Module 8 - Answers |
Lesson 1 |
Answer 1 |
8.1.1 The slopes of the secant lines appear to converge to 1, so the slope of the tangent line is probably 1. |
Answer 2 |
8.1.2 Approximately [-2.12, 4.12] x [-1, 2]. |
Lesson 2 |
Answer 1 |
8.2.1 The function is not differentiable at x = 0 because it is not locally linear there. The function is differentiable at x = 0 because it is locally linear there. |
Lesson 3 |
Answer 1 |
8.3.1 x = -2: dy/dx = -4; x = -1: dy/dx = -2; x = 0: dy/dx = 0; x = 1: dy/dx = 2; x = 2: dy/dx = 4 |
Answer 2 |
8.3.2 The derivative is 2.4, which represents the slope of the tangent line to f(x) = x^{2} at x = 1.2. |
Answer 3 |
8.3.2
The derivative of f(x) = 3x^{2} + 4x is the function g(x) = 6x + 4. |
Self Test |
Answer 1 |
Answer 2 |
Answer 3 |
Answer 4 |
Answer 5 |
The tangent line is y = 6(x – 1) + 2, which is shown with the function in a [–3, 3.32] x [–10, 10] window.
The graph is shown after zooming in three times centered on (1, 2) with zoom factors xFact = yFact = 10. |
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