# Prooblem 39 Sec 2.6 Calculus with Analytic Geometry by George Simmons

The problem statement is as follows:

Let $$y = f(x)$$ be a continuous function defined on the closed interval $$[0, b]$$ with the property that $$0 < f(x) < b$$ for all $$x \in [0, b]$$. Show that there exists a point $$c \in (0, b)$$ with the property that $$f(c) = c$$. Hint: Consider the function $$g(x) = f(x) - x$$.

I have wrapping my head around this problem since a couple of days in my free time. Initially it seemed easy. Proving that for some $$x \in (0, b)$$, $$f(x) = c$$, is a no brainer. But actually proving that $$f(c) = c$$ has baffled me. So I thought pitching this question here.

Given $$0 < f(x) < b$$.

Notice that if $$g(x) = x - f(x)$$ then

$$g(0) = - f(0) < 0$$

while

$$g(b) = b - f(b) > 0 \; \; \; \; ( since \; \; \; \; f(x) < b \; \; \forall x \in [0,b])$$

Now, use the IVT to conclude!

• Thanks a lot. Attention to details. Kudos. – python_noob Oct 3 '18 at 18:39
• But why did we choose this particular g(x) = f(x) - x? – python_noob Oct 3 '18 at 18:42
• because it works! it does the trick! – Mikey Spivak Oct 3 '18 at 18:44