Let \(f: R \backslash\{-1,1\} \rightarrow R, f(x)=\dfrac{x^3+x^2-2 x}{1-x^2}\).
- Show that \(f(x)\) can be written in the form \(f(x)=-x-1+\dfrac{1}{x+1}\), for \(x \in R \backslash\{-1,1\}\). (2 marks)
--- 8 WORK AREA LINES (style=lined) ---
- Consider the function with rule
- \begin{align*}
g(x)=\left\{\begin{array}{cl}
\dfrac{x^3+x^2-2 x}{1-x^2}, & x \in R \backslash\{-1,1\} \\
k, & x \in\{1\}
\end{array}\right.
\end{align*}
- \begin{align*}
- Find the value of \(k\) such that the graph of \(g\) is continuous at \(x=1\). (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
- Sketch the graph of \(y=f(x)\) on the axes below.
- Label the asymptotes with their equations. (3 marks)
--- 3 WORK AREA LINES (style=lined) ---
--- 0 WORK AREA LINES (style=lined) ---
Show Answers Only
| a. | \(f(x)\) | \(=\dfrac{x^3+x^2-2 x}{1-x^2}\) |
| \(=\dfrac{x\left(x^2+x-2\right)}{(1-x)(1+x)}\) | ||
| \(=\dfrac{-x(x+2)(1-x)}{(1-x)(1+x)}\) | ||
| \(=\dfrac{-x(x+2)}{x+1}\) | ||
| \(=\dfrac{-x^2-2 x}{x+1}\) | ||
| \(=\dfrac{-(x+1)^2+1}{x+1}\) | ||
| \(=-x-1+\dfrac{1}{x+1}\) |
b. \(k=-\dfrac{3}{2}\)
c.
Show Worked Solution
| a. | \(f(x)\) | \(=\dfrac{x^3+x^2-2 x}{1-x^2}\) |
| \(=\dfrac{x\left(x^2+x-2\right)}{(1-x)(1+x)}\) | ||
| \(=\dfrac{-x(x+2)(1-x)}{(1-x)(1+x)}\) | ||
| \(=\dfrac{-x(x+2)}{x+1}\) | ||
| \(=\dfrac{-x^2-2 x}{x+1}\) | ||
| \(=\dfrac{-(x+1)^2+1}{x+1}\) | ||
| \(=-x-1+\dfrac{1}{x+1}\) |
b. \(\text{Find \(k\) such that \(g(x)\) is continuous:}\)
| \(k\) | \(=\dfrac{k^3+k^2-2 k}{1-k^2}\) |
| \(k-k^3\) | \(=k^3+k^2-2 k\) |
| \(1-k^2\) | \(=k^2+k-2\) |
| \(0\) | \(=2 k^2+k-3\) |
| \(0\) | \(=(2 k+3)(k-1)\) |
\(\therefore k=-\dfrac{3}{2}\ \ (k \neq 1)\)
♦ Mean mark (b) 46%.
c. \(\text{Intercepts where}\ \ -x-1+\dfrac{1}{x+1}=0:\)
\((x+1)^2=1 \ \ \Rightarrow \ \ x^2+2 x=0 \ \ \Rightarrow \ \ x=0 \ \ \text{or} \ -2\)
\(\text{Asymptotes:} \ \ y=-1-x, \ x=-1\)
\(\text{Hole at} \ \left(1,-\dfrac{3}{2}\right)\)
♦♦ Mean mark (c) 37%.