- Relating to inverses, I believe that I understood the concept of one-to-one rather clearly. I learned that in order for a function and its inverse to be "one-to-one", the graph of the function must pass the horizontal line test, which means that if a horizontal line is passed over the graph of a function, the horizontal line intersects the graph at one point at a time. This horizontal line test determines if the inverse exists as a function. If the graph of the function passes the horizontal line test, then the function is "one-to-one".
- I understand that the inverses of a parent function are symmetrical to the parent function itself. If you fold the paper between both lines, such as folding the paper of the function f(x)=3 and f^-1(x) at the line y=x, the lines will lay right on top of each other.
- I also understand how to find the inverse (f^-1(x)) of a function (f(x)) and how to verify that (f o f^-1)(x)=(f^-1 o f)(x) = x. When you multiply the function by its inverse, you should get "x", and vise versa. In order to demonstrate how to find the inverse of a function, I will use the function f(x)=x^2 (x< (or equal to) 0). This function is also equal to y=x^2. First you change x to y, and y to x in the equation. You get x=y^2. The goal is to leave y alone, so you take the square root of both sides of the equation and end up getting sq root of x=y, also known as f^-1(x)=sq root of x. --- To verify that it is the inverse, you plug in sq root of x into the original function and get (f(x)=sq root of x^2). When you solve it by cancelling out the sq root and the ^2, you get f(x)=x. Also if you input x^2 into the inverse function, you get (f^-1(x)=sq root of x^2) also, which is equal to f^-1(x)=x. This proves that f^-1(x)=sq root of x.
- Today in class (Thursday), as we were solving logarithm equations for x, I believe that I understood how to solve most of them easily. For example, when solving the equation "7-3e^-x=2, when I got to the point of "e^-x=5/3", I knew that I was supposed to take the natural log of both sides so "e" could cancel out and so I could be left with "-x=ln5/3"--> "x= -ln5/3". Even though sometimes I may get confused or forget how to do a step, I know that with a little more practice, I will grasp the concept more fully.
2. Write about what you did NOT understand completely.
- Relating to Logarithms, I can honestly say that I am not so sure how to graph them. I know that their inverses are functions, but I still cannot grasp the concept of graphing them without using a graphing calculator, unless I input the values for x to get the y output. I do not know how to get the inverse function of a logarithm function that is complicated. Sometimes the log functions are kind of complex and.. yeah I need help. :D
- Even though I understand, in general, how to solve the logarithm functions for x, most of the problems on the homework C2 were a little more tricky and I did not understand them. For example, number 35 (e^x+e^-x=3) may look simple at first, but once I tried to solve it, my answer looked nothing like the one in the back of the book. This was "mind-boggling"! I tried to take the ln of both sides to cancel out the e's but I ended up getting "x-x=ln3" --> "0=ln3". Im not sure how to solve this or anything in relation to that...
your's is long. i wish i could help you on these problems but i myself do not understand how to graph this. what i did was plug in numbers in the calculator lol. and another way i did was find the inverse of the function. and basically the log is just the other graph but symmetrical.. i sound confusing but here.. remember how the inverse of f(x)^2 is log base2(x)and the symmetry line is y=x. then we do the shifting and stuff from there.
ReplyDeleteI dont quite get it myself lol.
I had problems with the whole homework c2 as well. but after i heard the davy crockett story i learned more about log. and it helped me on some of the problems from hw c2.
still i didnt understand 35,36,38.
:)
i do not know much about it. but i believe the equation was f(x)= 2^x which the inverse was f(x) = log base2 x. graphing the equation would need the original equation. idk it is confusing
ReplyDeletegraphing logs without a graphing calculator is possible, but i myself can't do it without plugging in 1 point. if its a normal log function (no transformations), you just need to figure out what x is when y is 2 (in other words, the base to the second power is what). without transformations, (1,0) should always be a point. once you have have (1,0) and (x,2) plugged in, you can pretty much figure out what the rest should look like (if it sounds funny, try it, it should work out just fine, because that's what i do). if it's a log function with transformations...yea, i have no trick to those, i just try to picture the original log function and take the transformation process one step at a time.
ReplyDelete#35: haha, i also tried to make it part of a natural log and also ended up getting 0=ln 3. after a couple tries, i actually figured out. simplify e^-x (make it 1/e^x) and after some manipulation, you end up with e^2x-3e^x+1=0. then just solve for e^x by completing the square (you treat e^x as the variable x in a quadratic equation). lastly, you make everything part of a natural log to cancel out the e and you have your answer (this is the general outline, i leave you to figure out the numerous gaps in between). yes, it doesn't at all sound pretty, but that's the way it's done (#36 in C2 works in a similar matter in case you had difficulties with that too).
Not sure about graphing Log functions but, I believe you just make a chart of inputs and insert it into the Equation. First using 0 as a point since all functions at least go through (1,0). then using other points that make sense.
ReplyDeleteThats all I know lol. Sorry :D
Forget my above comment.
ReplyDelete-I'll show you how to get up to the point where Jesus got e^2x-3e^x+1=0. Then I'll show you how to complete the square from there to get the answer.
First you have to change the given equation a bit. Do the reciprocal of e ^ -x. Then find the common denominator (leave the three on the right of the equal sign). After that, multiply by e^x to both sides (you'll know why when you are solving it). Thenn, you should have e^2x+1 = 3e^x. When you get to that point, change 3e^x in terms of e. SO you should now have: e^2x+1 = e^3e^x. Next, divide by e^3e^x to both sides. Now you should have e^2x+1/e^3e^x = 0. Since you have the same base for the quotients, you subtract and get: e^2x-3e^x+1= 0. Does that look familiar?
Ok now we complete the square, as Jesus said. (This part could be the hardest part to understand for some).
-first, understand that e^x will be like "x"
when you complete the square... you should have x^2 - 3x + 1 = 0. Now do the quadratic formula.
After the quadratic formula you should have:
3 +/- (square root) 5 / 2. <-- Exactly like the back of the book [except for the ln].
After that "you make everything part of a natural log to cancel out the e and you have your answer"
Keep in mind it is just as hard to explain it as it is to do the problem. Haha
heyy did you solve 36 i got stuckk
ReplyDeleteI had the same problems as you!!
ReplyDeletei also do not know how to graph logarithmic functions or how to solve #35. Sadly, i think that means i cannot help you! All i know is that a graphing calculator is definitely useful when graphing logarithmic functions!