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Grade 12
Mathematics
Online
Classroom
Rules
Find the derivative of a function at a given point
Learning Objectives
Sketch the graph of a function using the graph of its derivative
Understand the relationship between continuity and differentiability
Determine the differentiability of a function at a given point
Estimate the derivative at a given point using tables
f ‘(x) = 9x2
+ 2.
f ‘(1) = 9 + 2 = 11
f ‘(2) = 9(4) + 2 = 38
and
f ‘(3) = 9(9) + 2 = 83.
Note:
Keep in mind that,
the value of the derivative function at a point is the slope of the tangent line at that
point.
Rather than worrying about exact values of f (x),
we only wish to find the general shape of its
graph.
The graph indicates a sharp corner at x = 2, so you might expect
that the derivative does not exist.
To verify this, we investigate the derivative by evaluating one-sided
limits.
For h > 0, note that (2 + h) > 2 and so, f (2 + h) = 2(2 + h).
Likewise, if h < 0, (2 + h) < 2 and so, f (2 + h) =
4.
Figures 2.19a–2.19d show a variety of functions for which f (a) does
not exist.
In each case, convince yourself that the derivative does not exist.
There are many times in applications when it is not possible or practical to compute derivatives symbolically.
This is frequently the case where we have only some data (i.e., a table of values) representing an otherwise
unknown function.
Solution The instantaneous velocity is the limit of the average velocity as the time interval shrinks.
We first compute the average velocities over the shortest intervals given, from 5.9 to 6.0 and from
6.0 to 6.1.
Since these are the best individual estimates available from the data, we could just split the
difference and estimate a velocity of 35.1 ft/s.
However, there is useful information in the rest of the data. Based on the accompanying table, we
can conjecture that the sprinter was reaching a peak speed at about the 6-second mark.
Thus, we might accept the higher estimate of 35.2 ft/s.
We should emphasize that there is not a single correct answer to this question, since the data are
incomplete (i.e., we know the distance only at fixed times, rather than over a continuum of times).
C3L2_The Derivative_G12A.pptx
C3L2_The Derivative_G12A.pptx
C3L2_The Derivative_G12A.pptx
C3L2_The Derivative_G12A.pptx
C3L2_The Derivative_G12A.pptx

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C3L2_The Derivative_G12A.pptx

  • 3.
  • 4. Find the derivative of a function at a given point Learning Objectives Sketch the graph of a function using the graph of its derivative Understand the relationship between continuity and differentiability Determine the differentiability of a function at a given point Estimate the derivative at a given point using tables
  • 5.
  • 6.
  • 7. f ‘(x) = 9x2 + 2. f ‘(1) = 9 + 2 = 11 f ‘(2) = 9(4) + 2 = 38 and f ‘(3) = 9(9) + 2 = 83.
  • 8.
  • 9.
  • 10. Note: Keep in mind that, the value of the derivative function at a point is the slope of the tangent line at that point.
  • 11. Rather than worrying about exact values of f (x), we only wish to find the general shape of its graph.
  • 12.
  • 13.
  • 14.
  • 15. The graph indicates a sharp corner at x = 2, so you might expect that the derivative does not exist. To verify this, we investigate the derivative by evaluating one-sided limits. For h > 0, note that (2 + h) > 2 and so, f (2 + h) = 2(2 + h). Likewise, if h < 0, (2 + h) < 2 and so, f (2 + h) = 4.
  • 16. Figures 2.19a–2.19d show a variety of functions for which f (a) does not exist. In each case, convince yourself that the derivative does not exist.
  • 17. There are many times in applications when it is not possible or practical to compute derivatives symbolically. This is frequently the case where we have only some data (i.e., a table of values) representing an otherwise unknown function.
  • 18. Solution The instantaneous velocity is the limit of the average velocity as the time interval shrinks. We first compute the average velocities over the shortest intervals given, from 5.9 to 6.0 and from 6.0 to 6.1. Since these are the best individual estimates available from the data, we could just split the difference and estimate a velocity of 35.1 ft/s. However, there is useful information in the rest of the data. Based on the accompanying table, we can conjecture that the sprinter was reaching a peak speed at about the 6-second mark. Thus, we might accept the higher estimate of 35.2 ft/s. We should emphasize that there is not a single correct answer to this question, since the data are incomplete (i.e., we know the distance only at fixed times, rather than over a continuum of times).