Functions, modulus and partial fractions – Year 13 A level maths revision notes to download as PDF

Worked example 1

Find the range of \(f(x)=x^2-4x+7\), \(x\in\mathbb{R}\).

Complete the square: \((x-2)^2+3\), so the range is \(f(x)\ge3\).

Worked example 2

\(f(x)=2x+3\) and \(g(x)=x^2\). Find \(fg(x)\) and \(gf(x)\).

\(fg(x)=2x^2+3\) while \(gf(x)=(2x+3)^2\). They are different, so order matters.

Worked example 3

\(f(x)=3x-2\) and \(g(x)=\dfrac{1}{x+1}\), \(x\ne-1\). Solve \(fg(x)=1\).

\(fg(x)=\dfrac{3}{x+1}-2=1\) gives \(\dfrac3{x+1}=3\), so \(x+1=1\) and \(x=0\), which is in the domain of \(g\).

Worked example 4

\(g(x)=x^2\) for all real \(x\) and \(f(x)=\ln x\), \(x>0\). Find the domain and range of \(fg\).

\(fg\) is only defined for \(x\ne0\), because \(g(0)=0\) is not in the domain of \(f\). The range is all real numbers, since \(x^2\) takes every positive value.

Worked example 5

Find the inverse of \(f(x)=\dfrac{x+1}{x-2}\), \(x\ne2\).

\(y(x-2)=x+1\), so \(yx-2y=x+1\), \(x(y-1)=2y+1\) and \(f^{-1}(x)=\dfrac{2x+1}{x-1}\), \(x\ne1\).

Worked example 6

Sketch \(y=|2x-4|\).

Sketch \(y=2x-4\) (through \((2,0)\) and \((0,-4)\)) and reflect the part below the \(x\)-axis. The result is a V-shape with vertex \((2,0)\) crossing the \(y\)-axis at \((0,4)\). Label both intercepts and the vertex.

Worked example 7

Solve \(|2x-3|=5\) and \(|x-1|=2x-5\).

First: \(2x-3=5\) or \(-5\), so \(x=4\) or \(x=-1\). Second: \(x-1=2x-5\) gives \(x=4\) and \(2x-5=3>0\), valid. \(1-x=2x-5\) gives \(x=2\), but \(2x-5=-1<0\), so reject. The only solution is \(x=4\).

Worked example 8

Solve \(|x+1|<|x-3|\).

Square: \((x+1)^2<(x-3)^2\), so \(8x<8\) and \(x<1\).

Worked example 9

Express \(\dfrac{5x+1}{(x-1)(x+2)}\) in partial fractions.

\(5x+1=A(x+2)+B(x-1)\). \(x=1\): \(6=3A\), so \(A=2\). \(x=-2\): \(-9=-3B\), so \(B=3\). The answer is \(\dfrac{2}{x-1}+\dfrac{3}{x+2}\).

Worked example 10

Express \(\dfrac{3x+5}{(x+1)^2}\) in partial fractions.

\(3x+5=A(x+1)+B\). \(x=-1\) gives \(B=2\); comparing the coefficient of \(x\) gives \(A=3\). So \(\dfrac3{x+1}+\dfrac2{(x+1)^2}\).

Worked example 11

Express \(\dfrac{x^2+3x+3}{(x+1)(x+2)}\) in partial fractions.

Improper: \(x^2+3x+3=(x+1)(x+2)+1\), so the fraction is \(1+\dfrac1{(x+1)(x+2)}=1+\dfrac{1}{x+1}-\dfrac{1}{x+2}\).

Worked example 12

A cup of tea cools according to \(T=20+60e^{-0.1t}\) (\(^\circ\)C, \(t\) in minutes). Find the initial temperature, the long-term temperature and the time to reach \(50^\circ\)C.

At \(t=0\), \(T=80^\circ\)C. As \(t\to\infty\), \(T\to20^\circ\)C (room temperature). \(T=50\) gives \(e^{-0.1t}=0.5\), so \(t=10\ln2\approx6.93\) minutes. The inverse \(t=-10\ln\!\big(\tfrac{T-20}{60}\big)\) gives the time for any \(T\) between \(20\) and \(80\). The model is valid only for \(t\ge0\).

Worked example 13: range of a rational function

Find the range of \(f(x)=\dfrac{2x+1}{x-1}\), \(x\ne1\).

