Spivak - Proof of f(x) = c on [a, b]

[Simpl0S]

In Spivak's Calculus, on page 121 there is this theorem

Then he generalizes that theorem:

I tried proving theorem 4 on my own, before looking at Spivak's proof. Thus I let c = 0 and then by theorem 1, my proof would be completed. Is this a correct proof?

Spivak's proof for theorem 4:

And also can someone explain to me the choice of why Spivak has chosen the equation g = f - c. I understand his proof but the choice would have never occurred to me intuitively.

 

[fresh_42]

In Spivak's Calculus, on page 121 there is this theorem

View attachment 105918

Then he generalizes that theorem:
View attachment 105920

I tried proving theorem 4 on my own, before looking at Spivak's proof. Thus I let c = 0 and then by theorem 1, my proof would be completed. Is this a correct proof?

Spivak's proof for theorem 4:

View attachment 105921

And also can someone explain to me the choice of why Spivak has chosen the equation g = f - c. I understand his proof but the choice would have never occurred to me intuitively.

Your mistake has been, that you are not allowed to set $c=0$ because it has to be proven for arbitrary $c$.
So setting $g=f-c$ makes so to say the arbitrary $c$ for $f$ a $c=0$ for $g$. What is implicitly used here is the fact, that the sum or (difference) of two continuous functions - here $f$ and $c \cdot identity$ - is continuous again. The arbitrariness of $c$ goes entirely into the arbitrariness of $g$.

 

[PeroK]

And also can someone explain to me the choice of why Spivak has chosen the equation g = f - c. I understand his proof but the choice would have never occurred to me intuitively.

I'm not sure why you say you would never have thought of it.

If you'd drawn a diagram for a function with $c =2$, say, then you could make it into a function satisfying $c=0$ simply by moving it vertically.

Isn't that rather obvious? Especially now you've seen it.

 

[JonnyG]

And also can someone explain to me the choice of why Spivak has chosen the equation g = f - c. I understand his proof but the choice would have never occurred to me intuitively.

There are a lot of "tricks" and techniques that you will pick up along the way during your math studies. Some things that don't seem obvious at first will become more obvious later down the road and you'll be able to apply these tricks and techniques yourself.

 

[Ssnow]

The trick is quite usual, in the proof you translated the function $f$ subtracting $c$ defining a new function $g$ that satisfy conditions of theorem 1. So you are in the condition of a well understood case and you can conclude that exists a point $x\in [a,b]$ such that $g(x)=0$ that is $f(x)=c$.