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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.

3 Answers 3

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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: enter image description here

  • 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.

enter image description here

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    @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.2011-12-27
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$F(x)=x^2+a/x$

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

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.2011-12-28