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.

Corrected worked example

For $p(x)=x^3-4x^2+x+6$, $p(2)=0$, so $x-2$ is a factor. Division gives $p(x)=(x-2)(x^2-2x-3)=(x-2)(x-3)(x+1)$. The roots are $2$, $3$ and $-1$; none is repeated.

For this cubic, the roots sum to $2+3-1=4=-(-4)/1$ and their product is $2\cdot3\cdot(-1)=-6=(-1)^3(6/1)$. Use the factor test at the candidate value, then verify the full factorization and Vieta relations.

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

For $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0=0$, the sum of all roots (with multiplicity) is $-a_{n-1}/a_n$ and their product is $(-1)^na_0/a_n$. A repeated root requires a repeated factor; one successful factor test alone does not prove multiplicity.

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.

Complete the graph-feature check for $f(x)=\frac{x^2+1}{x-2}=x+2+\frac5{x-2}$. Besides vertical asymptote $x=2$ and oblique asymptote $y=x+2$, the y-intercept is $f(0)=-1/2$. There are no real x-intercepts because $x^2+1=0$ has no real solution. Record all intercepts, asymptotes, holes and excluded inputs before sketching.

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.

Periodic and self-inverse examples: $\cos x$ is even and $2\pi$-periodic, while $\sin x$ is odd and $2\pi$-periodic. The function $f(x)=1/x$ on $x\ne0$ is odd and self-inverse because $f(f(x))=1/(1/x)=x$. Self-inverse means $f^{-1}=f$; it does not mean every input is fixed by $f$.

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.

Further transformations: $y=[f(x)]^2$ makes outputs non-negative and retains the zeros of $f$; $y=f(ax+b)$ applies horizontal scaling and translation through the input. For example, $f(2x-4)=f(2(x-2))$ is horizontally compressed by factor $1/2$ and shifted right $2$. Modulus example: $|2x-3|\le5$ is equivalent to $-5\le2x-3\le5$, giving $-1\le x\le4$.