## University Calculus: Early Transcendentals (3rd Edition)

$(1,2)$ is the critical point of $y$. There is no endpoint. At $x=1$, function $y$ has the absolute and local minimum value of $2$, but it has no absolute and local maximum value.
$y=4-2x$ for $x\le1$ and $y=x+1$ for $x\gt1$ - Domain: $R$. Therefore, there is no endpoint. 1) Find the derivative of function $y$: - For $x\le1$: $y'=-2$ - For $x\gt1$: $y'=1$ We can see here that as $x\to1^-$, $y'=-2$, while as $x\to1^+$, $y'=1$. This means $y'$ is not defined at $x=1$. In other words, there is a critical point of $y$ at $x=1$. 2) Evaluate $y$ at the critical points: - For $x=1$, we would use the function $y=4-2x$. $$y=4-2\times1=2$$ So $(1,2)$ is the critical point of $y$. Now look at the graph of $y$. At $x=1$, function $y$ has the absolute and local minimum value of $2$, but it has no absolute and local maximum value.