If $f(x)$ and $g(x)$ are continuous on $[a,b]$ with the same range $[0,1]$, then $f(c)=g(c)$ for some $c\in [a,b]$ 
If $f(x)$ and $g(x)$ are continuous on $[a,b]$ with the same range $[0,1]$, then $f(c)=g(c)$ for some $c\in [a,b]$.

Let $h(x)=f(x)-g(x)$, $x\in [a,b]$. I want to use IVP. But I am in trouble to use it as $h(a)h(b)$ is not given as $<0$. 
What to do? Please help.
 A: You can suppose without loss of generality that $[a,b]=[0,1]$. So, there are basically three cases: 
$(1)$ $\:\:f(0)=g(0)$. In this case, we are done.
$(2)$ $\:\:f(1)=g(1)$. In this case, we are done. 
$(3)$ Now, study the function $f-g$. By the Intermediate Value Property, if $f(0)<g(0)$ and $g(1)<f(1)$, we are done. Maybe this isn't the case, however. Then, suppose that $f(0)<g(0)$ and $f(1)<g(1)$. Well, the only way that $f-g$ never becomes nonnegative is if $f<g$ on the whole interval $[0,1]$. This implies that $f(x)\ne 1$ for all $x\in [0,1]$, however, because if $f(x)=1$, then $g(x)\le f(x)$. 
So, we conclude that there must exist some point $y\in [0,1]$ for which $g(y)\le f(y)$. The result follows by the Intermediate Value Property. Note that the case where $f,g$ are flipped follows from the exact same argument by symmetry.
A: Ad absurdum if there is no $c$ such that $f(c)=g(c)$, since $f$ and $g$ are continuous then with the Intermediate Value Property, $\forall x \in [a,b], f(x)>g(x)$ or $f(x)<g(x)$, without loss of generalty we will suppose the first case.  
Since $g$ has range $[0,1],\forall x \in [a,b],g(x)\geq0$, so $\forall x \in [a,b],f(x)>g(x)\geq0$, finally $\forall x \in [a,b],f(x)>0$. So the range of $f$ is not $[0,1]$, we have a contradiction.
So the assertion is true.
A: Let $h$ defined by $h(x)=f(x)-g(x)$. Since the range of $f$ is $[0,1]$ there exists $u\in [a,b]$ with $f(u)=0$, $h(u)=-g(u)\leq 0$. There exists $v\in [a,b]$ with $f(v)=1$, $h(v)=1-g(v)\geq 0$. If $u\leq v$, the intermediate value theorem implies there exists $c\in [u,v]\subset [a,b]$ with $h(c)=f(c)-g(c)=0$. Similar argument if $v<u$.
A: Suppose, w.l.o.g., that $h(a)\geq0$. If we can find $x\in[a,b]$ such that
$h(x)\leq0$, we can conclude the desired result by the IVP. To arrive
at a contradiction, suppose
$$
h(x)=f(x)-g(x)>0\text{ for all }x\in[a,b].
$$
Now, pick $y\in[a,b]$ such that $g(y)=1$ to obtain $h(y)=f(y)-1>0$. Therefore, $f(y)>1$, a contradiction.
