Let $A = \{x ∈R:x ≥1/2\}$ and $B=\{x∈R: x≥3/4\}$. If $f: A→B$ is defined as $f(x) = x^2-x+1$, then the solution set of the equation $f(x) = f^{-1}(x)$ is

{1}

Clearly, $f: A → B$ is a bijection. This fact can also be observed from the graph of f(x) as it represents an arc of the parabola $y = x^2 - x+1$ lying on the right side of the vertex (1/2, 3/4).

We know that the curves $y = f (x)$ and $y = f^{-1}(x)$ are mirror images of each other in the line mirror $y = x$. This means that the two curves intersect at points lying on the line $y = x$.

$∴ f(x)=f^{-1}(x)$

$⇒ f(x) = x = x^2-x+1=x⇒ (x-1)^2=0⇒ x=1$

Hence, the solution set of the equation $f (x) = f^{-1}(x)$ is $\{1\}$.