Decomposition of a polynomial function $f:[a,b]\to \mathbb{R}$ as the sum of two monotonous functions. Prove or disprove. 

Let a arbtrary polynomial function $f:[a,b]\to \mathbb{R}$   such that $f(a)<0<f(b)$. There are  functions $u:[a,b]\to\mathbb{R}$ and $v:[a,b]\to \mathbb{R}$ satisfying the conditions 
  
  
*
  
*$u$ is continuous, 
  
*$v$ is continuous,
  
*$u$ is strictly monotone (increasing or decreasing), 
  
*$v$ is strictly monotone (increasing or decreasing,
  
  
  and that fulfills equality
  $$
f=u+v \qquad (\ast)
$$

My attempt. Set $f(x)=a_0+a_1\cdot x+a_2\cdot x^2+\ldots +a_n\cdot x^n$. Let's look for polynomial functions 
$$
u(x)=\alpha_0+\alpha_1 x+ \alpha_2 x^2+\ldots +\alpha_n x^n 
\\
v(x)=\beta_0+\beta_1 x+ \beta_2 x^2+\ldots +\beta_n x^n 
$$
that satisfy the equality $(\;\ast\;)$. So we have to solve the system
$$
\alpha_0+\beta_0=a_0\\
\alpha_1+\beta_1=a_1\\
\vdots \\
\quad\\
\vdots \\
\alpha_n+\beta_n=a_n\\
$$
with the restriction that for all $x\in [a,b]$ we have
$$
u^\prime(x)\neq 0, \mbox{ that is, } \alpha_1+ 2\cdot\alpha_2 x^{1}+\ldots+ i\cdot\alpha_i x^{i-1} +\ldots +n\alpha_n x^{n-1}\neq  0\\
v^\prime(x)\neq 0 \mbox{ that is, } \beta_1+ 2\cdot\beta_2 x^{1}+\ldots+ j\cdot\beta_j x^{j-1} +\ldots +n\beta_n x^{n-1}\neq 0\\
$$
In other words, we have to show that the set
$$
C[x]=
\left\{
(\beta_0, \beta_1, \ldots, \beta_n)\in\mathbb{R}^n 
\left|
\begin{array}{l}
\beta_1+ 2\cdot\beta_2 x^{1}+\ldots+ i\cdot\beta_i x^{i-1} +\ldots +n\beta_n x^{n-1}\neq  0
\\
\\
(a_1-\beta_1)+ 2\cdot(a_2-\beta_2) x^{1}+\ldots+ j\cdot(a_j-\beta_j) x^{j-1} +\ldots +n(a_n-\beta_n) x^{n-1}\neq 0
\end{array}
\right.
\right\}
$$
is not empty for all $x\in[a,b]$. 

But this approach to the problem is not looking very promising. Is there a smarter way to tackle this problem?


Your approach is much too complicated.
Hint: take $u(x) = c x$ where $c > \max_{a \le x \le b} f'(x)$.