Rewrite: \(f(x)=\dfrac{2(x-1)+3}{x-1}=2+\dfrac3{x-1}\). The fraction \(\dfrac3{x-1}\) can be any real number except 0, so \(f(x)\) takes every value except 2. Range: \(f(x)\in\mathbb{R}\), \(f(x)\ne2\). The same horizontal asymptote \(y=2\) shows on the graph.

Worked example 14: range of an exponential model

Find the range of \(f(x)=5-3e^{-x}\), \(x\ge0\).

At \(x=0\), \(f=5-3=2\). As \(x\) increases, \(e^{-x}\) decreases towards 0, so \(f\) increases towards 5 but never reaches it. Range: \(2\le f(x)<5\).

Worked example 15: domain of a composite

\(f(x)=\sqrt{x}\), \(x\ge0\) and \(g(x)=x-4\), \(x\in\mathbb{R}\). Find \(fg(x)\) and state its domain.

\(fg(x)=\sqrt{x-4}\). The square root needs \(x-4\ge0\), so the domain is \(x\ge4\) (not all of \(\mathbb{R}\), even though \(g\) is defined everywhere). Always check that the output of the inner function lies in the domain of the outer one.

Worked example 16: inverse with a logarithm

Find the inverse of \(f(x)=3\ln(2x-1)+4\), \(x>\tfrac12\), and state the domain and range of \(f^{-1}\).

Let \(y=3\ln(2x-1)+4\). Then \(\ln(2x-1)=\dfrac{y-4}3\), so \(2x-1=e^{(y-4)/3}\) and \(x=\dfrac12\left(e^{(y-4)/3}+1\right)\). So \(f^{-1}(x)=\dfrac12\left(e^{(x-4)/3}+1\right)\). The range of \(f\) is all real numbers, so \(f^{-1}\) has domain \(x\in\mathbb{R}\) and range \(f^{-1}(x)>\tfrac12\).

Worked example 17: testing for a self-inverse

Show that \(f(x)=\dfrac{4}{x}\), \(x\ne0\), is self-inverse.

\(ff(x)=\dfrac{4}{4/x}=x\). Since \(ff(x)=x\), \(f=f^{-1}\). The graph \(y=\dfrac4x\) is symmetrical in the line \(y=x\), which is exactly the geometric meaning of a self-inverse function.

Worked example 18: modulus equation solved by sketching

Solve \(|2x-4|=x+1\).

Case 1: \(2x-4=x+1\) gives \(x=5\) (RHS \(=6>0\)). Case 2: \(4-2x=x+1\) gives \(x=1\) (RHS \(=2>0\)). Both lie on the graph above, so the solutions are \(x=1\) and \(x=5\).

Worked example 19: sketch of the modulus of a quadratic

Sketch \(y=|x^2-4|\) and solve \(|x^2-4|=5\).

Sketch \(y=x^2-4\) (roots \(\pm2\), minimum \(-4\) at \((0,-4)\)) and reflect the part below the \(x\)-axis. The reflected hump has maximum \((0,4)\). To solve \(|x^2-4|=5\): \(x^2-4=5\) gives \(x=\pm3\); \(x^2-4=-5\) gives \(x^2=-1\), no real solutions. So \(x=\pm3\). The line \(y=5\) is above the hump of height 4, so it only meets the outer branches.

Worked example 20: an irreducible quadratic

Express \(\dfrac{3x^2+5x+7}{(x+1)(x^2+4)}\) in partial fractions.

The factor \(x^2+4\) has no real roots, so use \(\dfrac A{x+1}+\dfrac{Bx+C}{x^2+4}\). Then \(3x^2+5x+7=A(x^2+4)+(Bx+C)(x+1)\). Put \(x=-1\): \(5=5A\), so \(A=1\). Compare \(x^2\): \(3=A+B\), so \(B=2\). Compare constants: \(7=4A+C\), so \(C=3\). Answer: \(\dfrac1{x+1}+\dfrac{2x+3}{x^2+4}\). Check at \(x=0\): \(\dfrac74=1+\dfrac34\).

Worked example 21: a delivery charge

A courier charges \(C(d)=4+0.5d\) pounds for a delivery of \(d\) kilometres. VAT of 20% is then added, giving \(V(c)=1.2c\). Find \(VC(d)\) and the distance for a total of £12.

\(VC(d)=1.2(4+0.5d)=4.8+0.6d\). Setting \(4.8+0.6d=12\) gives \(0.6d=7.2\) and \(d=12\) km. The inverse of the combined function is \(d=\dfrac{y-4.8}{0.6}\) for \(y\ge4.8\).