3) f(x) = 1/(x²-1) 7) a) First factorize each of the quadratic polynomials that are in the expression. (I find integer roots by trying out the factors of the constant term): x²+6x+5 = (x+1)(x+5) x²+7x+12 = (x+3)(x+4) x²+2x-8 = (x-2)(x+4) x²-25 = (x-5)(x+5) x²-x-2 = (x+1)(x-2) x²-2x-15 = (x+3)(x-5) Since division by zero is not allowed, we note that the expression is undefined for: x=-3, x=-4,x=5 or x=-5. Now you can also rewrite the big quotient as a multiplication of the big numerator with the inverse of the big denominator.
(x+1)(x+5) / (x+3)(x+4) * (x-2)(x+4)/(x-5)(x+5) * (x+3)(x-5) / (x+1)(x-2) And you can also rewrite the multiplication of several quotients as the multiplication of all the numerators divided by the multiplication of all the denominators. So you get: (x+1)(x+5)(x-2)(x+4)(x+3)(x-5) / (x+3)(x+4)(x-5)(x+5)(x+1)(x-2) Simplify by removing factors common to the numerator and denominator, and you find that there is nothing left (1 / 1), so: the expression is undefined when x is in {-5,-4,-3,5} as stated earlier and 1 otherwise.
3) y= 1/(x^2 -1) 7) a) ans 1 b) (1/(2x-3))(((x+1)/(x-2))-((x-3)/(x+1))... (1/(2x-3))((7x-5)/(x-2)(x+1)) (7x-5)/((2x-3)(x-2)(x+1)).
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