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

1. Sep 14, 2016

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.

2. Sep 14, 2016

Staff: Mentor

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 $g$.

3. Sep 14, 2016

PeroK

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.

4. Sep 14, 2016

JonnyG

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.

5. Sep 19, 2016

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 $f(x)=c$.