# Finding the local maxima/minima of the following function

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

• Use the 2nd derivative. – Wuestenfux Oct 8 '18 at 7:05
• Local max/min: second derivative. Global Max/Min: compare values or find inequality. – Andreas Oct 8 '18 at 7:05
• Have used the second derivative. So are all of them local solutions? I think all of them are global solutions as $x \in \mathbb{R}$. – 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.
• Without drawing the graph, how would I figure out that $x=1$ and $x=3$ are global minimum points? – OGC Oct 8 '18 at 7:19
• 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. – Robert Z Oct 8 '18 at 7:23