stressedmathkid123

Okay, so let's break this down.

Note: != means not equal to. This would be an equal sign with a slash between it in written form. 

1) We have the equation (x-1)/(x-2) = (x-k)/(x+6) 

2) We can identitify the non permissible values, aka the values that make the denominator equal to 0 (we cannot divide by 0, gives us an error). Now, if we set both denominators to equal 0 and solve, we will get the following: x-2 = 0; x != 2 and x+6 = 0; x != -6. This means that x cannot equal -6 or 2

3) Okay, now we know what x cannot equal to. But the question asks what are the non permissible values of k. We can solve for x in this case to create an equation that will allow us to find the non-permissible values of k. 

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1) We start with the equation (x-1)/(x-2) = (x-k)/(x+6). 

2) We can cross multiply which will give us (x+6)(x-1) = (x-2)(x-k) 

3) Foil and combine like terms for both sides. x^2 + 5x - 6 = x^2 - xk - 2x + 2k 

4) Subtract x^2 from both sides. x^2 + 5x - 6 - x^2 = -xk - 2x + 2k

5) We can then isolate the other x^2 on the left side (+ x^2 - x^2). Then we get 5x - 6 = -xk - 2x + 2k 

6) We can add xk to both sides. Then we get 5x - 6 + xk = -2x + 2k 

7) Then we can add 2x to both sides. We then get 5x + 2x - 6 + xk  = 2k

8) Combine 5x and 2x. We get 7x - 6 + xk = 2k

9) Add 6 to both sides. We get 7x + xk = 2k + 6

10) Factor 7x + xk. We get x(k+7) = 2k + 6. Or you could arrange it like x(7+k) = 2k + 6. Whatever works for you. 

11) We can now comfortably solve for x. Divide both sides by (k+7). We get x = 2k+6/k+7.

12) Now we can find the non permissible values of k. Set the denominator to equal to 0. k + 7  = 0 and solve. Subtract 7 from both sides and we get k = -7. 

Solution (word it however you want but these would work): 

1) k != -7 

2) The non permissible values of k for the equation (x-1)/(x-2) = (x-k)(x+6) is k != -7

3) The equation (x-1)/(x-2) = (x-k)(x+6) has no solution for k when k = -7

Yeah, sure! 

1) Since we know that f(x+T) = f(x)

2) and f(x) = pi(x)^2 - cos (x), we can plug in f(x+T) for anywhere there is an f(x) for our manipulated side (left). 

Since this is a proof, there are going to be two sides. 

3) Then we get pi(x+T)^2 - cos(x+T) = pi(x)^2 - cos(x). Our left side going to be manipulated while our right side is going to be our "constant" with the original expression pi(x)^2 - cos(x). 

4) We can expand the expression which gives us pi(x^2+2(x)T+T^2) - cos(x+T) = pi(x)^2 - cos(x)

5) Then we can foil the pi, which gives us pi(x)^2+ 2pi(x)T + pi(T)^2 - cos(x+T) = pi(x)^2 - cos(x)

6) We can then subtract pi(x)^2 from both sides which will leave us with 2pi(T)x + pi(T)^2 - cos(x+T) = -cos(x)

I might've switched the T and the x in some places (such as 2pi(x)T) compared from my previous answer (where it was 2pi(T)x), but it doesn't affect the actual proof. We can still test our proof just fine in this case. Where you arrange the x and T's is mostly whatever is easier to understand. It's multiplication, so it's fine to flip them generally. It's like 4x5 is the same as 5x4. 

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I already explained this earlier but here's a more detailed explanation for the contradiction of the proof. 

7) Testing the proof: 

Now since we are given the hint that f(x) is a periodic function of period T when T > 0 for all x in the domain of f. We can test T = 0. Plug in 0 where ever there's a T and we get -cos(x) = -cos(x). This isn't what we are looking for because this shows that f(x) is a periodic function when T = 0 and is also a valid proof. Not a contradiction. 

So since we need to show that f(x) is not a periodic function, we need to make the proof impossible. Since we tried T = 0, we can also test x = 0. Plug in 0 where ever there's an x, and we get pi(T)^2 - cos(T) = -1. This shows that f(x) is not a periodic function when x = 0 and is not a valid proof. It's a contradiction. 

Solution: The function is not periodic when x = 0. Unless when T = 0, the function is periodic. 

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I actually forgot to show the expanding and foiling part on my actual assignment since I did it on a separate piece of paper. So it looked like I skipped two steps lol.  

I actually figured it out. My phrasing is probably a bit off but it makes sense (to me at least) 

We end up with 2pi(x)T + pi(T)^2 - cos(x+T) = -cos(x) after some simplifying, and distributing/expanding.  

If we let x = 0 

Then we end up with pi(T)^2 - cos(T) = -1. This makes our proof impossible. It's a contradiction because when we let x = 0, the function is not periodic. 

UNLESS, if we let T = 0, then our proof is valid. We end up with -cos(x) = -cos(x) which then makes our function periodic. However, this isn't what we are looking for, as we aren't trying to find what makes the function periodic.