how do you find extrema?

y = 2^(−|x|) where −6 < x <= 5

Whenever you have an absolute value in a problem you need to break the problem into two ranges: where what's in absolutes is positive and where it's negative.

Case #1: x >= 0, therefore |x| = x, so y = 2^(−|x|) = 2^x where 0 <= x <= 5

dy/dx = [ln(2)] 2^x = 0

So x has no real solution because dy/dx -> 0 when x -> negative infinity

So there are no local extrema anywhere for x >= 0

Now the global maximum would correspond to the greatest value of y, which is y = 2^5 = 32 for the range 0 <= x <= 5.

The global minimum for this range would be y = 2^0 = 1

Now let's look at case 2: x <= 0, therefore |x| = - x, so y = 2^(−|x|) = 2^(- x) where - 6 < x <= 0

dy/dx = [- ln(2)] 2^(- x) = 0

So x has no real solution because dy/dx -> 0 when x -> positive infinity

So there are no local extrema anywhere for x <= 0

Now the global maximum would correspond to the greatest value of y, which is y = 2^(- - 6) = 64- for the range - 6 < x <= 0. Note that y = 64 is not actually included because it approaches 64 from the left.

The global minimum for this range would be y = 2^0 = 1

Putting the two cases together we have for the range - 6 < x <= 5

that there are no local extrema, but we have a global minimum at [0,1] and a global maximum at

(- 6, 64). Use the set notation your teacher wants you to use. Note [0,1] is inclusive in the solution set, whereas (- 6, 64) is not in the solution set for the function domain specified.

If I remember correctly, an absolute extremum is the highest (or lowest) peak (or valley) on the curve - the min or max - while the local extrema are all the peaks and valleys of the curve - they may not be the highest or lowest points on the whole graph, but they're higher/lower than the points immediately near them. So, no - it is not necessary for a local extremum to be the absolute extremum... but an absolute extremum is always a local extremum, as well. Hope this isn't confusing...