Answer
$(1,2)$ 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.
Work Step by Step
$y=4-2x$ 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.
