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

$(0,3)$ and $(1,4)$ are the critical points of $y$. At $x=0$, function $y$ has the local minimum value of $3$, and at $x=1$, it has the local maximum value of $4$. There is neither absolute minimum nor maximum value.
$y=3-x$ for $x\lt0$ and $y=3+2x-x^2$ for $x\ge0$ - Domain: $R$. Therefore, there is no endpoint. 1) Find the derivative of function $y$: - For $x\lt0$: $y'=-1$ - For $x\ge0$: $y'=2-2x$ We have $y'=0$ when $2-2x=0$, or $x=1$. So there is a critical point at $x=1$. - Also, at $x=0$, we have $y'(0)=2-2\times0=2$. Since $2\ne-1$, $y$ is not differentiable at $x=0$, meaning $y'$ is not defined at $x=0$ as well. Therefore, there is one more critical point at $x=0$. 2) Evaluate $y$ at the critical points: - For $x=0$, we would use the function $y=3+2x-x^2$. $$y=3+2\times0-0^2=3$$ - For $x=1$, we would again use the function $y=3+2x-x^2$. $$y=3+2\times1-1^2=3+2-1=4$$ So $(0,3)$ and $(1,4)$ are the critical points of $y$. Now look at the graph of $y$. At $x=0$, function $y$ has the local minimum value of $3$, and at $x=1$, it has the local maximum value of $4$. There is neither absolute minimum nor maximum value.