7.3 and part of 7.5Rates of Change, Slopes, and DerivativesA. What algebra skill you’ll needB. Where average rate of change (ARC) is on a graphC. How to find it algebraically (without a graph)D. How we get the difference quotients definition of ARCE. Where instantaneous rate of change (IRC) is on a graphF. How we get the difference quotient definition of IRCG. How to find the IRCH. What a derivative isI. What Leibniz’s Notation looks likeJ. Word Problems
A. What algebra skills you’ll need• Negative, positive slope• Slope formula• Combining rational expressions
B. Where average rate of change (ARC) is on a graph The average rate of change(ARC) between x = 2 and x = 5 is the SLOPE of that red dotted line. [a secant line] Let’s find it:
C. How to find it algebraically (without a graph)• Well, since the ARC is really a SLOPE, let’s recall the slope formula: rise y2 − y1 f ( x2 ) − f ( x1 ) = = run x2 − x1 x2 − x1• We can use that f(x) notation because we’ll be dealing with functions. Recall that f(x) simply means the y value that corresponds to x.• Let’s apply this to a problem:
" If f ( x) = 3 x + 5, find the average 2rate of change between x = 1 and x = 2."
You try : " If f ( x) = x + 1, find the average 2rate of change between x = 3 and x = 5."
D. How we get the difference quotients definition of ARC• You may or may not have been introduced to the famous “difference quotient” in your algebra class. Here is what it looked like: f ( x + h) − f ( x) h• First, I’ll show you where this came from.• Then , we’ll discover what is so great about it.
Let’s take this curve, and label a fixed point x. Itwould have corresponding y-value labeled f(x). If I jog over a certain distance h on the x-axis. What could I call this new fixed point?
x + h would be that newly created point on the x-axis. Its corresponding y-value would be called f(x + h).
Here is the red, dotted secant line. We need the slope of it.
RISE = RUN =That’s where the difference quotient came from.Now, I’ll show you what is so great about it.
E. Where instantaneous rate of change (IRC) is on a graph• It’s the slope of the tangent line. I’ll draw it:
F. How we get the difference quotient definition of IRC Remember that secant line that was h wide? • It was h wide. Imagine you could make h shrink (“approach zero”).
That would get closer and closer to the slope of the tangent line! The IRC! f ( x + h) − f ( x) IRC = lim h →0 h This will be useful because we can use this formula algebraically (no graphing necessary to find or guess at the IRC.) Let’s see if we can work one…
G. How to find the IRC• Find the instantaneous rate of change of this function at x = 1. f ( x) = x 2
1Find the slope of the tangent line to f ( x) = at x = 2. x
2You try : Find the slope of the tangent line to f ( x) = at x = 3. x
You try : Find the instantaneous rateof change of f ( x) = 2 x at x = 3. 2
Sometimes they ask you whether the slope ofthe tangent line is positive, negative, or zero.
H. What a derivative is• We can evaluate the IRC at any point we want algebraically, but wouldn’t it be easier if we could create a function for the IRC, and then we could plug in any value. It would be more efficient.• The DERIVATIVE is just that. Basically, it is a function for the IRC.• When you see “IRC,” think “Derivative.”
The derivative, noted by f ′( x ) , can be found by f ( x + h) − f ( x)f ′( x ) = lim . h →0 hFind the derivative of f ( x) = x 2 − 7 x + 150.
f ( x + h) − f ( x)You try : f ′( x ) = lim h →0 hFind the derivative of f ( x) = x 2 − 2 x + 41.
I. What Leibniz’s Notation looks like dInstead of writing f ′( x ) , Leibniz wrote f ( x ). dx dyInstead of writing y′, Leibniz wrote . dx
J. Word Problems• First of all, the units of the ARC or IRC is always consistent with it being a rate. [Like, “something” per “something”]• More particularly, the first “something” is the units of the function f(x), and the second “something” is the units of the x.• “something” per “something”
Refining crude oil requires heating and cooling at differentrates. Suppose the temperature of the oil at time x hours is f ( x) = x − 7 x + 150 degrees Fahrenheit. Find the IRC of 2temperature at time x = 6, and interpret it.
You try : Suppose the temperature of the oil at time x hours is f ( x) = x − 2 x + 12 degrees Fahrenheit. Find the IRC of 2temperature at time x = 2, and interpret it.
Use the limit definition to find an EQUATION(y=mx+b) of the tangent line to the graph of f at the given point:f ( x ) = − x ; ( − 1,− 1) 2
The word “differentiable” means…_________________________________
• If it is differentiable at a point, then it is continuous at that point.• BUT• Just because it is continuous at a point doesn’t necessarily mean it is differentiable there.• Example: