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Enter the function y1 = abs(ln(x)) on the "y =" screen, and graph it in a standard window. |
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The screen shot at right shows what you (probably) get. If you are alert, you are asking yourself, "The graph in the first quadrant looks ok, but what's going on in the second quadrant? The domain of y = ln(x) is x > 0, so y = |ln (x)| should also be undefined when x = 0, right?" Yes, you are right - we don't expect to see the part of the graph in II. Read on. |
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Set your calculator's angle |
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This time, you get the "correct" graph. (If you got this graph initially, switch to radian mode to see the graph above.) Since we usually (always?) work in radian mode in AP Calculus, this is a little disturbing. What's going on? |
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Go back to the |
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Now, some theory. Do you remember from pre-algebra (the "good old days"!) the equation involving the polar form of a complex number: If you let This is a form of the Euler equation This is the result that the TI-89 gives you if you evaluate ln(-3). |
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OK, so what about |ln(-3)|? Well, in "complex number land", the "| |" symbol means "magnitude", not "absolute value"! Remember graphing complex numbers on the complex (Argand) plane? The magnitude of the complex number ln(-3) is: which is a real number! (approximately 3.328)! |
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The screen shot at right shows that the calculator is
indeed plotting the point (-3, 3.32815)! So, the calculator
isn't wrong - it is just applying a little fancier math than
we really want to see in this case. Remember that you can
shut this off by switching to DEGREE is only valid if theta is in radians. |
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