Proof that an injective function is strictly monotonic
Solution 1:
Note that$$\phi(0)=f(b)-f(a)>0\quad\text{and}\quad\phi(1)=f(y)-f(x)<0.$$So, since $\phi$ is continuous, it must be equal to $0$ somewhere between $0$ and $1$.
Note that$$\phi(0)=f(b)-f(a)>0\quad\text{and}\quad\phi(1)=f(y)-f(x)<0.$$So, since $\phi$ is continuous, it must be equal to $0$ somewhere between $0$ and $1$.