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

  1. $f(x)=12x^5-45x^4+40x^3+6$
  2. $f(x)=x+\frac{1}{x}$
  3. $f(x)=(2x+4) (x^2-1)$

Give a little explanation or procedural details if possible

  • 2
    $\begingroup$ @ShuvroShuvro: Welcome to MSE! It would be helpful to tell us what you have attempted and where you are confused so we can help correct the issue. If we just solve these problems, it really does you no good. Regards $\endgroup$
    – Amzoti
    Mar 5, 2013 at 20:59
  • $\begingroup$ @Amzoti: thanks for ur suggestion . I will try to follow it in future $\endgroup$ Mar 5, 2013 at 21:21
  • $\begingroup$ So this is the question and at the bottom in pencil is how far I got. I’m just not sure where to go next. I’ve had a mind blank. Cheers $\endgroup$
    – Anielle
    Sep 2, 2018 at 11:12

2 Answers 2


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.

  • $\begingroup$ @Jared: thanks , now I understand . But what is inflection/saddle point? $\endgroup$ Mar 5, 2013 at 21:18
  • $\begingroup$ again, suppose if I get more than 2 value or a single value for setting the first derivative equal to 0 , what should I do then ? And let , I have found both the values positive or both negative , what should I do then ? $\endgroup$ Mar 5, 2013 at 21:25
  • $\begingroup$ @Jared Thanks... (my bad) $\endgroup$
    – apnorton
    Mar 5, 2013 at 22:01
  • $\begingroup$ @ShuvroShuvro If you get more than 2 values, check all of them with the second derivative. Re: inflection points: Basically, this is when the first derivative has a max or min. More info can be seen here: en.wikipedia.org/wiki/Inflection_point Re: Both values positive or negative: That's ok--they are the $x$ coordinates of possible extrema that just means that you don't have an extreme value with a negative $x$ coord (for example, the function $f(x) = (x-2)^2$). Just plug them in like normal points. $\endgroup$
    – apnorton
    Mar 5, 2013 at 22:10
  • $\begingroup$ thanks a lot for your help $\endgroup$ Mar 6, 2013 at 10:03

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.


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