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

$f(t)$ has the absolute maximum value of $2$ at $t=7$ and the absolute minimum value of $0$ at $t=5$ in the defined domain.
$$f(t)=|t-5|\hspace{1cm}4\le t\le7$$ 1) For $t\in[4,5)$, we have $(t-5)\lt0$, so $$f(t)=5-t$$ While for $t\in[5,7]$, we have $(t-5)\ge0$, so $$f(t)=t-5$$ So here we have 2 different functions for$f(t)$ in 2 separate intervals $[4,5)$ and $[5,7]$. Therefore, we need to consider $t=5$ as one of the endpoints besides $t=4$ and $t=7$. 2) We can skip the lengthy work of deciding whether $t=5$ is a critical point or not by checking the differentiability of $f'(t)$ at $t=5$, since we will eventually check $f(t)$ for $t=5$ as an endpoint. Find the derivative $f'(t)$ for $t\ne5$: - For $t\in[4,5)$: $f'(t)=-1$ - For $t\in[5,7]$: $f'(t)=1$ None of the values of $t$ could get $f'(t)=0$ or undefined. So $f(t)$ has no critical points. 2) Evaluate $f(t)$ at the endpoints: - For $t=4$, we would use $f(t)=5-t$: $$f(4)=5-4=1$$ - For $t=5$, we would use $f(t)=t-5$: $$f(5)=5-5=0$$ - For $t=7$, we would use $f(t)=t-5$: $$f(7)=7-5=2$$ This means $f(t)$ has the absolute maximum value of $2$ at $t=7$ and the absolute minimum value of $0$ at $t=5$ in the defined domain.