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https://youtu.be/mcckcS8z9sU

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This video is about AP Calculus BC L'Hospital's Rule

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Antiderivatives and indefinite integrals | AP Calculus AB | Khan Academy

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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
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If the nth term is given by Xn then the next approximation is given by the formula listed underneath.
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Newton's Method / Newton-Raphson Method 

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Newton's Method / Newton-Raphson Method 

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It is a method for finding successively better approximations to the zeros of a real-valued function.
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If f(x) has a local maximum or minimum at c, and if f'(c) exists, then f'(c) = 0
AASG

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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

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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

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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

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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]
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When checking for maximums and minimums, don't forget to check the endpoints.
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What we KNOW:
dx/dt = 2
h = 25ft
x^2 + y^2 = h^2
We are all beautiful.

What we NEED:
dy/dt
AASG

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Strategy for solving related rates problems

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Strategy for solving related rates problems

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Identify what you KNOW and what you NEED
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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

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I am a quirky relatable teen just looking to explore my identity and grow as a person! :^) <33333

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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