Question

find the absolute maximum and minimum values of the following function on the given interval. then graph the function

$f(t)=2 - |t|$, $-1\leq t\leq7$

find the absolute maximum value. select the correct choice below and, if necessary, fill in the answer boxes to complete your choi

a. the absolute maximum value $\square$ occurs at $t = \square$.
(simplify your answers. use a comma to separate answers as needed.)

b. there is no absolute maximum.

Explanation:

Step1: Analyze the function \(f(t)=2 - |t|\)

The function \(f(t)=2 - |t|\) is a V - shaped graph. The vertex of \(y = |t|\) is at \((0,0)\), and for \(y=2 - |t|\), the vertex is at \((0,2)\). The function \(y = 2-|t|\) can be written as a piece - wise function:
\(f(t)=

$$\begin{cases}2 + t, & t<0\\2 - t, & t\geq0\end{cases}$$

\)
The derivative of \(f(t)\) (using the definition of the derivative for piece - wise functions):
For \(t < 0\), \(f^\prime(t)=\lim\limits_{h
ightarrow0}\frac{f(t + h)-f(t)}{h}=\lim\limits_{h
ightarrow0}\frac{(2+(t + h))-(2 + t)}{h}=1\)
For \(t>0\), \(f^\prime(t)=\lim\limits_{h
ightarrow0}\frac{f(t + h)-f(t)}{h}=\lim\limits_{h
ightarrow0}\frac{(2-(t + h))-(2 - t)}{h}=-1\)
At \(t = 0\), the left - hand derivative \(f^\prime(0^{-}) = 1\) and the right - hand derivative \(f^\prime(0^{+})=-1\). The derivative does not exist at \(t = 0\). Critical points occur where \(f^\prime(t)=0\) or \(f^\prime(t)\) does not exist. Since \(f^\prime(t)
eq0\) for \(t
eq0\) and \(f^\prime(0)\) does not exist, \(t = 0\) is a critical point.

Step2: Evaluate the function at critical points and endpoints

We need to evaluate \(f(t)\) at the critical point \(t = 0\) and the endpoints \(t=-1\) and \(t = 7\)

• When \(t=-1\), \(f(-1)=2-|-1|=2 - 1=1\)
• When \(t = 0\), \(f(0)=2-|0|=2\)
• When \(t = 7\), \(f(7)=2-|7|=2 - 7=-5\)

Answer:

A. The absolute maximum value \(2\) occurs at \(t = 0\)