Verify that f(x)= (1-x^2) - (2+x) can be written as: -x + 2 - 3/(2+x)

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The discussion centers on verifying the equation f(x) = (1-x^2) - (2+x) and whether it can be expressed as -x + 2 - 3/(2+x). Attempts to manipulate the equation include factoring and finding a common denominator, but these efforts have not yielded a solution. A key observation is that the right-hand side is undefined for x = -2, while the left-hand side is defined, indicating they cannot be equal for all x. Further evaluations at specific points, such as x = 0 and x = 1, show differing results for both sides of the equation. Therefore, the two expressions are not equivalent across their domains.
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Homework Statement



verify that f(x)= (1-x^2) - (2+x) can be written as: -x + 2 - 3/(2+x)

Homework Equations





The Attempt at a Solution



i've tried manipulating as 1/(2+x) -x^2/(2+x), multiplying the orig. equation by (2-x)/(2-x), factoring, etc. nothing's worked, and i can't think of another approach. can anyone steer me in the right direction? thank you!
 
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suegee3000 said:

Homework Statement



verify that f(x)= (1-x^2) - (2+x) can be written as: -x + 2 - 3/(2+x)

Homework Equations





The Attempt at a Solution



i've tried manipulating as 1/(2+x) -x^2/(2+x), multiplying the orig. equation by (2-x)/(2-x), factoring, etc. nothing's worked, and i can't think of another approach. can anyone steer me in the right direction? thank you!

So basically you want to show:

(1-x^2) - (2+x) = -x + 2 - \frac{3}{2+x}<br />

I'd be inclined to put the RHS over a common denominator, and then look to see what the next steps would be to distribution things out and start looking for simplifications...
 
On obvious point should be that the right hand side is NOT defined for x= -2 while the left hand side is. They can't possibly be equal! At first I thought that you mean "show they are equal for all x except -2" but then I notice that if x= 0, the left hand side is 1- 2= -1 while the right hand side is 1/2. And if x= 1, the left side is -3 while the right side is 0. Do you see my point?
 
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