put $x=2$

$a=8$

$F"(x)=2+2a/x^3$ and let $F"(x)=0$ and put $x=1$

$a =-2$

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I believe your results $a=8$ and $a=-2$ are not correct. – Kavka Dec 28 '11 at 2:04