$h:[0,1] \to\mathbb{R}$ continuous, and ivt The question is as follows:
$$ \text{Supposd } h:[0,1] \rightarrow \mathbb{R} \text{ is continuous. Show that there exist } w \in [0,1] \text{ such that}
\\h(w)= \frac{w+1}{2}h(0)+\frac{2w+2}{9}h(\frac{1}{2})+\frac{w+1}{12}h(1)$$
I know that I have to use Intermediate Value Theorem and I have to show $h(w)$ lies between $h(0)$ and $h(1)$, but I have no idea how to prove it.
I have tried to separate $h(w)$ into $\frac{w+1}{2}h(0)+\frac{w+1}{9}h(\frac12)+\frac{w+1}{9}h(\frac12)+\frac{w+1}{12}h(1)$ and then use IVT twice on the interval $[0,\frac12]$ and $[\frac12,1]$ but it seems not working. I have also tried to see if $h$ is an interpolation of 3 points but it also fails.

Let  $f(x)=\frac {h(x)}{(x+1)}$ which is continuous on $[0,1]$ and $T=1/2f(0)+1/3f(1/2)+1/6f(1)$.
Then we want to prove ,$$f(w)=T \ \ \text{for some } \ w\in [0,1]$$
Let $M$ and $m$ be maximum and minimum value taken by $f(x)$.
Then note that, $m\leq T\leq M$.
