OKAYYY, so my blog wouldn't post lastnight so i just saved it and i'm posting
it nowwww.
This past week in calculus has been prettty good for me actually, i think i've made more A's this week alone than i did with all of advanced math and calculus combined, but that's just because i'm good a guessing multiple choiceeeee.
i'm just going to go over some old stuff incase it comes up soon because things tend to do that in calc.
discontinuities
1. Jump
-The limit does not exist
-The function is continuous everywhere except at the jump. Therefore, if we are
talking about the function as a whole we say that it is not continuous.
2. Removable- when the graph is not defined at a point..(open circle)
-The limit exists
-The function is continuous everywhere except at that point. Therefore, if we are
talking about the function as a whole we say that it is not continuous.
3. Infinate – An asymptote
-The limit cab exist.
-The function is continuous everywhere except at the asymptote.
4. Oscillation – An extreme oscillating graph
-The limit doesn't exist
-The function is not continuous.
WHAT TO DO WHEN YOU SEE THESE KEY WORDS:
Maximum
Minimum
Critical values
Increasing
Decreasing
Take derivative
set equal to 0
solve for x.
Set up intervalsusing these numbers.
Plug in numbers on each interval.
If it changes from positive to negative it is a maximum. If it changes from negative to positive it is a minimum. All potential maximums and minimums are called critical values. If the interval is positive the
interval is increasing. If the interval is negative then the interval is decreasing.
help?? um can someone explain linearization, i kinda remember what it is and what to do but not really so and explanation would be awesomee.
Linearization is kind of like equation of a tangent line except there is one more step to it. Most of the time they give you an equation and an x and y value. If no y value is given, plug the x value into the function and solve for y. Then take the derivative of the function and plug in x to find the slope. Then put everything into point slope form, y-y1=slope(x-x1). Then they will give you an extra x value for linearization that you plug into the extra x. Then solve. Hope this helps ya!
ReplyDeleteLinearization's key words include approximate or if you see decimals that should also make the light bulb go off in your head. Some problems will, however, directly state use linearization to find ...
ReplyDeleteFor linearization, you need a point, and a slope. The point or at least the x-value should be given. If you only have the x- value, plug in for x to find y. The slope can be found by taking the derivative or using the slope formula. Once you have a point and a slope, plug into the point-slope formula. Then plug in the number you are approximating for x and simplify.
Example: If f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to APPROXIMATE f(1.2).
Point (1,2) Slope(5)
Point-slope: y-2 = 5(x-1)
Plug in: y-2 = 5(1.2-1)
Simplify: y-2= 5(.1)
y= 1+2 = 3
For linerization, you just use the tangent line formula and work the problem like you would a tangent line. You will be given a decimal in the problem also and you will plug that decimal into x after you set up your equation in point slope form.
ReplyDeleteLinerization:
ReplyDeleteTreat the problem just like you would want to find the tangent line. Then they usually give you a decimal, and all you do is plug that in for x in the point slope formula from your tangent line.
you find the tangent line and plug in the decimal they gave your for ya x. how easy was that :)
ReplyDeletelinearization.
ReplyDeleteThe steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
linearization:
ReplyDeletefirst of all key word is approximate :)
like you already didn't know that aimmmmm.
treat the problem like a tangent line, then plug in the decimal you're given for x.
voila!