Add any questions/concerns over the 5.3 and 5.4 assignments here and I will try to respond in a timely manner. Feel free to post questions or comments about those posted here as well. Or if there are any I haven't addressed that you have questions about.
1st hour's questions are in gray type, my responses are highlighted
Questions from first hour:
Key themes seem to be making sure you remember how and when to use u-substitution, and remembering how to do some of the applications that we have done in the past such as finding extrema, area, particular solutions, etc.
If you have questions on these past topics here are links to videos that may help:
finding area - shout out to Calc/Phys at around the 3:20 mark! :-)
particular solutions - didn't find a video for this one, they key is that "find the solution" means work backward from a derivative to find the original function, and "particular" means find the C value rather than just leaving it as C.
p. 340 #77-80 an example of one
a.) First to find the domains we look at the functions and what possible x values we can use. For f(x) we can't take the square root of a negative so the domain is when x-4=0 or is greater than 0. That would be when x=4 or is greater than 4. For f^-1(x) the domain is set by the problem to be x greater than or equal to 0.
b.) For ranges we think of what possible answers we can get. For f(x) we can get 0 or any positive number as an answer to a square root. So our range is y greater than or equal to 0. For f^-1(x) the range would be y greater than or equal to 4 because no matter what values I put in for x the smallest answer we can get is 4.
c.) Do on your calculator
d.) Find f'(5) and (f^-1)'(1) when you do this you get .5(5-4)^-.5 =.5 and 2(1)=2 . 2 and .5 are reciprocals of each other so it checks out.
p. 347-350 #45, 65, 67, 71, 99, 105, 109, [110, 111, 112, 117, 118] examples of a few please (work for a sampling of these will be posted below)
where does the negative 2x come from on 110
110. The negative 2x in the answer comes from the middle term when we foil this. We would get -2e^x*e^-x and when we multiply like bases we add the exponents giving us -2 e^0. Since anything to the 0 power is 1 this is just -2. When we integrate -2 we get -2x.