$f: \mathbb{R}\rightarrow \mathbb{R}$. 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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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