Post #36

Tangent Lines:
To find a tangent line, you need a slope and a point.
The problem usually gives an equation and a x-value.
To find the slope, take the derivative of the equation and plug in the x-value given.
To find the point, use the x-value and plug it into the original equation to find the y-value.
(x,y) will be your point.
From there, put the point and slope in point-slope form, which is y-y1 = m (slope) (x-x1)

Example: Find the line tangent to x^3 - 2x^2 + 4x at x=1
Take derivative and plug in x: 3x^2 - 4x +4
3(1)^2 - 4(1) +4 = 3
The slope is m =3
Plug in x to find y: (1)^3 - 2(1)^2 +4(1) = 3
Point slope form: y-3 = 3 (x-1)
The answer choices may have the equation written in a different form.
Another option is solved for y: y-3 = 3x -3 = y = 3x

You can also be asked to just find the slope of the line tangent to a center graph. In this case the answer would be 3.

Normal lines:
The equation of a normal line is solve by using the same steps as the equation of the tangent line except instead of using the slope, you have to use the perpendicular slope, which is the negative reciprocal of the slope.
Using the same example:
Take derivative: 3x^2 - 4x +4
Plug in: 3(1)^2 -4(1) +4 = 3
Now take the negative reciprocal of the slope: -1/3
Your point will be the same as it was for the tangent line (1,3)
Now you plug into point slope using the negative reciprocal slope
y-3 = -1/3 (x-1)
Again, this problem may be written differently such as 3y-9 = -1 (x-1)
3y - 9 = -x +1
x+3y = 10

L'Hopital's Rule:
This is used to find the limit when it originally equals 0/0. The mistake is often made by saying the limit is nonexistent, but L'hopital's rule states the take the derivative of the top and the derivative of the bottom then plug in the number given to find the limit. If it still does not work, you have to continue to take the derivative of each separately until it does.
Example: Lim as y->0 sin y / y
Sin (0) / 0 = 0/0 so L'hopital's rule can be applied
the derivative of sin y is cos y
the derivative of y is 1
now it is the lim y ->0 cos y/ 1
cos 1 = 1
1/1 = 1
The limit is 1.

I can use help on finding limits as they approach the left or the right only because I seem to have forgotten how to do those.