# Math Help - Max and min of absolute value interval

1. ## Max and min of absolute value interval

Let $f(t)=3-|t-3|$ on the interval $[-1,5]$. Find the absolute maximum and absolute minimum of $f(t)$ on this interval.

The absolute max occurs at $t = 3$
The absolute min occurs at $t = -1$

I think I am doing this wrong. Those are the correct answers but I have no idea how 3 is the max. fuggled to me.

2. Originally Posted by pseizure2000
Let $f(t)=3-|t-3|$ on the interval $[-1,5]$. Find the absolute maximum and absolute minimum of $f(t)$ on this interval.

The absolute max occurs at $t = 3$
The absolute min occurs at $t = -1$

I think I am doing this wrong. Those are the correct answers but I have no idea how 3 is the max. fuggled to me.
why does that "fuggle" you? since whatever is in absolute value signs is nonneggative, the maximum value of the function is when what is in the absolute value signs is zero. so it would be 3 - 0 = 3. this happens on the interval when t = 3

what specifically is your concern? how did you work out the problem?

3. I'd do it like this---I will test the f(t) for every t in the given interval.

f(t) = 3 -|t -3|, in the interval [-1,5]

f(-1) = 3 -|-1 -3| = -1 <--------------minimum
f(0) = 3 -|0 -3| = 0
f(1) = 3 -|1 -3| = 1
f(2) = 3 -|2 -3| = 2
f(3) = 3 -|3 -3| = 3 <-----------------maximum
f(4) = 3 -|4 -3| = 2
f(5) = 3 -|5 -3| = 1

4. Originally Posted by ticbol
I'd do it like this---I will test the f(t) for every t in the given interval.

f(t) = 3 -|t -3|, in the interval [-1,5]

f(-1) = 3 -|-1 -3| = -1 <--------------minimum
f(0) = 3 -|0 -3| = 0
f(1) = 3 -|1 -3| = 1
f(2) = 3 -|2 -3| = 2
f(3) = 3 -|3 -3| = 3 <-----------------maximum
f(4) = 3 -|4 -3| = 2
f(5) = 3 -|5 -3| = 1
well, t does not have to be an integer it seems. t can be a real number, in which case someone could argue that: "perhaps a t you neglected to check gives the absolute max"

5. Originally Posted by Jhevon
well, t does not have to be an integer it seems. t can be a real number, in which case someone could argue that: "perhaps a t you neglected to check gives the absolute max"
I see.

Okay, say t = 2.99
f(2.99) = 3 -|2.99 -3|= 2.99

Say again t = 3.001
f(3.001) = 3 -|3.001 -3| = 2.999

So, where would f(t) be greater than 3 in the given interval?

6. fuggled because my teach told me i need to evaluate all critical points (where f'(x) DNE and/or f'(x) = 0 and endpoints

maybe i'm not doing it right but I couldn't find 3 as a critical point

7. Originally Posted by pseizure2000
fuggled because my teach told me i need to evaluate all critical points (where f'(x) DNE and/or f'(x) = 0 and endpoints

maybe i'm not doing it right but I couldn't find 3 as a critical point
your instructor is correct, the thing is, f'(x) does not exist at 3 (the graph is a cusp at t = 3, we have the same problem when we try to evaluate the derivative of |x| at x = 0, for instance), that's why you can't find it. you are dealing with absolute values here, you need to split the interval in two and work on it piece by piece. find where the turning point is, and then split the interval at that turning point. here the turning point is 3, so consider the interval [-1,3] and [3,5]. on both interval, the absolute max is 3, so we conclude that the absolute max is 3 for the entire interval

8. Originally Posted by pseizure2000
fuggled because my teach told me i need to evaluate all critical points (where f'(x) DNE and/or f'(x) = 0 and endpoints

maybe i'm not doing it right but I couldn't find 3 as a critical point
I suspected using the first derivative of f(t) is one way, but I forgot how to differentiate an absolute quantity.

So I thought of doing it graphically.
The graph of f(t) = |t| is a V that opens upward, whose vertex is at (0,0), and the slopes of the two rays are -1 or 1, or they are at 45 degrees with the horizontal.

You make that f(t) = -|t|, then you "invert", or turn upside down, the graph of f(t) = |t|

Then you add 3 to the -|t| to make it f(t) = 3 -|t|, and you shift the vertex of f(t) = -|t| vertically upward by 3 units.

Then you subtract 3 from the |t| to make it |t -3|, and you shift the graph 3 units to the right.

You graph the final f(t) = 3 -|t-3|. You'd see that its vertex is a maximum and it is at (3,3).

9. We have $-1\leq t\leq 5$.
Substract 3 from all members: $-4\leq t-3\leq 2\Rightarrow 0\leq|t-3|\leq 4$.
Multyply all members by -1: $0\geq-|t-3|\geq -4$.
Add 3 to all members: $3\geq 3-|t-3|\geq -1$.

10. Originally Posted by red_dog
We have $-1\leq t\leq 5$.
Substract 3 from all members: $-4\leq t-3\leq 2\Rightarrow 0\leq|t-3|\leq 4$.
Multyply all members by -1: $0\geq-|t-3|\geq -4$.
Add 3 to all members: $3\geq 3-|t-3|\geq -1$.
Very nice, simple yet very impressive, as always