# Math Help - Easy Analysis IVT

1. ## Easy Analysis IVT

Help!

Q: 1) State the intermediate value theorem
2) Suppose $f:[0,1] \rightarrow \mathbb{R}$ is continous on $[0,1]$, $f(0) = 0$ and $f(1)=1$.
Prove that there exists a $c \in (0,1)$ such that $f(c) = 1-c$.

My solution:

A: 1) Suppose $f:[a,b] \rightarrow \mathbb{R}$ is continuous on $[a,b]$ with $f(a) < 0 < f(b)$.

Then $\exists c \in (a,b)$ such that $f(c) = 0$ .

2) ????????????????

Any help would be appreciated thanks

2. HINT for #2.
Is $g(x)=x+f(x)-1$ a continuous function? WHY?
$g(0)=?~\&~g(1)=?$

3. g is continous on [0,1] because its limit as x tends to c is g(c).

thanks !