$f: [a,b] \to \Bbb R$ is continous on $[a,b]$ then $f([a,b])=[m,M]$ s.t $f(a) \neq m$ , $f(a) \neq M$ then $ \exists c \in (a,b)$ s.t $f(c)=f(a)$.

Question

i was doing a qus. and while doing that a a concept struck in my mind. I dont know its right or wrong. Can anyone please verify it.

As witten in qus.$f: [a,b] \to \Bbb R$ is continous on $[a,b]$ then $f([a,b])=[m,M]$ s.t neither $f(a) \neq m$ nor $f(a) \neq M$ then $ \exists c \in (a,b]$ s.t $f(c)=f(a)$.
i tried in this way. Let $g(x)=f(x)-f(a)$ is continous on $[a,b]$ then as $f([a,b])=[m,M]$ then $\exists x_1,x_2 \in (a,b]$ s.t $f(x_1)=m$ and $f(x_2)=M$ then $g(x_1)=f(x_1)-f(a) \lt 0$ and $g(x_2)=f(x_2)-f(a) \gt 0$ so $\exists c \in (x_1,x_2)$ s.t $g(c)=0$ i.e $\exists c \in (a,b)$ s.t $f(c)=f(a)$.
is it right?

Answer 1

Or you might reason as follows:

• $f(a) \neq M, \;f(a) \neq m \stackrel{f([a,b])=[m,M]}{\Rightarrow} m < f(a) <M$
• $f$ is continuous, so according to MVT $\exists c \in (a,b]: f(c) = f(a)$