Let $\displaystyle f$ be a continuous function from $[0,4]$ to $[3,9].$ I came across the following problem that says:  

Let $\displaystyle f$ be a continuous function from $[0,4]$ to $[3,9].$ The which of the following options is correct?
  $1.$ there must be an $x$ such that $f(x)=4$
  $2.$ there must be an $x$ such that $3f(x)=2x+6$
  $3.$ there must be an $x$ such that $2f(x)=3x+6$
  $4.$ there must be an $x$ such that $f(x)=x$  .

I do not know how to progress with it.Can someone point me in the right direction? Thanks in advance for your time.  
EDIT: Using the hint given by @Sampath ,I see that   
Since "A continuous function $f:[a,b] \to [c,d]$ must intersect the line joining $(a,c)$ and $(b,d)$", so here the line joining the  points $(0,3)$ and $(4,9)$ is given by $y=f(x)=\frac {3x+6}{2} \implies 2f(x)=3x+6.$
 A: Hint:
For the "no" answers, give a counter example. For the "yes" answers, apply the intermediate value theorem.
A: Hint A continuous function $f:[a,b] \to [c,d]$ must intersect the line joining joining $(a,c)$ and $(b,d)$. 
A: Hint: Try constant functions, as they behave the most easiest.
Use that $4 \leq f(x) \leq 9$ and use the intermediate value theorem.
A: First, we prove that if $f:[a,b]\to[a,b]$ is a continous function then there's a fixed point i.e. $x\in[a,b]$ such that $f(x)=x$. Indeed, let the function $h$ defined by $h(x)=f(x)-x$, then $h$ is continous on $[a,b]$ and $h(a)h(b)\leq0$ so by mean value theorem, there's $x\in[a,b]$ such that $h(x)=0$.
Now, we return to the question. In 3. we define the function $h$ by $h(x)=\frac{2f(x)-6}{3}$, so $h:[0,4]\to [0,4]$ is continous. We can conclude.
A: A related problem. Here is a useful result 
" Every continuous function $f$ from a convex compact subset $K$ of a Euclidean space to $K$ itself has a fixed point".
The above result is known as Brouwer fixed-point theorem.
