Page 127 - Calculus Workbook For Dummies

111
Chapter 7: Analyzing Those Shapely Curves with the Derivative

4
t + 4
d Using the first derivative test, determine the local extrema of s = 2 . Local maxes at
- t 2
` - , 2 - 2j and ` , 2 - 2j.
1. Do the differentiation thing.
4
t + 4
s = 2
- t 2
2 l
4
2
2
3
4
4
4 -
4 -
_ t + 4 _i l - t 2 i - _ t + i _ t 2 i _ t 4 i _ - t 2 i - _ t + i ^ t 4 h t - + 4
4
s = l 2 = 4 = 3
2
_ - t 2 i t 4 t
2. Find the critical numbers.
4
t - + 4 = 0
t 3
4
4 - t = 0
2
2
_ 2 - t _ i 2 + t i = 0
2
` 2 - t `j 2 + t _ j 2 + t i = 0
t = 2 or - 2
So - 2 and 2 are two critical numbers of s.
t = 0 is a third important number because t = 0 makes the derivative’s denominator equal zero,
and so you need to include zero on your sign graph in order to define test regions. Note, how-
ever, that t = 0 is not a critical number of s because s is undefined at t = 0. And because there is
no point on s at t = 0, there cannot be a local extremum at t = 0.
3. Test values: You’re on your own.
4. Make a sign graph (see the following figure).
increasing decreasing increasing decreasing
+ – + _
0
She loves me; she loves me not; she loves me; she loves me not.
5. Find the y-values.
4
j
` - 2 + 4
s - 2 = 2 = 4 + 4 = - 2 You climb up the hill to - , 2 - 2j, then down, so there’s a
`
j
`
2 -
- ` 2j - 4
local max there.
4
0 + 4
s 0 = 2 = undefined (which you already knew). Therefore, there’s no local extremum
^ h
- 2 0 ^ h
at t = 0. Remember that if a problem asks you to identify only the x-values and not the
y-values of the local extrema, and you only consider the sign graph, you would incorrectly
conclude — using the current problem as an example — that there’s a local min at t = 0.
So you should always check where your function is undefined.
4
2 + 4
j
s ` 2 = 2 = - 2. Up then down again, so there’s another local max at ` , 2 - 2j.
- 2 2
As always, you should check out this function on your graphing calculator.