This is Beautiful. You're Beautiful. AP Cal BC Review is Beautiful. by Aamirzanus

First Derivative Test

aaawesome117 — 2016-12-05 15:37:57 UTC

You can use the first derivative to find maximums and minimums
if f'(x) = 0
then x is a critical point
If the slope goes from positive to negative across the point, it is a max. If the slope goes from negative to positive, it is a min.
AASG

aaawesome117 — 2016-12-05 16:01:21 UTC

Second Derivative Test

aaawesome117 — 2016-12-06 15:08:52 UTC

You can use the second derivative to find maximums and minimums as well.
if f''(x) > 0
x is a minimum
if f''(x) < 0
x is a maximum
if f''(x) = 0
you must use the first derivative test
AASG

Strategy for solving related rates problems

aaawesome117 — 2016-12-06 15:20:46 UTC

Identify what you KNOW and what you NEED
AASG

Example:

aaawesome117 — 2016-12-06 15:23:23 UTC

What we KNOW:
dx/dt = 2
h = 25ft
x^2 + y^2 = h^2
We are all beautiful.

What we NEED:
dy/dt
AASG

Remember...

aaawesome117 — 2016-12-06 15:31:24 UTC

When checking for maximums and minimums, don't forget to check the endpoints.
AASG

The Extreme Value Theorem

aaawesome117 — 2016-12-06 15:45:41 UTC

If f(x) is continuous on closed interval [a,b], then f(x) attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers c and d in [a,b]
AASG

The Mean Value Theorem

aaawesome117 — 2016-12-07 23:26:49 UTC

If f(x) is a differentiable function on the interval [a,b], then there exists a number c between a and b such that
f'(c) = (f(b) - f(a))/(b - a)
AASG

Determining Concavity

aaawesome117 — 2016-12-07 23:30:04 UTC

a) If f''(x) > 0 for all x in interval I, then the graph of f(x) is concave up (like a bowl) on I.
b) If f''(x) < 0 for all x in interval I, then the graph of f(x) is concave down (like an upside down bowl) on I.
AASG

Finding Maximums and Minimums

aaawesome117 — 2016-12-07 23:39:17 UTC

We want to find the absolute maximum and minimum values of y=x^2 on the interval [1,2].

First, take the derivative and set to zero.
y'=2x
0=2x
Critical point at x = 0
Then find the value of y at x = 0
y = (0)^2 = 0
Check the endpoint values as well
y = (1)^2 = 1
y = (2)^2 = 4
The smallest value (0) is the absolute minimum, the largest value (4) is the absolute maximum
AASG

Fermat's Theorem

aaawesome117 — 2016-12-07 23:59:46 UTC

If f(x) has a local maximum or minimum at c, and if f'(c) exists, then f'(c) = 0
AASG

aamirz18 — 2016-12-08 23:08:51 UTC

aamirz18 — 2016-12-09 00:15:04 UTC

Newton's Method / Newton-Raphson Method

aamirz18 — 2016-12-09 00:41:45 UTC

It is a method for finding successively better approximations to the zeros of a real-valued function.
ZA

The Next Term in the Approximation

aamirz18 — 2016-12-09 00:46:34 UTC

If the nth term is given by Xn then the next approximation is given by the formula listed underneath.
ZA

Remember...

aamirz18 — 2016-12-09 00:51:48 UTC

In Some Cases X2 is a worse approximation than X1. This can occur if f'(x1) is close to 0. It could even happen that an approximation falls outside the domain of f. In these cases Newton's Method fails and a better initial approximation (x1) should be chosen
ZA

Newton's Method Video

aamirz18 — 2016-12-09 01:01:22 UTC

Antiderivatives

aamirz18 — 2016-12-09 01:14:48 UTC

Theorem #1 for Antiderivatives

aamirz18 — 2016-12-09 01:26:37 UTC

aamirz18 — 2016-12-09 01:35:48 UTC

Antiderivative Video

aamirz18 — 2016-12-09 01:38:52 UTC

aamirz18 — 2016-12-09 05:11:02 UTC

Newton's Method

aaawesome117 — 2016-12-09 14:45:04 UTC

https://youtu.be/mcckcS8z9sU