Functions, Domain, Range and Inverses

Worked example 1: reading function notation

Task. Let \(f(x)=3x+2\).

1. Find \(f(-2)\).

2. Find \(f(a+1)\).

3. Solve \(f(x)=17\).

Solution

Replace every \(x\) by the input, using brackets when necessary.

(a) \[\begin{aligned}f(-2)&=3(-2)+2\\&=\boxed{-4}\end{aligned}.\]

(b) \[\begin{aligned}f(a+1)&=3(a+1)+2\\&=\boxed{3a+5}\end{aligned}.\]

(c) The output is given; find the input: \[3x+2=17\quad\Longrightarrow\quad3x=15\quad\Longrightarrow\quad\boxed{x=5}.\] Thus \(f(5)=17\): input \(5\) gives output \(17\).

Worked example 2: reading inputs and outputs from a curve

Task. No formula for \(f\) is given. Use the marked points on \(y=f(x)\).

1. Find \(f(3)\).

2. Solve \(f(x)=7\).

Solution

(a) From input \(3\), move vertically to \((3,4)\), then across to output \(4\). Thus \(\boxed{f(3)=4}\): the point \((3,4)\) is the pair (input, output).

(b) The line \(y=7\) meets the curve at \((-2,7)\), \((1,7)\) and \((4,7)\). Read the three inputs: \[\boxed{x=-2,\quad x=1\quad\text{or}\quad x=4}.\] Here \(f(-2)=f(1)=f(4)=7\): three different inputs give the same output.

Worked example 3: a restricted graph

Task. Let \(f(x)=(x-1)^2\), with \(-1\leq x\leq3\).

1. State the domain.

2. State the range.

Solution

(a) The allowed inputs are stated: \[\boxed{D=[-1,3]}.\] (b) A square cannot be negative. The smallest output is \(f(1)=0\). At the endpoints, \[f(-1)=(-2)^2=4,\qquad f(3)=2^2=4.\] All outputs between \(0\) and \(4\) occur, so \[\boxed{R=[0,4]}.\]

The filled endpoints show that \(x=-1\) and \(x=3\) are included. Checking only the endpoints would miss the minimum at \(x=1\).

Worked example 4: a square root

Task. Let \(g(x)=\sqrt{5-2x}\).

1. Find the domain.

2. Find the range.

Solution

(a) The expression inside a real square root must be non-negative: \[5-2x\geq0\quad\Longrightarrow\quad -2x\geq-5\quad\Longrightarrow\quad x\leq\frac52.\] The inequality reverses when we divide by \(-2\). Thus \(\boxed{D=(-\infty,\frac52]}\).

(b) A square root is non-negative, and \(g(\frac52)=0\). As \(x\) becomes more negative, \(5-2x\) grows without bound. Every non-negative output occurs, so \(\boxed{R=[0,\infty)}\).

Worked example 5: combining restrictions

Task. Find the largest real domain of \(h(x)=\dfrac{\sqrt{x+2}}{x-1}\).

Solution

Both conditions must hold at the same time: \[\underbrace{x+2\geq0}_{\text{square root defined}}\ \Longrightarrow\ x\geq-2, \qquad\underbrace{x-1\neq0}_{\text{denominator nonzero}}\ \Longrightarrow\ x\neq1.\] Therefore \(\boxed{D=[-2,1)\cup(1,\infty)}\). The symbol \(\cup\) joins the two allowed intervals. The input \(-2\) is allowed because it gives \(0/(-3)=0\); input \(1\) is not allowed.

Worked example 6: a logarithm (later-course connection)

Task. Find the largest real domain of \(p(x)=\ln(4-x)\).

Solution

The logarithm’s argument must be strictly positive: \[4-x>0\quad\Longrightarrow\quad x<4.\qquad\boxed{D=(-\infty,4)}.\] Zero is allowed inside a square root, but not inside a real logarithm.

Worked example 7: a discrete cost model

Task. A school trip costs a fixed booking fee of € 24 plus € 6 per student. There can be at most 30 students. Let \(C(n)\) be the total cost for \(n\) students, including a booking with no students yet registered.

