Texas Instruments TI-84 Plus Manual page 53

TI-83, TI-83 Plus, TI-84 Plus Guide
Suppose you want to find the slope of the secant line between
the points (0, f(0)) and (2, f(2)). That is, you are finding the
slope of the secant line between the points (a
(a + k, f(a + k)) for a = 1 and k = 1.
Type the expression on the right and then press
with
nDeriv(
Remember that when you are on the home screen, you can recall previous instructions with the
keystrokes
2ND
Recall the last entry and edit the expression so that k changes
from 1 to 0.1. Press
edit the expression so that k changes from 0.1 to 0.01. Press
ENTER .
Repeat the process for k = 0.001 and k = 0.0001. Note how the
results are changing. As k becomes smaller and smaller, the
secant line slope is becoming closer and closer to 1.
•
A logical conclusion is that the tangent line slope is 1. However, realize that we have
just done another type of numerical investigation, not an algebraic proof.
In the table below, the first row lists some values of a, the input of a point of tangency, and
the second row gives the actual slope (to 7 decimal places) of the tangent line at those values.
The algebraic method gives the exact slope of the line tangent to the graph of f at these input
values.
Use your calculator to verify the values in the third through sixth rows that give the slope
of the secant line (to 7 decimal places) between the points (a
for the indicated values of k. Find each secant line slope with
a = input of point of tangency
slope of tangent line = f
slope of secant line, k = 0.1
slope of secant line, k = 0.01
slope of secant line, k = 0.001
slope of secant line, k = 0.0001
You can see that the slope of the secant line is very close to the slope of the tangent line for
small values of k. For this function, the slope of the secant line does a great job of approxi-
mating the slope of the tangent line when k is very small.
Now repeat the process, but do not include k in the instruction. That is, find the secant line
slope by calculating
slope of secant line
Copyright © Houghton Mifflin Company. All rights reserved.
MATH 8 [nDeriv(].
ENTER (ENTRY).
Again recall the last entry, and
ENTER .
'
0.6593805
(a)
0.6595364
0.6593820
0.6593805
0.6593805
nDeriv(Y1, X, a, k).
0.6593805
− −
k)) and
k, f(a
Access
ENTER .
−
k, f(a
nDeriv(Y1, X, a, k).
2.3 5
0.4472136
0.4472360
0.4472138
0.4472136
0.4472136
Did you obtain the following (to 7 decimal places)?
0.4472136
−
k)) and (a + k, f(a + k))
12.82 62.7
0.2792904 0.1262892
0.2792925 0.1262892
0.2792904 0.1262892
0.2792904 0.1262892
0.2792904 0.1262892
0.2792904 0.1262892
53