This week in calculus we learned about absolute MAXS and MINS.
1. first derivative test
2. plug the critical values into origional function to get y-values.
3. plug endpoints in to origional function to get y-values.
4. highest y-value is absolute max.
5. lowest y-value is absolute min.
-absolute maxs or mins or written as a point or simply as the y-value.
For example:find the absolute max or min of f(x)=3x^4-4x^ on [-1,2].
f1(x)=12x^3-12x^2=0
12x^2(x-1)=0
x=1,0
(-1, 0)U(0,1)U(1,2)
f1(-.5)=-ve f1(.5)=-ve f1(1.5)=+ve
min @x=1
3(1)^4-4(1)^3=-1 (1,-1)
3(-1)^4-4(-1)^3=-7 (-1,-7)
3(2)^4-4(2)^3 (2,16)
-32=16
abs min:(1,-1)or -1
-1 at x=1
i actually understand the max and min thing but doing it with the trig functions like that long problem we did in class confuses me. all the steps are throwing me way off so if someone could kind of help me with that it would be nice.
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For trig functions, you really just have to remember everything we learned last year. The unit circle, the chart, how to convert from degrees to radians (multiply by pi/180), and you have to remember what quadrants are sin, cos, and tan positive and negative.
ReplyDeletesin- positive is positive in the 1st and 2nd quadrant and negative in the 3rd and 4th
cos- positive in the 1st and 4th quadrants and negative in the 2nd and 3rd
tan- positive in the 1st and 3rd quadrants and negative in the 2nd and 4th
They trig problems are not really that hard if you really think about them but they are time consuming.