Find the points where $g(x)$ touches the x axis

Question

Let $f(x)$ be defined as $$f(x)=\int_0^1|x-t|dt$$

and $g(x)$ be defined as $$g(x)=f(x)-x$$

then find the points in (0,1) where $g(x)$ touches the x axis.

My attempt:

I need to find where $g(x)=0$, i.e. $f(x)=x$

I can split the integral as,

$$f(x)=\int_0^x(x-t)dt+\int_x^1(t-x)dt$$

and on putting $f(x)=x$, I get $x=\frac{1}{4}$.

Am I solving it right?

Answer 1

The approach is correct, but I think you have a problem when calculating $f(x)$, this is the explicit result

$$ f(x) = \frac{1}{2} - x(1-x) $$

So the solution is ($0<x<1$)

$$ x = 1 - 2^{-1/2} $$

Answer 2

\begin{align} f(x) &= \int_0^x(x-t)dt+\int_x^1(t-x)dt \\ &= \left. \left (xt - \dfrac 12t^2 \right)\right|_{t=0}^{t=x} +\left. \left (\dfrac 12t^2 - xt \right)\right|_{t=x}^{t=1} \\ &= \left(x^2 - \dfrac 12x^2 \right) +\dfrac 12 - x - \dfrac 12x^2 + x^2 \\ &= x^2 - x + \dfrac 12 \end{align}