1. Let $f$ be defined by\(f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\)Calculate \(f^{-1}(0)+f^{-1}(6).\)
2. There are numbers A and B for which\(\frac A{x-1}+\frac B{x+1}=\frac{x+2}{x^2-1}\)for every number \(x\neq\pm1\). Find A-B.
3.The greatest integer function, \(\lfloor x\rfloor\), denotes the largest integer less than or equal to x. For example, \(\lfloor3.5\rfloor=3\), \(\lfloor\pi\rfloor=3\) and \(\lfloor -\pi\rfloor=-4\). Find the sum of the three smallest positive solutions to \(x-\lfloor x\rfloor=\frac1{\lfloor x\rfloor}.\) Express your answer as a mixed number.
thank you in advance
+128474 2) x + 2 A B
______ = ____ + ____
x^2 - 1 x - 1 x + 1
Factor the denominator of the first fraction as ( x -1) (x + 1) and multiply through by this factorization and we get that
x + 2 = A ( x + 1) + B ( x - 1) simplify
1x + 2 = (A + B)x + ( A - B) equate terms and we have that
A + B = 1
A - B = 2
+128474 1)
f-1 (0) means that when we plug an original x value into one of the functions, then it returns 0
Note that in the first function if we put 3 into it for x, we get 3 - x = 3 - 3 = 0
And x = 3 is in the domain of this function ...... so
f-1(0) = 3
f-1(6) means that when we plug an original x value into one of the functions, then it returns 6
Note that if we put -3 into the first function for x, we get that 3 - x = 3 - (-3) = 6
And x = -3 is in the domain of this function, as well....so
f-1(6) =-3
So
f-1(0) + f-1(6) = 3 + (-3) = 0
