f(0) = 0, f(1) = 1, prove f(x) + 2019 = 2019f'(x) + x have at least one root on (0, 1)?

Question

Suppose that f(x) is continuous on [0, 1], differentiable on (0, 1) and f(0) = 0, f(1) = 1. Prove that: f(x) + 2019 = 2019f'(x) + x have at least one root on (0, 1).

I think this problem should be solved by using Lagrange theorem, but don't know where to start, can anybody give me a hint?

Answer 1

Define $g(x)=f(x)-x$, $g(0)=g(1)=0$. The problem requires us to prove that there exists a $c \in (0,1)$ such that $g(c)=2019g'(c)$.

Now, for any $\alpha\in\mathbb{R}$, let $h(x)=e^{\alpha x}g(x)$.

By applying Lagrange's Theorem on $[0,1]$, we get $0=e^{\alpha c}(g'(c)+\alpha g(c))$ for some $c \in (0,1)$.

That is, there exists a $c\in (0,1)$ such that $-\alpha g(c)=g'(c)$ for any $\alpha\in\mathbb{R}$.

Simply put $\alpha=\frac{-1}{2019}$.

Answer 2

The problem is equivalent to prove that $g(x)=2019g'(x)$ has a solution with $g(1)=g(0)=0$.

$\bullet$ Suppose $g'(x)<g(x)/2019$ for all $x$:

Let $a$ a maximum point for $g$(note $g(a)\geq 0$), then $$g(a)=\int_0^{a}g'(t)dt\leq \int_0^{a}g(t)/2019 dt\leq \int_0^{a}g(a)/2019dt= ag(a)/2019\leq g(a)/2019$$ A contradiction (unless $g\equiv 0$, but in this case the proposition is obvious)

$\bullet$ Suppose $g'(x)>g(x)/2019$ for all $x$: Similarly, Let $b$ a minimum point for $g$(note that $g(b)\leq 0$), then $$g(b)=\int_0^{b}g'(t)dt\geq\int_0^{b}g(t)/2019 dt\geq \int_0^{b}g(b)/2019dt= bg(b)/2019\geq g(b)/2019$$ A contradiction as well

Hence we proved that $g'(x)-g(x)/2019$ has to change signs eventually and then by Intermediate value theorem we get that it has to be $0$ for some $x$.

Answer 3

Here’s a slightly different/elementary approach towards this problem.

$f(x) + 2019 = 2019f’(x) + x$

This is a linear differential equation whose general solution is y = x + C. Applying initial value condition, we obtain that C = 0.

EDIT: This solution can even be obtained by just rearranging terms . $f(x) - x = 2019(f’(x) - 1)$ clearly suggests that solution is indeed of the form $f(x) = x$.

Now, this problem requires us to prove that there exists a $z \in (0, 1)$ s.t f’(z) = 1 or f(z) = z (you can check why).

Applying Mean’s value theorem on $[0, 1]$, we obtain that there does exist $c \in (0, 1)$ s.t. f’(c) = 1 or f(c) = c. Put c = z and you’re done.