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.
Best Answer
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.