Mathematics - Quadratic Equation Question with Solution | TestHub

Let $y=x^2+1$. Since $x\in \mathbb{R}$, $x^2\ge 0$, which implies $y\ge 1$. The given expression can be written as a quadratic in $y$: $g(y)=y^2-4y+7$. This is a parabola opening upwards, so its minimum occurs at the vertex. The vertex of $g(y)$ is at $y=-(-4)/(2\cdot 1)=2$. Since $y=2$ is within the domain $y\ge 1$, the minimum value of the expression occurs at $y=2$. Substituting $y=2$ into $g(y)$, we get $g(2)=(2{)}^2-4(2)+7=4-8+7=3$. Thus, the minimum value of the expression is 3.