Mathematics - Functions Question with Solution | TestHub

First, we find the values of $y$ such that $f(y)=2$. The function can be written as:$f(x)=\{\begin{array}{ll}(x-2{)}^2+2& \text{if }x<2\\ -x+6& \text{if }2\le x<4\\ x-2& \text{if }x\ge 4\end{array}$ If $y<2$, $(y-2{)}^2+2=2⟹(y-2{)}^2=0⟹y=2$. This contradicts $y<2$, so no solution here. If $2\le y<4$, $-y+6=2⟹y=4$. This contradicts $y<4$, so no solution here. If $y\ge 4$, $y-2=2⟹y=4$. This is a valid solution. Thus, $y=4$ is the only solution for $f(y)=2$. Now, we need to solve $f(x)=4$. If $x<2$, $(x-2{)}^2+2=4⟹(x-2{)}^2=2⟹x-2=\pm \sqrt{2}⟹x=2\pm \sqrt{2}$. Since $x<2$, $x=2-\sqrt{2}$ is a solution. If $2\le x<4$, $-x+6=4⟹x=2$. This is a valid solution. If $x\ge 4$, $x-2=4⟹x=6$. This is a valid solution. The distinct real solutions are $2-\sqrt{2}$, $2$, and $6$. Therefore, there are 3 distinct real solutions.