1. Write a model for \(C(n)\).

2. State its domain.

3. State its range.

4. Find \(C(12)\) and interpret your answer.

5. Find the number of students when the total cost is € 126.

Solution

(a) Fixed cost \(+\) cost per student \(\times\) number of students gives \[\boxed{C(n)=24+6n}.\] (b) Students are counted in whole numbers, and the capacity is 30: \(D=\{0,1,2,\ldots,30\}\).

(c) \(R=\{24,30,36,\ldots,204\}\). These are separate costs, not every value in \([24,204]\).

(d) \(C(12)=24+6(12)=96\). The total cost for 12 students is € 96.

(e) Set the cost equal to 126 and solve: \[24+6n=126\quad\Longrightarrow\quad6n=102\quad\Longrightarrow\quad\boxed{n=17}.\] Seventeen students is an allowed input, so this solution makes sense.

Worked example 8: a continuous model

Task. A tank contains 80 litres and drains at a constant rate of 4 litres per minute.

1. Write a volume model \(V(t)\), where \(t\) is the time in minutes.

2. Give a sensible domain while the tank drains.

3. Give the corresponding range.

Solution

(a) After \(t\) minutes, \(4t\) litres have left, so \(\boxed{V(t)=80-4t}\) litres.

(b) The tank is empty when \[80-4t=0\quad\Longrightarrow\quad t=20.\] Hence \(\boxed{0\leq t\leq20}\) minutes; negative time is excluded.

(c) The volume decreases from 80 litres to 0 litres, so \(\boxed{0\leq V(t)\leq80}\) litres.

Worked example 9: finding an inverse step by step

Task. Let \(f(x)=2x+1\), for \(x\in\mathbb{R}\).

1. Find \(f^{-1}(x)\).

2. Find \(f^{-1}(7)\).

Solution

(a) Write the original equation: \(y=2x+1\).

Step 1: swap \(x\) and \(y\).\(x=2y+1\).

Step 2: make \(y\) the subject.\[\begin{aligned}x-1&=2y\\y&=\frac{x-1}{2}\end{aligned}.\]

Step 3: define the inverse.\(\displaystyle\boxed{f^{-1}(x)=\frac{x-1}{2}}\).

The inverse subtracts 1, then divides by 2: it undoes the original operations in reverse order.

(b) \(\displaystyle f^{-1}(7)=\frac{7-1}{2}=\boxed{3}\).

Check: \(f(3)=2(3)+1=7\), so the inverse correctly takes 7 back to 3.

Worked example 10: build the inverse sketch in three stages

Task. Step 1 shows the complete graph of a one-to-one function \(f\), with three labelled points and included endpoints. No formula for \(f\) is given.

1. Sketch \(y=f^{-1}(x)\).

2. State the domain of \(f^{-1}\).

3. State the range of \(f^{-1}\).

Solution

(a) Step 1: draw the mirror line.

Copy the given graph and draw the dashed line \(y=x\). Use the same scale on both axes so that this is a \(45^\circ\) line.

The given points are \(P(-1,0)\), \(Q(1,2)\) and \(R(3,4)\). The graph increases from \(P\) to \(R\), so \[D_f=[-1,3],\qquad R_f=[0,4].\]

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Step 2: swap each point’s coordinates.

Plot the corresponding points on the inverse: \[\begin{aligned} P(-1,0)&\longmapsto P'(0,-1),\\ Q(1,2)&\longmapsto Q'(2,1),\\ R(3,4)&\longmapsto R'(4,3). \end{aligned}\] Each pair of points lies at equal distances on opposite sides of \(y=x\).

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Step 3: reflect the rest of the curve.

Draw a smooth reflection through the new points, keeping the endpoints filled.

The brick-coloured curve is \(y=f^{-1}(x)\); the pale dashed curve is the original.

(b) Use the original range: \[\boxed{D_{f^{-1}}=[0,4]}.\] (c) Use the original domain: \[\boxed{R_{f^{-1}}=[-1,3]}.\]

The vertical line test

A graph represents \(y\) as a function of \(x\) if no vertical line intersects it more than once.

