2.2 Functions - AHL content

Use factor and remainder theorems to test polynomial roots.

For a polynomial p(x), (x−a) is a factor exactly when p(a)=0; the remainder on division by (x−a) is p(a). The coefficients and roots are linked by Vieta relationships.

Worked example

For p(x)=x³−4x²+x+6, p(2)=0, so x−2 is a factor. Division gives (x−2)(x²−2x−3)=(x−2)²(x+1).

The root 2 appears twice, so it is a repeated root; the factor test is evaluated at the candidate value, not at the factor expression itself.

A zero of p is a number a with p(a)=0; x=0 is not the same statement as the factor x.

Read asymptotes from a rational function’s leading structure.

Vertical asymptotes occur at non-cancelled denominator zeros; horizontal or oblique behaviour comes from comparing numerator and denominator degrees after simplification.

Worked example

For f(x)=(x²+1)/(x−2), x=2 is a vertical asymptote. Polynomial division gives f(x)=x+2+5/(x−2), so y=x+2 is the oblique asymptote.

A cancelled factor creates a hole, not a vertical asymptote; always simplify and record excluded domain values.

An asymptote describes limiting behaviour, not a value the function reaches at the asymptote.

Use symmetry and domain restrictions to analyse inverses.

An even function satisfies f(−x)=f(x), an odd function satisfies f(−x)=−f(x). An inverse exists as a function only after the original is one-to-one on its chosen domain.

Worked example

f(x)=x² is even but not one-to-one on ℝ. Restricting to x≥0 gives f⁻¹(x)=√x; the graph reflects across y=x.

Check the domain before finding an inverse; the same formula can produce different inverse branches.

Symmetry does not imply invertibility: even functions usually map two inputs to one output.

Solve an inequality by comparing graphs or sign intervals.

To solve g(x)≥f(x), find where h(x)=g(x)−f(x) is non-negative. Intersections split the number line into intervals whose signs must be checked.

Worked example

For x²≥2x, x(x−2)≥0, so x≤0 or x≥2. The graph gives the same result because the parabola lies above the line outside the intersections.

Include equality at roots for ≥ or ≤, and exclude points where an expression is undefined.

The intersection points are boundaries; they are not automatically the only solutions.

Apply modulus and reciprocal transformations in the correct order.

For y=|f(x)|, negative parts reflect above the x-axis; y=f(|x|) mirrors the right-hand graph into x<0; y=1/f(x) keeps zeros as excluded inputs and swaps large/small values.

Worked example

If f(x)=x−1, then |f(x)| has a corner at x=1, while f(|x|)=|x|−1 has a V-shaped graph with a corner at x=0.

Transform the graph in stages and preserve domain restrictions; the two modulus forms are not interchangeable.

Absolute value outside changes outputs; absolute value inside changes which inputs are used.