Calculus I by Paul Dawkins

By Paul Dawkins

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Extra resources for Calculus I

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In the end, regardless of the angle chosen, we’ll get the same list of solutions, but the value(s) of n that give the solutions will be different depending on our choice. Also, in the above example we put in a little more explanation than we’ll show in the remaining examples in this section to remind you how these work. Example 2 Solve -10 cos ( 3 t ) = 7 on [-2,5]. Solution Okay, let’s first get the inverse cosine portion of this problem taken care of. 3462 è 10 ø Don’t forget that we still need the “3”!

Here is the solution. 0689 for the second angle. As usual for these notes we’ll use the positive one. 0689 + 2p n n = 0, ±1, ±2,K Now, we still need to find the actual values of x that are the solutions. These are found in the same manner as all the problems above. We’ll first add 1 to both sides and then divide by 2. 5345 + p n n = 0, ±1, ±2,K So, in this example we saw an argument that was a little different from those seen previously, but not all that different when it comes to working the problems so don’t get too excited about it.

Since 2 raised to 4 is 16 we get, log 2 16 = 4 because 2 4 = 16 We’ll not do the remaining parts in quite this detail, but they were all worked in this way. [Return to Problems] (b) log 4 16 log 4 16 = 2 because 4 2 = 16 Note the difference the first and second logarithm! The base is important! It can completely change the answer.

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