Functions - Domain and Range

Replace every x with (-2), keeping the brackets so that the square of -2 comes out positive.

1. \mathrm{f}(-2)=4(-2)^2-3(-2)
2. =4\times4+6
3. =16+6
4. Answer: 22

\mathrm{g}(x)=-6 means the output is -6, so set 3(x-4) equal to -6 and solve for x.

1. \mathrm{g}(x)=-6 means 3(x-4)=-6 (it does not mean put x=-6 into g)
2. Divide by 3: x-4=-2
3. Answer: x=2

3^x increases as x increases, so put the two ends of the domain into f, and remember what a negative power means.

1. 3^x is increasing, so the smallest output is at x=-2 and the largest at x=2.
2. \mathrm{f}(-2)=3^{-2}=\dfrac{1}{9} (a negative power gives a reciprocal, not a negative number).
3. \mathrm{f}(2)=3^2=9, but x=2 is not in the domain (x<2), so 9 is not reached: use <.
4. x=-2 is included, so use \leqslant at the bottom.
5. The range is written in terms of \mathrm{f}(x), not x.
6. Answer: \dfrac{1}{9}\leqslant \mathrm{f}(x)<9

f increases as x increases, so work out the output at the smallest value of x in the domain.

1. The gradient, 5, is positive, so f increases as x increases
2. The smallest x in the domain is -1, and it is included
3. \mathrm{f}(-1)=-5+2=-3
4. As x gets larger, \mathrm{f}(x) gets larger without limit
5. The range is written in terms of \mathrm{f}(x), not x
6. Answer: \mathrm{f}(x)\geqslant -3

A fraction cannot be worked out when its denominator is zero.

1. \mathrm{f}(x) cannot be worked out when the denominator is zero: x^2-9=0
2. x^2=9, so x=3 or x=-3 (both square to 9)
3. Every other value of x gives a non-zero denominator, including x=0 and x=9
4. Answer: for all x except x=3 and x=-3

Replace \mathrm{f}(x) with \dfrac{x-4}{3} in the inequality and solve it for x.

1. -3\leqslant \dfrac{x-4}{3}<1
2. Multiply through by 3: -9\leqslant x-4<3
3. Add 4: -5\leqslant x<7
4. The signs stay the same: -3 is in the range, 1 is not
5. Answer: -5\leqslant x<7

2x^3+1 increases as x increases, so the least and greatest values come from the ends of the domain.

1. x^3 increases as x increases, so h does too
2. Least value: \mathrm{h}(-2)=2(-8)+1=-15
3. x=-2 is in the domain, so -15 is reached: use \leqslant
4. Greatest value: \mathrm{h}(1)=2+1=3
5. x=1 is not in the domain (x<1), so 3 is not reached: use <
6. Answer: -15\leqslant \mathrm{h}(x)<3

n can be negative, zero or positive, so think about the smallest value a square can take as well as the values at the ends.

1. n=0 is allowed, and a square is never negative, so the least value of n^2 is 0
2. At the ends: (-6)^2=36 and 2^2=4
3. n=-6 is not allowed (-6<n), so 36 is not reached: use <
4. Every value from 0 up to (but not including) 36 is possible
5. Answer: 0\leqslant n^2<36

Think about what happens to 3^x when x is a large negative number: it gets close to a value but never reaches it.

1. 3^x increases as x increases, so the greatest value is at x=2
2. \mathrm{h}(2)=9-4=5, and x=2 is in the domain: use \leqslant
3. As x becomes a large negative number, 3^x gets closer and closer to 0 but is always positive
4. So \mathrm{h}(x) gets closer to -4 but never equals it: use <
5. Answer: -4<\mathrm{h}(x)\leqslant 5

Work out \mathrm{f}(4), then think about what happens to \dfrac{12}{x} as x gets larger and larger.

1. \mathrm{f}(4)=\dfrac{12}{4}=3, and x=4 is in the domain
2. As x increases, \dfrac{12}{x} decreases, so 3 is the greatest value
3. \dfrac{12}{x} stays positive for positive x and never reaches 0
4. Answer: 0<\mathrm{f}(x)\leqslant 3

Find where the turning point of the graph is and check whether it lies inside the domain.

1. x^2-6x+1=(x-3)^2-8, so the turning point is at x=3
2. x=3 is not in the domain 4<x\leqslant 7, so -8 is not part of the range
3. For x>3, f increases, so use the ends of the domain
4. \mathrm{f}(4)=16-24+1=-7, but x=4 is not included: use <
5. \mathrm{f}(7)=49-42+1=8, and x=7 is included: use \leqslant
6. Answer: -7<\mathrm{f}(x)\leqslant 8

Write \mathrm{f}(x) in the form a(x+b)^2+c and think about the smallest value a square can take.

1. 2x^2-12x+23=2(x^2-6x)+23
2. =2(x-3)^2-18+23
3. =2(x-3)^2+5
4. (x-3)^2\geqslant 0, so \mathrm{f}(x)\geqslant 5, with \mathrm{f}(3)=5
5. There is no upper limit because the domain is all real values
6. Answer: \mathrm{f}(x)\geqslant 5

Complete the square to find the turning point and check whether it lies inside the domain before using the end values.

1. Complete the square: x^2-4x+7=(x-2)^2+3
2. The minimum point is at x=2, which is inside the domain, so the least value is \mathrm{h}(2)=3 (and it is reached, so \leqslant).
3. Check the ends: \mathrm{h}(-1)=1+4+7=12 and \mathrm{h}(3)=9-12+7=4
4. The largest value would be 12, at x=-1, but x=-1 is not in the domain (-1<x), so use <.
5. Answer: 3\leqslant \mathrm{h}(x)<12

Form a quadratic equation and solve it, then check each solution against the domain.

