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1. The graph of the equation $y = \frac{x}{x^3 + Ax^2 + Bx + C}$, where $A,B,C$ are integers, is shown below. Find $A + B + C$.

2. \[f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\] Calculate $f^{-1}(0)+f^{-1}(6)$.

Guest Apr 26, 2018
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1) There is one characteristic about the given graph that will allow one to determine the equation of the original function: the vertical asymptotes. Vertical asymptotes are vertical lines that a function approaches but never reaches or crosses. I like to call this type of asymptote an absolute asymptote since the function will never cross this. Horizontal asymptotes are not absolute; there are times where the function crosses a horizontal asymptote. In fact, there is a horizontal asymptote located at \(y=0\) here, but that is beyond the scope of this particular problem. There are three vertical asymptotes on the graph. They are located at \(x=-2,x=1,\text{ and }x=2\). The function never reaches this point because of a division-by-zero error. 

 

\(x=-2\) \(x=1\) \(x=2\)
\(x+2=0\) \(x-1=0\) \(x-2=0\)
     

 

Let's multiply these factors together. We know that their product equals 0 since the right-hand side of all the equations equal zero.

 

\((x+2)(x-2)(x-1)=0\)

 

Let's expand. I decided to rearrange to make the difference of squares more obvious. Let's expand. 

 

\((x+2)(x-2)(x-1)=0\) Let's do the first two binomials first. Since they will result in a difference of squares, we can take the shortcut.
\((x^2-4)(x-1)=0\) Now, let's expand completely.
\(x^3\textcolor{red}{-\hspace{1mm}1}x^2\textcolor{blue}{-\hspace{1mm}4}x\textcolor{green}{+\hspace{1mm}4}=0\\ x^3\textcolor{red}{+A}x^2\textcolor{blue}{+B}x\textcolor{green}{+C}\\ A=-1,B=-4,C=4\) This is the denominator of our function! Notice how this function perfects aligns with the question. The only thing left to do is find their sum.
\(\hspace{3mm}A+B+C\\ -1-4+4\\ -1\)  
TheXSquaredFactor  Apr 26, 2018

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