Now, the MVT (Mean Value Theorem) has 3 basic premises that apply.
1. The Increasing/Decreasing Rule
So this rule basically states that, and I quote, "If, over the course of the interval, the secant line's slope is positive, then the tangent line's slope will also be positive."
In literal terms, this means that when the function is increasing, the secant line, and therefore the tangent line's slopes will be positive.
What this means to us is that functions increase when f'(x) > 0 and they decrease when f'(x) <0
The second rule is that a line with a derivative of zero is constant, so x=5 is just a horizontal line.
The third rule of the MVT is that "3. Functions with the same derivative only differ by a constant." Therefore, if f'(x) = g'(x), then f(x) = g(x) + constant
Here is an example for you: we have both f'(x) and g'(x) equaling 2x
You know that f(x) must = x^2 and that g(x) must also = x^2. However, g(x) could also equal x^2 + 121. That would still result in a derivative of 2x, the same as for f'(x), proving that functions with identical derivatives only differ by a constant.
Now that we have the MVT explained, we can move onto the First Derivative Test. What this rule states is that "When the derivative goes from positive to negative, you have a max. When the derivative goes from negative to positive, you have a min. And finally, if the derivative doesn't change signs, then there is no max or min." Pretty self-explanatory.
In this graph right here, you can see that, at -2, the derivative is negative and then switches to positive. This shows that there is a relative minimum at that point. At -1, the derivative is positive and goes to negative, leading there to be a maximum at that point.
To put all of this together, there is a good example that Mr. Marchetti showed to us on the board. You have the function y= x^3 - 4x. You need to find the maximum(s) and minimum(s). Ready go. So to solve this, you need to first find the equation's derivative. That is y' = 3x^2 - 4. You make that equation equal to 0 in order to determine where the critical points are in this specific function.
0 = 3x^ - 4
4 = 3x^2
4/3 = x^2
+/- (4/3)^.5 = x (this would be a lot less cluttered if I could insert stuff...)
+/- 2 / (3)^.5 = x
Now that you have the critical points of the function (where the derivative changes signs, which results in a max or a min as stated in the First Derivative Test), you can figure out how the function's derivative behaves in each interval. The intervals that you will have in this instance will be (- infinity, -2/(3)^.5), (-2/(3)^.5, 2/(3)^.5), and (2/(3)^.5, infinity). What you do is pick any value within each interval and plug that into the derivative equation that you found earlier (3x^2 - 4) to find how the derivative in that interval behaves. Remember, the endpoints of each interval are 0's, so the derivative crosses the x axis at those points and changes signs.
(- infinity, -2/(3)^.5)...: 3(-5)^2 - 4 = 71, positive in this interval
(-2/(3)^.5, 2/(3)^.5)...: 3(0)^2 - 4 = -4, negative in this interval
(2/(3)^.5, infinity)...: 3(5)^2 - 4 = 71, positive in this interval
now, to find which of these critical points is the maximum and which is the minimum, you refer again to the First Derivative Test. It states that when the derivative goes from positive to negative, you have a maximum. Therefore, at the point -2/(3)^.5), you have a maximum. Also, because the First Derivative Test states that when the derivative goes from negative to positive, you have a minimum, you know that your minimum is at 2/(3)^.5.
That is all that was talked about in class on Wednesday. Again, we'll see if the inserting pictures problem resolves itself or not.
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