Uniform convergence on $[0,1]$ of $f_{n}(x)=x^{n}-3x^{n+2}+2x^{n+3}$ I have sequence:
$f_{n}(x)=x^{n}-3x^{n+2}+2x^{n+3},x\in [0,1]$  
Need to check uniform convergence  
My steps:


*

*Find $f(x)$
$\lim_{n\rightarrow \infty}{ f_n(x)} = f(x) = 0$  

*$f_n(x)$ converge uniformly to the function f(x) if  
$\lim_{n\rightarrow \infty}{\sup|f_n(x)-f(x)|}=0$  
$g(x)=f_n(x)-f(x) = x^{n}-3x^{n+2}+2x^{n+3}$  
I tried to find the extremum, but it's too difficult.
Is there another way of finding the supremum of $|g(x)|$?
 A: $f_n(x) = x^n (1-x)^2(1+2x)$ leads to $|f_n(x)|\leq 3x^n(1-x)^2$. On the other hand, over the interval $[0,1]$ the non-negative function $x^n(1-x)^2$ attains its maximum at $x=\frac{n}{2+n}$, and
$$\left(\frac{n}{n+2}\right)^n\cdot \left(\frac{2}{n+2}\right)^2\leq\frac{4}{(n+1)^2}, $$
thus $|f_n(x)|\leq\frac{12}{(n+1)^2}$ implies that the convergence towards zero is uniform.
A: Hint: $ f_n(x) = x^n (1 - 3 x^2 + 2 x^3)$.  Given $\epsilon > 0$, take $\delta>0$ so $|1 - 3 x^2 + 2 x^3 |< \epsilon$ for $1-\delta \le x \le 1$, and take $N$ such that $|f_n(x)| \le \epsilon$ when $n \ge N$ and $0 \le x \le 1 - \delta$. 
A: From $f_n(x)=2x^{n+3}-3x^{n+2}+x^n$, it's straightforward to obtain
$$f_n'(x)=2(n+3)x^{n+2}-3(n+2)x^{n+1}+nx^{n-1}=x^{n-1}(x-1)Q_n(x)$$
where $Q_n(x)=2(n+3)x^2+nx-n$. Since $Q_n(0)=-n\lt0$ and $Q_n({1\over2})={n+3\over2}+{n\over2}-n={3\over2}\gt0$, we see that the supremum of $|f_n|$ occurs at a value of $x$ between $0$ and $1\over2$. But for $0\lt x\lt{1\over2}$ we have
$$|f_n(x)|=x^{n-1}(x-1)^2(2x+1)\le2x^{n-1}\le{2\over2^{n-1}}={1\over2^{n-2}}$$
and thus $|f_n(x)|\le1/2^{n-2}$ for all $x\in[0,1]$.
