Composition function exercise to find $ f(0)$

Exercise:

Let $f:\mathbb{R}\to \mathbb{R}$ be a function defined to be $\left( f\circ f \right)\left( x \right)={{x}^{2}}-x+1$

Determine the form of $f\left( x \right)$ then deduce the value of $f\left( 0 \right)$

Solution: we have $\left( f\circ f \right)\left( x \right)=f\left( f\left( x \right) \right)={{x}^{2}}-x+1$

As this expression is quadratic so take $f\left( x \right)=ax+b$ where $a,b\in \mathbb{R}$

Hence $f\left( f\left( x \right) \right)=f\left( ax+b \right)=ax\left( ax+b \right)+b={{a}^{2}}{{x}^{2}}+abx+b$

but $f\left( f\left( x \right) \right)={{x}^{2}}-x+1={{a}^{2}}{{x}^{2}}+abx+b$

by compression we get ${{a}^{2}}=1\,\,,\,\,\,ab=-1\,\,\And \,\,b=1$ hence $a=-1$

so $f\left( x \right)=-x+1$ hence $f\left( 0 \right)=1$