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I'm just having little bit of difficulty with the following question:

Find the local maxima and minima of $f : [0, 1] \rightarrow \mathbb{R}$ defined by $$ f(x)=x^4(1-x)^6 $$

So we know the function is differentiable on the open interval $(0,1)$ and hence we can test the nature of the critical points on $(0,1)$. Also, since $f$ is continuous on the closed, bounded interval we can use the Extreme Value Theorem on the end points.

But, when we take the first derivative of this function we find that critical points occur at

$$x=0,\frac{2}{5},1$$

ie two of the critcal points occur at the end points.

The second derivative test shows that the critical point at $\frac{2}{5}$ is a maximum so thats all well and good. However, my question is, how do I test the nature of the two other critical points? Since by the definition of differentiability we can not use the first and second derivative tests on the end points.

Is it merely just a matter of arguing that since $f(0)=f(1)=0$ and since by the definition of EVT we know that on a closed, bounded interval a continuous function has both a maximum and a minimum value on that closed, bounded interval and since $x=\frac{2}{5}$ attains the maxima then by EVT $x=0$ and $x=1$ must attain the minima?

I would like to make my argument as rigorous as possible.

Thanks.

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  • $\begingroup$ Since $f(x)$ is non-negative, and $f(0)=f(1)=0$, I think you can be confident it achieves minimum values there $\endgroup$ – Henry May 13 '15 at 5:04
  • $\begingroup$ It seems to me that you cannot make a claim as to whether or not the end points are local extrema. You certainly can claim they are absolute extrema (in this case they are both the absolute minimum), but that's it. I would approach this by looking at the "snapshot" of that function. Indeed there is no reason to believe, that even though the function appears to be $f(x) = x^4(1 - x)^6$ on your interval, that it continues to be that function outside of the interval (or is even defined outside of that interval). $\endgroup$ – Jared May 13 '15 at 5:04
  • $\begingroup$ Yes, a differentiable function on $[a,b]$ will attain its extreme values at the endpoints or a critical value. Your list is exhaustive, so $f(1)=f(0)=0$ is the minimum. Also, check your algebra as the critical points for $f(x)=x^4(1-x)^6$ is not 2/3 ! $\endgroup$ – matt biesecker May 13 '15 at 5:07
  • $\begingroup$ @Jared. For this function, it happens that 0 and 1 are critical points. But you're right, it's more relevant that they are the endpoints. $\endgroup$ – matt biesecker May 13 '15 at 5:09
  • $\begingroup$ @mattbiesecker Actually that's not correct, for the function $f(x) = x^4(1 - x)^6: \mathbb{R} \mapsto \mathbb{R}$ it is true that $x = 0, 1$ are critical point (assuming all the math is correct). This is not true for the function $f(x) = x^4(1 - x)^6: [0, 1] \mapsto \mathbb{R}$ $\endgroup$ – Jared May 13 '15 at 5:11
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I am viewing $[0,1]$ as the entirety of the domain.

Observe that $f'(x)=2x^3(x-1)^5(5x-2).$ We have $f'(x)=0$ at $x=0,2/5,1.$ Also, $f'(x)>0$ on $(0,2/5),$ $f'(x)<0$ on $(2/5,1).$ Therefore $f(2/5)$ is a local maximum. Moreover, $f'(x)>0$ on $(0,2/5)$ implies that $f(x)\geq f(0)$ on some (relatively) open neighborhood $[0,c)$ about $0.$ We conclude that $f(0)$ is a local minimum. Similarly, $f(x)\geq f(1)$ on a (relatively) open neighborhood about 1 and consequently $f(1)$ is a local minimum.

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  • $\begingroup$ Thanks! So we don't need to apply EVT at all? $\endgroup$ – sho May 13 '15 at 5:46
  • $\begingroup$ Yeah. You were asked about local min/max, not global. $\endgroup$ – matt biesecker May 13 '15 at 5:47
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It can be solved by using the AM-GM inequality : $\dfrac{f(x)}{4^4\cdot 6^6}= \left(\dfrac{x}{4}\right)^4\cdot \left(\dfrac{1-x}{6}\right)^6 \leq \left(\dfrac{4\cdot \dfrac{x}{4}+6\cdot \dfrac{1-x}{6}}{10}\right)^{10}=\dfrac{1}{10^{10}} \to f(x) \leq \dfrac{4^4\cdot 6^6}{10^{10}}= f_{\text{max}}$,and this maximum occurs when $\dfrac{x}{4}=\dfrac{1-x}{6} \iff x = \dfrac{2}{5}$. Also $f_{\text{min}}=0$ at $x = 0,1$.

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