Critical Numbers and f following g

How to find THE DERIVATIVE

Suppose a driver is on a highway with mile markers.
And also suppose that we would be able to know
the mile position that paired up with the elapsed
time on the highway.
For example:

after 1 hour the mile marker was 111
at 2 hours the marker was 171
at 1.1 hours the marker was 117

Moving into the world of Math:
( 1, 111 )
( 1.1, 117 )
( 2, 171 )

What y value would you assume
went with x = 0 ?
( 0, __?__ )

How about y = 51?
What assumption are you making?
(The speed was a constant 60 mph.)
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Given a function f(x)

Suppose you want to know the INSTANTANEOUS
RATE OF CHANGE in the function at a particular
value of x. Let's say when x = a.
This is called the DERIVATIVE of f(x) @ x=a.
f ' (a) = ?

If f(x) is a LINEAR FUNCTION, then the slope
of the line gives the RATE OF CHANGE of y with
respect to a change in x. This ratio (Delta y)/ (Delta x)
is the constant m in the form
y = m x + b
You can check my blog for more on SLOPE!

In general, the SLOPE OF THE TANGENT
LINE TO THE CURVE at the particular value
of x ( x = a ) gives the INSTANTANEOUS
RATE OF CHANGE at this point.

In CALCULUS we look at the AVERAGE RATE
OF CHANGE to find the INSTANTANEOUS.

The AVERAGE RATE OF CHANGE is found by picking
two points on the curve and finding the slope
of the SECANT LINE that passes thru them.
Suppose we are looking for the SLOPE of the TANGENT
LINE (INSTANTANEOUS RATE OF CHANGE)
at the point on the curve where x=a. Let's call it point P.
Using one point on the curve as P ( a , f(a) )
We choose a point very close to P calling it Q.
Q can and should be allowed to be on either side of P.

After we find the SECANT SLOPE we make point Q
get very close to point P. As this happens we look for
the limit of these SECANT SLOPES.



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