1. x^2-2x=15
2. x^2-2x-15=0
3. (x-5)(x+3)=0, so x=5 or x=-3
4. x=-3 is not in the domain (x>1), so reject it
5. Answer: x=5

The graph of 5x-x^2 is an upside-down curve: find where its turning point is and whether it lies inside the domain.

1. 5x-x^2=x(5-x) is zero at x=0 and x=5, so the turning point is halfway, at x=2.5
2. x=2.5 is inside the domain, so the greatest value is \mathrm{g}(2.5)=12.5-6.25=6.25
3. 6.25 is reached, so use \leqslant
4. Ends: \mathrm{g}(1)=5-1=4 and \mathrm{g}(4)=20-16=4
5. Neither end is in the domain, so 4 is never reached: use <
6. Answer: 4<\mathrm{g}(x)\leqslant \dfrac{25}{4}

As a is negative, f is decreasing, so the greatest value of \mathrm{f}(x) comes from the smallest value of x. Answer part (a) before attempting part (b).

Part (a)

1. a<0, so f is decreasing: the greatest output comes from the smallest x
2. \mathrm{f}(1)=10: a+b=10
3. \mathrm{f}(5)=-2: 5a+b=-2
4. Subtract: 4a=-12
5. Answer: a=-3

Part (b)

1. a+b=10
2. -3+b=10
3. Answer: b=13

Part (c)

1. 13-3x>4
2. 9>3x, so x<3
3. The domain starts at x=1 (included)
4. Answer: 1\leqslant x<3

Put each point into y=a\times b^x to get two equations, then divide one by the other to get rid of a. Answer part (a) before attempting part (b).

Part (a)

1. At (-1,\ 20): a\times b^{-1}=20, so \dfrac{a}{b}=20
2. At (1,\ 5): ab=5
3. Divide: ab\div\dfrac{a}{b}=b^2, so b^2=\dfrac{5}{20}=\dfrac{1}{4}
4. b>0, so b=\dfrac{1}{2}
5. Answer: b=\dfrac{1}{2}

Part (b)

1. ab=5
2. a\times\dfrac{1}{2}=5
3. Answer: a=10

Part (c)

1. \mathrm{g}(x)=10\times\left(\dfrac{1}{2}\right)^x, which decreases as x increases
2. a\times b^x means a times b^x: work out the power first
3. Greatest: \mathrm{g}(-2)=10\times 2^2=40, but x=-2 is not in the domain: use <
4. Least: \mathrm{g}(3)=10\times\dfrac{1}{8}=\dfrac{5}{4}, and x=3 is in the domain: use \leqslant
5. Answer: \dfrac{5}{4}\leqslant \mathrm{g}(x)<40

Work out 8^{\frac{2}{3}} by taking the cube root first and then squaring. Answer part (a) before attempting part (b).

Part (a)

1. \mathrm{f}\left(\dfrac{2}{3}\right)=8^{\frac{2}{3}}=\left(\sqrt[3]{8}\right)^2
2. =2^2=4
3. \mathrm{g}(4)=\sqrt{12+k}
4. \sqrt{12+k}=4, so 12+k=16
5. Answer: k=4

Part (b)

1. \mathrm{g}(x)=\sqrt{3x+4}
2. The square root of a negative number cannot be worked out, so 3x+4\geqslant 0
3. 3x\geqslant -4
4. Answer: x\geqslant -\dfrac{4}{3}

Part (c)

1. \sqrt{3x+4} increases as x increases
2. \mathrm{g}(0)=\sqrt4=2 and \mathrm{g}(15)=\sqrt{49}=7
3. Both ends are in the domain: use \leqslant at both
4. Answer: 2\leqslant \mathrm{g}(x)\leqslant 7

The lowest value of a quadratic on a domain is at its turning point if the turning point is inside the domain; the highest value is at one of the ends.

Part (a)

1. Complete the square: x^2-2x+c=(x-1)^2+c-1
2. The minimum point is at x=1, which is inside the domain because -2\leqslant 1\leqslant k (as k>1).
3. So the least value of f is c-1, and this must equal 2.
4. c-1=2
5. Answer: c=3

Part (b)

1. \mathrm{f}(x)=x^2-2x+3
2. At the left end: \mathrm{f}(-2)=4+4+3=11, which is less than 18.
3. So the greatest value, 18, must happen at the other end, x=k.
4. k^2-2k+3=18
5. k^2-2k-15=0
6. (k-5)(k+3)=0, so k=5 or k=-3
7. k>1, so k=5
8. Answer: k=5

Differentiate to find where the gradient is zero, and check that each solution lies in the domain. Answer part (a) before attempting part (b).

Part (a)

1. \mathrm{f}'(x)=3x^2-12
2. At a stationary point 3x^2-12=0, so x^2=4
3. x=2 or x=-2, and x=-2 is not in the domain
4. \mathrm{f}(2)=8-24=-16
5. Answer: (2,\ -16)

Part (b)

1. \mathrm{f}''(x)=6x, and \mathrm{f}''(2)=12>0, so (2,\ -16) is a minimum
2. It is the only stationary point in the domain, so the least value is -16 (reached)
3. Ends: \mathrm{f}(-1)=-1+12=11 and \mathrm{f}(5)=125-60=65
4. The greatest value would be 65, but x=5 is not in the domain: use <
5. Answer: -16\leqslant \mathrm{f}(x)<65