Q: Suppose that f: [0,2] -> [0,4] is continuous. Show that there is an x$\displaystyle \epsilon$[0,2] such that $\displaystyle f(x) = 2x$.
Proof: All I know is use the intermediate value theorem...
That is the whole problem. But there is a hint here provided by my professor:
Hint: Consider the function g: [0,2] -> R defined by g(x) = f(x) - 2x and use the intermediate value theorem.
Now, again, we have not learn about using derivatives in proof yet.
Thank you!
Define a function on the interval $\displaystyle [0,2]$ as,
$\displaystyle g(x)=f(x)-2x$
Since, $\displaystyle f(x),2x$ are countinous on $\displaystyle [0,2]$ so too is $\displaystyle g(x)=f(x)-2x$ continous on $\displaystyle [0,2]$.
Now,
$\displaystyle g(0)=f(0)-0=f(0)$
$\displaystyle g(2)=f(2)-4$
By definition $\displaystyle f(0)\geq 0$
If $\displaystyle f(0)=0$ then the proof is complete.
Because $\displaystyle x=0\in [0,1]$ is one such point. So there is no harm in assuming that $\displaystyle f(0)>0$
By definition $\displaystyle f(2)\leq 4$
If $\displaystyle f(2)=4$ then the proof is complete.
Because $\displaystyle x=2\in [0,2]$ is one such point. So there is no harm in assuming that $\displaystyle f(2)<4$
Thus, $\displaystyle g(0)=f(0)>0$ and $\displaystyle g(2)=f(2)-4<0$.
Since $\displaystyle g(x)$ is continous on $\displaystyle [0,2]$ there is a point $\displaystyle c\in [0,2]$ such that $\displaystyle g(c)=0$ by the intermediate value theorem.
Thus, $\displaystyle g(c)=f(c)-2c=0$. Thus, $\displaystyle f(c)=2c$ for some point in $\displaystyle [0,2]$.
Q.E.D.