Science:Math Exam Resources/Courses/MATH104/December 2015/Question 10 (d)/Solution 1

We first find the x-intercepts/roots of $f(x):$

$\begin{array}{rl}{x}^2+x-\frac{1}{x}-1& =0\\ x^3+x^2-1-x& =0\\ (x-1)(x+1{)}^2& =0\\ x& =\pm 1\end{array}$

(In fact, since $x=-1$ is a root of multiplicity 2, we know that the graph of $f(x)$ just 'touches' the x-axis at $x=-1,$ whereas at the simple (i.e., multiplicity 1) root $x=1$ it crosses the x-axis.)

We plot these on our grid:

From 10 (b), we know that $f(x)$ is increasing on $[-1,\mathrm{\infty })\setminus \{0\}$ and decreasing on $(-\mathrm{\infty },-1].$

From 10 (c), we know that $f(x)$ is concave up on $(-\mathrm{\infty },0)\cup [1,\mathrm{\infty })$ and concave down on $(0,1].$

From 10 (a), we know the behaviour of $f$ around its vertical asymptote $x=0.$ Namely, $\lim {}_{x\to 0^{-}}f(x)=\mathrm{\infty }$ and $\lim {}_{x\to 0^{+}}f(x)=-\mathrm{\infty }.$ Combining all the above information, we obtain the following sketch:

Note: It is helpful to evaluate $f$ at integer values (which should be relatively easy to do), e.g., $x=-3,-2,2$ to guide your sketch.

Finally, the complete sketch: