Chapter 4 - Section 4.1 - Extreme Values of Functions - Exercises - Page 216: 45

critical point will be (3,-2)

Given $y=x^2-6x+7$ Critical point occurs where first derivative is zero: so$\frac{dy}{dx}=\frac{d{(x^2-6x+7)}}{dx}$ ${\frac{dy}{dx}}=2x-6$ so the critical point is: ${\frac{dy}{dx}}=0$ $f'(x)=2x-6=0$ $x=3$ find the critical value: $y=3^2-6\times3+7=-2$ thus, the final critical point will be: (3,-2)