Expert: Scotto - 10/19/2009

Question
I just don't know where to start on these questions just some advanced questions that I didn't have to do but I want to know how to solve cause they might be on a test. Please reply as soon as you can. Hopefully you can find the answers Thanks a lot.

1)Find a number δ such that if | x - 2 | < δ, then | 2x - 4 | < ε, where ε = 0.1. (Round all answers to four decimal places.)


(b) Repeat part (a) with ε = 0.01

2)How do you find the inverse of y= 1 + x^2 + tan((pi/2)*x))

3)Find all real solutions to an equation how do i do this? Whats real solutions if its like x+5/x-2 = 9/x+2 + 28/x^2-4

Thanks so much really appreciate it.


Answer
1) If |x-2|<δ, then |2x-4|<ε, ε = 0.01.
Looking at |x-2| and |2x-4|, it can be seen that |2x-4| = |2(x-2)|, so the 2nd is twice the 1st.
If ε = 0.01, δ is the is half of ε, so δ is 0.05.

b) If ε = 0.01, then δ = ε/2, so δ = 0.005.

2) I know of know way to find the inverse directly.
However, if x is close to 1, then we approximately have y = 1 + x² + x, and there is no solution.
If we take y = x² + ex + 1 where e = tan(x), the solution realm is for e>=2, for the quadratic has a √(b²-4ac) = √(e²-4) in it.

If you have the equation (x+5)/(x-2) = 9/(x+2) + 28/(x²-4),
note that x²-4 = (x+2)(x-2).  Multiplying both sides by x²-4 gives us (x+5)(x+2) = 9(x-2) + 28.
Multiplied out, this is x²+7x+10 = 9x-18 + 28.
Combining terms gives us x² - 2x = 0, so we have x(x-2) = 0, so x=0 or x=2.
It looks like we have two answers, however, one is not a solution.

Substituting these back in to the original equation tells us that x=2 is not valid,
for we divide by x-2, and that is division by 0.  If we put in 0, though, we get
5/(-2) = 9/2 + 28/(-4), which multiplies out to -5/2 = 9/2 - 7, or -5/2 = 9/2 - 14/2,
which is true.