Why hello there, it's Kristin_R here, one of the few non-AP students left in the calculus class.
Today in class we learned about the first and second derivative tests.
The first derivative test:
c is a critical point (meaning the inputs where the derivative is zero or undefined, which is a horizontal tangent line) somewhere over the interval (a,b).
If f'(x) > 0 over the interval (a,c) and f'(x) < 0 for all values of x in the interval (c, b), then that function has a maximum over that interval.
If f'(x) ,0 over the interval (a,c) and f'(x) . 0 over the interval (c,b), then that function has a minimum over that interval.
Here's an example...
f(x) = x^2-8x=4
therefore, f'(x) = 2x-8
If we let the derivative equal zero, the root is at x=4.
When we plug four back into the original function, we get a y-value of -12. .
This means that the root is at (4, -12).
- ----- -12 ------ - --------4-------- +
By the first derivative test, x=4 is a minimum because to the left of 4, f is decreasing , and to the right of four, f is increasing.
The second derivative test:
When the parent function is concave up (valley), f'' is postive.
When the parent function is concave down (hill), f'' is negative.
When f' has a root, the parent function has a point of inflection, which is where the function changes concavity.
So, the first derivative test shows where the function is increasing or decreasing, as well as maximum or minimum extrema.
The second derivative test shows us the concavity of the graph.
"Just finding a root of your second derivative is not enough! You have to check your intervals!"
Just a quick little note - in grade eleven, you should have been taught about how to graph certain expontential functions. Remember that the degree of the given function is the same amount of possible roots in the graph.
So, thats it, thats the whole ball of wax...y'all dig?
Homework is up to and including page 64 in the white and orange book, as well as up to and including page 42 in the blue book.
The scribe for next class is....John. D!
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