Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
I have added the homework tag for you since you say that it's homework. Please do it yourself the next time. –  J. M. Dec 27 '11 at 2:00
    
@AndréNicolas Oops, thanks for catching that. In retrospect, I am completely at a loss, why I chose to correct it to "perimeter" when the question has nothing to do with geometry. :) –  Srivatsan Dec 27 '11 at 2:08
    
@AndréNicolas. "The second derivative is $1+2ax^{-3}$"?? –  Kavka Dec 27 '11 at 15:20
    
@Kavka: Thanks, had a little trouble differentiating $2x$. –  André Nicolas Dec 27 '11 at 15:23

3 Answers 3

(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 $a$.

share|improve this answer
  • 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

share|improve this answer
    
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 $x=1$

$a =-2$

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.