Topics:
•More on limits which involve infinity:
How to tell if a limit is $\infty$ or $-\infty$ without using the graph
Here are my handwritten (sloppy) notes from class:
Limits at infinity for polynomial and rational functions
• Derivatives: introduction to the idea
There are two ways of looking at what we are trying to achieve with the derivative:
One way : geometrically, thinking about the graph of a function f(x), we are trying to find something like a slope for functions that are not linear. Why would we want to do this? Because the slope of a line represents the rate at which y (or f(x)) is changing as x increases. If you think of x as representing time, that may be helpful. If the slope is 1.5, that means that if x increases by 1, the value of the function will increase by 1.5.
A line has the same slope (rate of change) at every point, but we can’t expect that to be true for a function whose graph is not a line. But hopefully we can find something like a slope at a single point on the graph. It makes sense that this should be the slope of the tangent line at that point, if you think about it. (That’s if there is a tangent line!)
Another way: Thinking about the function itself, the rate of change of the function may represent something we are very interested in. For example, if we have a position function as in the example, the rate of change of the function is the velocity. (Also, for a velocity function, the rate of change will give the acceleration!) We would like to be able to define velocity for position functions which are not linear, meaning that the velocity is not the same at every point.
Example: I basically used the same example that is used at the start of Chapter 2. The position function represents the height of an object which is dropped from a height of 150 feet, t seconds after it is dropped. The function is
$f(t) = -16t^{2} + 150$
For any function (not just this one), we define the average rate of change (here it’s the average velocity) on an interval $[a,b]$ to be the slope of the secant line that passes through the two points $\left(a, f(a)\right)$ and $\left(b, f(b)\right)$ as
$\frac{f(b) – f(a)}{b-a}$
In class we computed the average rate of change for our position function on the intervals [2, 3] and [2, 2.5] . On your own you can compute the average rate of change on the intervals [2, 2.1] and [2, 2.01] and more if you like… or you can look at the textbook!
Notice that the right-hand endpoints of these intervals are getting closer and closer to 2. The idea is that we want to look at what happens to these slopes (average velocities) as the second point approaches (2, 86). This will mean taking a limit.
Looking at the results from those average velocities, it appears that they do approach a limit and the limit is -64 (or very close to that).
Now we will rewrite the formula for the average velocity a bit. Think of taking the interval [2, 2+h] where h is some (small) number, possibly negative. So we would think of our intervals as [2, 2+1], [2, 2+0.5], [2, 2+0.1], etc. Then the slope of the secant line (the average velocity) would be written as
$\frac{f(2+h) – f(2)}{(2+h)-2} = \frac{f(2+h) – f(2)}{h}$
And we will define the instantaneous velocity at t=2 to be the limit of this as h goes to 0:
$\displaystyle \lim_{h\rightarrow 0} \frac{f(2+h) – f(2)}{h}$
Let’s compute that limit. We already know that $f(2) = 86$. To find $f(2+h)$:
\begin{align*}
f(2+h) = -16(2+h)^{2} + 150 & = -16(4 + 4h + h^{2}) + 150\\
& = -64 -64h -16h^{2} + 150\\
& = 86 -64h – 16h^{2}
\end{align*}
So $f(2+h) – f(2) = 86 -64h – 16h^{2} -86 = -64h – 16h^{2}$
\begin{align*}
\displaystyle \lim_{h\rightarrow 0} \frac{f(2+h) – f(2)}{h} & = \lim_{h\rightarrow 0} \frac{-64h – 16h^{2}}{h}\\
& =\lim_{h\rightarrow 0} \frac{h(-64 – 16h)}{h}\\
& = \lim_{h\rightarrow 0}(-64 – 16h)\\
& = -64 – 16(0) = -64
\end{align*}
So the slope of the tangent line is -64, as we predicted.
