Calculus I by Paul Dawkins

By Paul Dawkins

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So, if we can remember these rules we will be able to determine the remaining angle in [ 0, 2p ] that also works for each solution. aspx Calculus I As a final quick topic let’s note that it will, on occasion, be useful to remember the decimal representations of some basic angles. 5708. This will be of great help when we go to determine the remaining angles So, once again, we can’t stress enough that calculators are great tools that can be of tremendous help to us, but it you don’t understand how they work you will often get the answers to problems wrong.

Remember that all this says is that we start at then rotate around in the 6 6 counter-clockwise direction (n is positive) or clockwise direction (n is negative) for n complete rotations. The same thing can be done for the second solution. So, all together the complete solution to this problem is p + 2p n , n = 0, ± 1, ± 2, ± 3,K 6 11p + 2p n , n = 0, ± 1, ± 2, ± 3,K 6 p As a final thought, notice that we can get - by using n = -1 in the second solution. 6 Now, in a calculus class this is not a typical trig equation that we’ll be asked to solve.

2. f ( x ) ¹ 0 . An exponential function will never be zero. 3. f ( x ) > 0 . An exponential function is always positive. 4. The previous two properties can be summarized by saying that the range of an exponential function is ( 0, ¥ ) . 5. The domain of an exponential function is ( -¥, ¥ ) . In other words, you can plug every x into an exponential function. 6. If 0 < b < 1 then, a. f ( x ) ® 0 as x ® ¥ b. f ( x ) ® ¥ as x ® -¥ 7. If b > 1 then, a. f ( x ) ® ¥ as x ® ¥ b. f ( x ) ® 0 as x ® -¥ These will all be very useful properties to recall at times as we move throughout this course (and later Calculus courses for that matter…).

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