The circle \(x^2+y^2=1\) fails: the input \(x=\frac12\) gives two outputs. The whole circle is not the graph of a function \(y=f(x)\).

The horizontal line test

A function is one-to-one if no horizontal line intersects its graph more than once.

\(y=x^2\) passes the vertical test but fails the horizontal test: \(f(-1)=f(1)=1\). On \(\mathbb{R}\), it has no inverse function.

Worked example 11: deciding from a table

Task. The complete domain of \(f\) is \(\{-2,0,3\}\). The table gives its outputs. \[\begin{array}{c|rrr}x&-2&0&3\\\textbf{HL extension}ine f(x)&5&1&7\end{array}\]

1. Does an inverse function exist? Explain your answer.

2. If it exists, find \(f^{-1}(7)\).

3. State the domain of \(f^{-1}\).

4. State the range of \(f^{-1}\).

Solution

(a) Yes. The outputs \(5,1,7\) are all different, so \(f\) is one-to-one.

(b) Since \(f(3)=7\), we have \(\boxed{f^{-1}(7)=3}\).

(c) The original outputs become inputs: \(\boxed{D_{f^{-1}}=\{1,5,7\}}\).

(d) The original inputs become outputs: \(\boxed{R_{f^{-1}}=\{-2,0,3\}}\).

Worked example 12: translate an exponential graph upwards

Task. Let \(g(x)=2^x+3\).

1. State the domain.

2. State the range.

3. State the horizontal asymptote.

Solution

Start with the parent function \(f(x)=2^x\). Its domain is \(\mathbb{R}\), its range is \((0,\infty)\) and its horizontal asymptote is \(y=0\).

(a) Adding 3 translates the graph vertically upwards by 3 units (section 2.11). The inputs are unchanged, so \(\boxed{D_g=\mathbb{R}}\).

(b) Every output increases by 3, so \[(0,\infty)\ \xrightarrow{\text{add }3}\ \boxed{R_g=(3,\infty)}.\]

Dashed curve: \(y=2^x\).

Solid curve: \(y=2^x+3\).

For example, \((0,1)\) moves to \((0,4)\).

(c) The horizontal asymptote also moves up by 3 units: \[y=0\ \longmapsto\ \boxed{y=3}.\] The domain is unchanged because the input \(x\) has not changed.

Check the range algebraically: \(2^x>0\), so \(2^x+3>3\). The output 3 is approached but never reached, and there is no upper bound. Therefore the lower endpoint remains excluded.

Worked example 13: a translated, restricted function and its inverse

Task. Let \(f(x)=x^2+1\) for \(0\leq x\leq2\).

1. Find the range of \(f\).

2. Find the inverse rule \(f^{-1}(x)\).

3. State the domain of \(f^{-1}\).

4. State the range of \(f^{-1}\).

Solution

(a) Find the range by translating.

Start with \(y=x^2\) on the same restricted domain \([0,2]\). Its range is \([0,4]\).

Adding 1 translates the graph upwards by 1 unit. Therefore every output increases by 1: \[[0,4]\ \xrightarrow{\text{add }1}\ \boxed{R_f=[1,5]}.\] The domain stays \([0,2]\). Both endpoints of the range are still included.

Dashed: \(y=x^2\).

Solid: \(y=x^2+1\).

(b) Find the inverse. Write \(y=x^2+1\).

Step 1: swap \(x\) and \(y\).\(x=y^2+1\).

Step 2: make \(y\) the subject.\(y^2=x-1\), so \(y=\pm\sqrt{x-1}\).

The inverse must return an original input in \([0,2]\). Its outputs must therefore be non-negative, so choose \(y=\sqrt{x-1}\).

Step 3: define the inverse.\(\displaystyle\boxed{f^{-1}(x)=\sqrt{x-1}}\).

(c) The inverse’s domain is \(\boxed{D_{f^{-1}}=R_f=[1,5]}\).

(d) The inverse’s range is \(\boxed{R_{f^{-1}}=D_f=[0,2]}\). For example, \(f(2)=5\) and \(f^{-1}(5)=\sqrt4=2\).