What is the Proof for Local Minimum and Maximum at First Derivative Equals Zero?

[Karol]

Homework Statement

Homework Equations

At local minimum or maximum the first derivative equals zero

The Attempt at a Solution

a) $$f'=2x-\frac{a}{x^2},~~2\cdot 2-\frac{a}{4}=0~\rightarrow a=16$$
Near 0 from the left a/x gets large negative values, smaller than for a=16. that's my proof for the local minimum.
The same for b).
But how to prove c)?
If a>0, f'<0 for x<0. so f decreases. for x>0 we found a minimum, and f''>0, so it holds water.
Similar reasoning for a<0. is this the way?

 

[Mark44]

Homework Statement

View attachment 215032

Homework Equations

At local minimum or maximum the first derivative equals zero

The Attempt at a Solution

a) $$f'=2x-\frac{a}{x^2},~~2\cdot 2-\frac{a}{4}=0~\rightarrow a=16$$
Near 0 from the left a/x gets large negative values, smaller than for a=16. that's my proof for the local minimum.
The same for b).
But how to prove c)?
If a>0, f'<0 for x<0. so f decreases. for x>0 we found a minimum, and f''>0, so it holds water.
Similar reasoning for a<0. is this the way?

Part c doesn't ask you to prove anything -- just find the inflection point.
Part d is the "show" part. If there were a rel. maximum somewhere, what must be true?

 

[Karol]

If there were a rel. maximum somewhere, what must be true?

Then f'=0 and f''<0 there, and there would have been higher points.
The end points at infinity, both for x>0 and x<0 are higher.
$$f'=2x-\frac{a}{x^2},~~f'=0~\rightarrow x=\sqrt[3]{\frac{a}{2}}$$
$$f''=2x-\frac{1}{x^2},~~f''\left( \sqrt[3]{\frac{a}{2}} \right)=...=\sqrt[3]{\frac{4}{a^2}}(a-1)$$
For a>1 f''>0 at ##~\displaystyle x=\sqrt[3]{\frac{a}{2}} so it's a local minimum.
But for 0<a<1 f''<0

 

[Mark44]

Then f'=0 and f''<0 there, and there would have been higher points.
The end points at infinity, both for x>0 and x<0 are higher.
$$f'=2x-\frac{a}{x^2},~~f'=0~\rightarrow x=\sqrt[3]{\frac{a}{2}}$$

I almost agree with the above.
Suppose there is a relative maximum at x = c.
Then $f'(c) = 2c - \frac a {c^2} = 0$ and $f''(c) < 0$
Solving for c in the first equation gives $c =\sqrt[3]{\frac a 2}$, the same value you show.

$$f''=2x-\frac{1}{x^2},~~f''\left( \sqrt[3]{\frac{a}{2}} \right)=...=\sqrt[3]{\frac{4}{a^2}}(a-1)$$

This part (above) is incorrect. $f''(x) \ne 2x - \frac 1 {x^2}$
Redo your calculation for f''(x), and then evaluate f''(c) and see whether f''(c) can ever be negative.

For a>1 f''>0 at ##~\displaystyle x=\sqrt[3]{\frac{a}{2}} so it's a local minimum.
But for 0<a<1 f''<0

 

[Karol]

$$f''=2x-\frac{1}{x^3},~~f''\left( \sqrt[3]{\frac{a}{2}} \right)=...=6$$
So for every a, f holds water

 

[Mark44]

$$f''=2x-\frac{1}{x^3},~~f''\left( \sqrt[3]{\frac{a}{2}} \right)=...=6$$
So for every a, f holds water

$f''(\sqrt[3]{\frac a 2}) \ne 6$, so your conclusion doesn't hold water. Also, if it were true that $f''(\sqrt[3]{\frac a 2}) = 6$ (which isn't true), then that would mean that the graph of f would hold water.

 

[Karol]

$$f''=2x-\frac{1}{x^3},~~f''\left( \sqrt[3]{\frac{a}{2}} \right)=2\cdot \sqrt[3]{\frac{a}{2}}-\frac{2}{a}=2\left[ \sqrt[3]{\frac{a}{2}}-\frac{1}{a} \right]$$

 

[haruspex]

$$f''=2x-\frac{1}{x^3},~~f''\left( \sqrt[3]{\frac{a}{2}} \right)=2\cdot \sqrt[3]{\frac{a}{2}}-\frac{2}{a}=2\left[ \sqrt[3]{\frac{a}{2}}-\frac{1}{a} \right]$$

The problem is not that complicated. In part a) you supposed the function had a local extremum. How many values of a did you find?

 

[Mark44]

$$f''=2x-\frac{1}{x^3},~~f''\left( \sqrt[3]{\frac{a}{2}} \right)=2\cdot \sqrt[3]{\frac{a}{2}}-\frac{2}{a}=2\left[ \sqrt[3]{\frac{a}{2}}-\frac{1}{a} \right]$$

Your work is incorrect right from the start. For one thing, you lost the factor of a that was present in the original function.
$f(x) = x^2 + ax^{-1}$
$f'(x) = 2x - ax^{-2}$
$f''(x) = ?$
If f has a relative maximum, f''(a) must be positive for some value of a.

 

[Karol]

$$f''=2\left[ 1+\frac{a}{x^3} \right]$$
$$f''\left( \sqrt[3]{\frac{a}{2}} \right)=6$$

 

[Mark44]

$$f''=2\left[ 1+\frac{a}{x^3} \right]$$

This is f''(x)

$$f''\left( \sqrt[3]{\frac{a}{2}} \right)=6$$

So what do you conclude?

 

[Karol]

f''>0 → the function hods water, so at the point of of f'=0, f has a minimum
But according to the answer: