Find the absolute extrema of the function on the closed interval.

Question

y=1-|t-1| [-7,4]

I got the minimum correct (-7,-7); I got max wrong (4,-2)

Answer?

Answer 1

If you think of the graph, this absolute value function is an upside-down "V" with a vertex at (1, 1). If you were going to treat this like a calculus problem, you would need to break it into 2 pieces (a piece-wise function).

y = 1 - |t - 1| is the same as

y = 1 - (t - 1) for t > 1 (where the function inside the absolute value signs is positive) and y = 1 - -(t - 1) for t < 1 (where the function inside the absolute value signs is negative).

So f(t) = 2 - t for t > 1 and

f(t) = t for t < 1

To find local extremes, we take the derivative and set it equal to zero.

The derivative for t > 1 is -1 and setting -1 = 0 we see there are no solutions.

The derivative for t < 1 is 1 and setting 1 = 0 we see there are no solutions.

Now, we must determine if there are any places on the interval where the derivative does not exist. Since there is a sharp point at t = 1, the derivative does not exist there (despite the fact I "pretended" like it did when I said "the derivative for t > 1 is -1".

We also need to consider the endpoints of the interval, t = -7 and t = 4

t f(t)

-7 -7

1 1

4 -2

(notice that since there were no standard critical points, we only have to check the endpoints and the place where the derivative does not exist.

So the absolute min is at t = -7 and is -7 and the absolute max is when t = 1 and is 1.

Answer 2

Your Minimum is correct.

The Max, if you sketch the function, peaks at x=1 and y=1, thus the Max is (1,1)

I recommend you plot this on a TI-84 or equivalent, and get use to seeing what

functions look like, then you can determine everything about it.