I am having trouble with a maths problem, if anyone can at all help please try to do so:
Given f(x)=X^2+a/x

A)For what value(s) of a does this function have a local minimum at x=2?

B)Show that this function can not have a local maximum for any value of a.

I have spent a lot of time looking at youtube lectures, google, my study guide etc.
I am still unsure of the process,
any help would be really appreciated.
Thank you.

2. Originally Posted by paddy12321
I am having trouble with a maths problem, if anyone can at all help please try to do so:
Given f(x)=X^2+a/x

A)For what value(s) of a does this function have a local minimum at x=2?

B)Show that this function can not have a local maximum for any value of a.

I have spent a lot of time looking at youtube lectures, google, my study guide etc.
I am still unsure of the process,
any help would be really appreciated.
Thank you.

when a = 16

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

$0 = 2x -\frac{a}{x^2}$

plug in 2 for x and solve for a
when a = 16

3. For your function

$f(x)=x^2+ax^{-1}$

$f'(x)=0$ will locate the turing points

then you must find $f''(x)$ to determine the nature of the turning point and in this case $f''(x)>0$ for a minimum

I'll get you started...

I have $f'(x)=2x-ax^{-2}$

and $f''(x)=2+ax^{-3}$

4. Thank you very much for your responses, but it is more so the second part I am finding difficult. sorry for not specifying that. I get what I am supposed to do I just don't know what I can use to prove there is no maximum when there is a variable, but seriously thank you very much for responding, I really didn't expect to be answered so quickly. but if anyone could tell me what formula, theorem etc. to use to show there is no maximum, that would really help.
Thank you.

5. $f'(2)= 0 = 2\times 2 -\frac{a}{2^2}$

This gives $a = 16$ as '11rcd11' has suggested

For part b) you have to show $f''(x) \geq 0 ,~\forall a$ in $f(x)=x^2+ax^{-1}$