Problem Statement
Consider the rational function $f(x) = \frac{x^3 + 2x^2 - 1}{x^2 - 1}$ for $x \in \mathbb{R}, x \neq \pm 1$.
**(a)** [3 marks] Determine the equation of the oblique (slant) asymptote of the graph of $y = f(x)$.
**(b)** [5 marks] Find the exact coordinates of any stationary points and determine their nature using the second derivative test.
Verified Solution & Marking Scheme
Part (a) [M1]
Perform polynomial long division: $(x^3 + 2x^2 - 1) \div (x^2 - 1)$
Part (a) [A1]
Quotient is $x + 2$ with remainder $x + 1$
Part (a) [A1]
As $x \to \pm \infty, \frac{x+1}{x^2-1} = \frac{1}{x-1} \to 0$. Hence oblique asymptote is $y = x + 2$.
Part (b) [M1]
Differentiate simplified form $f(x) = x + 2 + \frac{1}{x - 1}$ for $x \neq -1$: $f'(x) = 1 - \frac{1}{(x - 1)^2}$
Part (b) [M1]
Set $f'(x) = 0 \implies (x - 1)^2 = 1 \implies x - 1 = \pm 1 \implies x = 2$ or $x = 0$
Part (b) [A1]
Stationary points are $(0, 1)$ and $(2, 5)$
Part (b) [M1]
Second derivative: $f''(x) = \frac{2}{(x - 1)^3}$
Part (b) [R1]
$f''(0) = -2 < 0 \implies (0,1)$ is a local maximum; $f''(2) = 2 > 0 \implies (2,5)$ is a local minimum.