This week we had exams, and starting on Thursday, we began notes on finding critical numbers, extreme values, maximums and minimums, and horizontal tangents all by taking the derivative, setting it equal to zero and solving for x.
Guidelines for finding extrema on a closed interval:
-to find the extrema of a continuous function f on a closed interval [a,b] use the following steps:
1. find the critical numbers of f in (a,b).
2. evaluate f at each critical number in (a,b).
3. evaluate f at each endpoint in (a,b).
4. the least of these values is the minimum and the greatest is the maximum.
5. anywhere the function is not differentiable is a critical number.
Key words:
Extreme values- maximums and minimums
Absolute max- highest point on an interval
Absolute min- lowest point on an interval
Relative max- any max not on the interval
Relative min- any min not on the interval
Critical numbers- any max or min typically referred to as x=c
**always check endpoints on an interval!
Ex: f(x) =2x-3x^ (2/3) on [-1, 3]
F prime(x) = 2-2x^ (-1/3) =0
2-2x^ (-1/3) =0
-2x^ (-1/3) =-2
(x^ (-1/3))^3 =-2/-2=1^ 3
x=1
f(-1)= 2(-1) -3(-1)^(2/3)= -5-> min
f(1)= 2(1)-3(1)^(2/3)= -1
f(3)= 2(3)-3(3)^(2/3)= -.24
f(0)= 2(0)-3(0)^(2/3)= 0-> max
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