1. ## Analysis problem

Let g: [a,b] --> [c,d] be a continuous function in $x_0\in$[a,b], and f: [c,d] --> R a continuous function in g( $x_0$) $\in$ [c,d]

Show that f o g: [a,b] --> R defined by (f o g)(x) = f(g(x)) $\forall$x $\in$ [a,b] is continuous in $x_0 \in$ [a,b]

2. Anyone have any ideas on solving this?

3. Hello,
Originally Posted by Cato
Let g: [a,b] --> [c,d] be a continuous function in $x_0\in$[a,b], and f: [c,d] --> R a continuous function in g( $x_0$) $\in$ [c,d]

Show that f o g: [a,b] --> R defined by (f o g)(x) = f(g(x)) $\forall$x $\in$ [a,b] is continuous in $x_0 \in$ [a,b]
Use the definition of continuity for f in $x_0$ :
$\forall \epsilon > 0,~ \exists \delta > 0,~ |x-x_0|< \delta \Rightarrow |f(x)-f(x_0)|< \epsilon$

4. Originally Posted by Cato
Show that f o g: [a,b] --> R defined by (f o g)(x) = f(g(x)) $\forall$x $\in$ [a,b] is continuous in $x_0 \in$ [a,b]
Let me rephrase the problem a little. Let $g: D_1 \to D_2$ and $f: D_2 \to \mathbb{R}$. Let $g$ be continous at $x_0 \in D_1$ and $f$ be continous at $g(x_0) \in D_2$. Let $\epsilon > 0$. There is $\delta > 0$ so that if $y\in D_2 \text{ and }|y-g(x_0)| < \delta \implies |f(y) - f(g(x_0))| < \epsilon$. There is $\delta_1 > 0$ so that $x\in D_1 \text{ and }|x-x_0| < \delta_1 \implies |g(x) - g(x_0)| < \delta \implies |f(g(x)) - f(g(x_0))| < \epsilon$. Thus, we see that $f\circ g$ is continous at $x_0$.