Differential problem, find the maximum and minimum value Find the maximum, minimum value and inflection/saddle point of the following function


*

*$f(x)=12x^5-45x^4+40x^3+6$

*$f(x)=x+\frac{1}{x}$

*$f(x)=(2x+4) (x^2-1)$


Give a little explanation or procedural details if possible
 A: I'll do the second one, and try to solve the other two based on how I did this one.  If you need more explanation, let me know.
(1) Set the first derivative equal to zero to find critical points:
$$f(x) = x+\frac{1}{x}$$
$$f'(x) = 1-\frac{1}{x^2}$$
$$0 = 1-\frac{1}{x^2}$$
Solving, we find that we have critical points at $x=\pm1$.
(2) Check the second derivative to determine max/min/unknown:
$$f''(x) = \frac{2}{x^3}$$
At $x=+1$, $f''(x) > 0$.  Thus we have a minimum.
At $x=-1$, $f''(x) < 0$.  Thus, we have a maximum.
A: Many questions!
$1.$ We have $f'(x)=60x^4-180x^3+120x^2=60x^2(x^2-3x+2)=60x^2(x-1)(x-2)$. 
Note that $(x-1)(x-2)\gt 0$ if $x\lt 1$ or $x\gt 2$. Ao $f(x)$ is increasing in $(-\infty,1]$, then decreasing in $[-1,2]$, then increasing in $[2,\infty)$. (It hesitates slightly at $x=0$, since the derivative is $0$ there, but then decides to keep on increasing for a while.)
So there is a local (relative) maximum at $x=1$, and a local minimum at $x=2$. There is no global maximum, since $f(x)$ is  large when $x$ is large positive or negative. But the local minimum at $x=2$ is also a global minimum.
Note that $f''(x)=60(4x^3-9x^2+4x)=60x(4x^2-9x+4)$. Set this equal to $0$. The solutions are $x=0$ and (by the Quadratic Fomula) $x=\frac{9\pm\sqrt{17}}{8}$.
Note that $f''(x)$ is negative for $x\lt 0$, positive between $0$ and the first root of the quadratic, then negative between the two roots of the quadratic, and finally positive. So there is a change of concavity at each of the $3$ roots, and therefore there are $3$ inflection points.
$2.$ This has been done by anorton. Please note that there is no (absolute) maximum, since $x+\frac{1}{x}$ blows up as we approach $0$ from the right. There is also no absolute minimum, for $x+\frac{1}{x}$ becomes very large negative as we approach $0$ from the left.
There is one local maximum, and one local minimum.  We have $f'(x)=0$ at $x=\pm 1$. Note also (very importantly) the singularity at $x=0$. So there are $3$ "critical points," $-1$, $0$ and $1$.  We examine the behaviour of the function in the four regions determined by the critical points.
For example, note that $f'(x) \gt 0$ if $x\lt -1$, and $f'(x)\lt 0$ if $-1\lt x\lt 0$.  So $f(x)$ is increasing in $(-\infty,-1]$ and decreasing in $[-1,0)$. It follows that there is a local maximum at $x=-1$. 
$3.$ This is less interesting. We have $f'(x)=6x^2+8x-2$, so we need to use the quadratic Formula to find the critical points. 
