this week in calculus we did the same thing as the past couple weeks, been working on practice ap tests, and going over the corrections, binder grades and such. i'm just gonna review today.
Optimization is used for finding maximum/minimum area.
STEPS FOR OPTIMIZATION:
1. Identify primary and secondary equations. **
2. Solve the secondary equation for any variable, then plug that back into primary equation.
3. Take the derivative of the primary equation after plugging in
4. set ^that equal to zero
5. solve for the OTHER variable.
6. Plug ^that back into the secondary equation.
7. solve for the last missing variable.
**Primary equation has variable that is being maximizied/minimized. Secondary equation is the other one, haha.
How to find area under a curve.
S (Top Equation)-(Bottom Equation) on interval [a,b].
To find a and b set 2 equations equal to each other and solve.
Also, if the area is on y, solve for X.
if it's on x, solve for Y.
ok so i have a couple questions.
-can someone go over the rules for a limit approaching zero?
-rules for integral of functions like sin cos tan cot sec csc?
-someone can you explain how to find points of inflection, max & min, concavity, all of the ways to find things for graphs like that.
-ok this might be a stupid question, but whenever it says find the slope of the equation of the tangent line to [a,b], idk what to do!? help!
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First Derivative Test
ReplyDeleteThe First Derivative Test is used for relative and absolute maximums and minimums. The Advanced Placement test is loaded with both multiple-choice and short answer questions that involve the First Derivative Test.
Example:
Find the absolute maximums and minimums of the equation (1/3)x^3-2x^2+3x.
1. Take the derivative of the equation.
x^2-4x+3
2. Set it equal to zero then solve for x.
x^2-4x+3=0 (x-3)(x-1) x=3 x=1
3. Set up intervals.
(-infinity,1) (1,3) (3, infinity)
4. Plug in a number found within all intervals into your first derivative (relative maximums and minimums).
(-infinity, 1)=positive number at x=1 there is a maximum
(1,3)=negative number at x=3 there is a minimum
(3,infinity)=positive number
5. Plug in x values to original equation to find absolute maximums and minimums.
x=1 is the absolute maximum (1,2/3)
x=3 is the absolute minimum (3,0)
Second Derivative Test
The Second Derivative Test is useful for finding concavity or absolute maximums and minimums.
Example:
Find the absolute maximum of the equation y=x^3-3x
1. Find first derivative.
x^3-3x
3x^2-3
2. Set equal to 0 then solve for x.
3x^2-3=0 3x^2=3 x^2=1 x=1 x=-1
3. plug into second derivative.
3x^2
6x 6(1)=6 6(-1)=-6 x=6 there is a absolute maximum.
On which interval(s) does the function f(x)=x^4-4x^3+4x^2+6 increase?
ReplyDeleteKey word: increase, which means to take first derivative test
x^4-4x^3+4x^2+6
4x^3-12x^2+8x
4x(x^2-3x+2)
(x-2)(x-1)
x=0, 1, 2
(-infinity, 0) (0,1) (1,2) (2, infinity)
-1: 4(-1)^3-12(-1)^2+8(-1) = - ve
.5: 4(.5)^3-12(.5)^2+8(.5) = + ve
1.5: 4(1.5)^3-12(1.5)^2+8(1.5) = - ve
3: 4(3)^3-12(3)^2+8(3) = + ve
Going from negative to postive means you are increasing. And I think you put it as a point, right?
First Derivative Test
ReplyDeleteSecond derivative Test
plug into calculator
These things can all help you find all max and mins relative or absolute. relative is just saying that its a min on an interval of graph. Absolute is the absolute lowest point on the graph.
the first derivative test is used to find absolute maximums and minimums. To do these, first you take the derivative of the given equation, then you set it equal to zero and solve for x, then set up intervals using the x values, then plug in numbers within the intervals into the derivative to find where your relative maximums and minimums are, then you plug the x values into the original equation to find the absolute max's and min's.
ReplyDeleteFind the absolute maximums and minimums of the function f(x)=x^2-8x+4
First, you take the derivative of it:
f'(x)= 2x-8
Then you set it equal to zero and solve for x.
2x-8=0 2x=8 x=4
Then you set up the intervals
(-infinity, 4) U (4, infinity)
then you plug in a number found within the interval into the first derivative to find relative max's and min's.
(-infinity, 4)= negative number
(4, infinity)= positive number
so at x=4 there is a minimum
after that, you plug in that x value into the original equation to find the absolute max's and min's. (in this case it is only a min)
x=4 is the absolute minimum (4,-12)