Blog 8

Soo, I did pretty well on this nine weeks exam :) hhaa, anyways. We started learning chapter three this week and it seems like its going to be pretty tough because it goes back to everything we learned so far in chapter’s one and two, and even stuff that we learned in advance math. Right now we are learning how to find extreme values. Extreme values are the max and mins of a function, or graph. The absolute max is the highest point on an interval. The absolute min is the lowest point on an interval. We also learned how to find relative maxs and mins, which are any max or min not on an interval. If it asks you to find a critical number, you’re looking for any max or min. **Keep in mind anywhere the function is NOT differentiable it is a critical value. **To find a critical number, extreme value, max, min, or horizontal tangent, take derivative and set equal to 0. **Also when dealing with a fraction you would take the derivative by doing the quotient rule, and then set the top equal to 0. The bottom doesn’t really matter in a fraction. **If it gives you an interval you will take the derivative of the function, set it equal to 0, then check to see if the numbers are in the interval. Then you will plug the numbers you found, and the interval numbers back into the original function. The highest point will be your abs. max, the lowest your abs. min.

Ex: Find the extrema of f(x)=3x^4-4x^3 on the interval [-1,2]
12x^3-12x//62=0
12x^2(x-1) x=0, x=1

*Now you’re going to plug back into the original equation.
F(-1)=3(-1)^4-4(-1)^2=7
F(0)=3(0)^4-4(0)^3=0
F(1)=3(1)^4-4(1)^3=-1
F(2)=3(2)^4-4(2)^3=16
**-1 Abs. min, 16 abs. max

Now for something I don’t really understand. I don’t understand the trig functions ones where you have to use the quadrants then convert to radians. I know how to use the quadrant and convert to radians, I just don’t know how to get there.