1
$\begingroup$

$f: \mathbb{R}\rightarrow \mathbb{R}$. Then determining if each of the solutions is a global maximum or a global minimum. $$f(x) = \frac{x^{4}}{4} - 2x^{3} + \frac{11}{2}x^{2} - 6x + 2$$ for all $x \in \mathbb{R}$.

I have used the synthetic division to find this equation: $$\left ( x^{2} - 5x + 6 \right )\left ( x - 1 \right ) = 0$$

So the critical points are $x^{*} = 1,2,3$.

I know that if $x^{*} = 1$, then ${f}''\left ( 1 \right ) = 2 > 0$, so this is minimum.

If $x^{*} = 2$, then ${f}''\left ( 2 \right ) = 1 < 0$, so this is maximum.

If $x^{*} = 3$, then ${f}''\left ( 3 \right ) = 2 > 0$, so this is minimum.

How do I determine whether these solutions are local max/min and global max/min?

Are all of them global solutions as $x \in \mathbb{R}$?

EDIT: I have plugged the values of the $x's$ in the main function.

For $x=1$ and $x=3$, $f(1)= -0.25 = f(3)$, and for $x=2$, $f(2)= 0$.

So is $x=1,3$ a local minimum and $x=2$ a global maximum?

$\endgroup$
  • $\begingroup$ Use the 2nd derivative. $\endgroup$ – Wuestenfux Oct 8 '18 at 7:05
  • $\begingroup$ Local max/min: second derivative. Global Max/Min: compare values or find inequality. $\endgroup$ – Andreas Oct 8 '18 at 7:05
  • $\begingroup$ Have used the second derivative. So are all of them local solutions? I think all of them are global solutions as $x \in \mathbb{R}$. $\endgroup$ – OGC Oct 8 '18 at 7:06
1
$\begingroup$

From the second derivative test the extremum points that you have found are all local. Note that $\lim_{x\to \pm\infty}f(x)=+\infty$, so $x=1$ is not a global maximum point. On the other hand, since $f(1)=f(3)=-1/4$, it follows that $x=1$ and $x=3$ are global minimum points.

$\endgroup$
  • $\begingroup$ Without drawing the graph, how would I figure out that $x=1$ and $x=3$ are global minimum points? $\endgroup$ – OGC Oct 8 '18 at 7:19
  • 1
    $\begingroup$ Since $f$ is differentiable in $\mathbb{R}$, and $\lim_{x\to \pm\infty}f(x)=+\infty$, any global minimum point is also a critical point. $\endgroup$ – Robert Z Oct 8 '18 at 7:23
  • $\begingroup$ I was actually thinking of the critical points being local max/min and then establishing if the local max/min solutions are also global max/min solutions. $\endgroup$ – OGC Oct 8 '18 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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