Teach Me Maths

Function Notation

Question 11 mark

f(x)=4x-3. Find f(5).

Hint

Replace each occurrence of the input letter by the given number.

Worked solution

f(x)=4x-3. Find f(5).

  1. f(5)=4\times5-3.
  2. This gives f(5)=17.

Question 21 mark

g(x)=x^2+6. Find g(-3).

Hint

Put the negative input in brackets before squaring.

Worked solution

g(x)=x^2+6. Find g(-3).

  1. g(-3)=(-3)^2+6.
  2. The square is 9, so g(-3)=15.

Question 31 mark

h(x)=\frac{x+7}{3}. Find h(2).

Hint

Substitute in the numerator before dividing.

Worked solution

h(x)=\frac{x+7}{3}. Find h(2).

  1. h(2)=\frac{2+7}{3}.
  2. Therefore h(2)=3.

Question 42 marks

f(x)=2x^2-x. Find f(-2).

Hint

Use brackets around each negative input.

Worked solution

f(x)=2x^2-x. Find f(-2).

  1. f(-2)=2(-2)^2-(-2).
  2. This is 8+2, so f(-2)=10.

Question 51 mark

p(x)=\frac{20}{x}, where x\ne0. Find p(-4).

Hint

The input replaces the whole denominator.

Worked solution

p(x)=\frac{20}{x}, where x\ne0. Find p(-4).

  1. p(-4)=\frac{20}{-4}.
  2. Therefore p(-4)=-5.

Question 62 marks

f(x)=3x+8. Find x when f(x)=29.

Hint

Set the rule equal to the required output and solve the equation.

Worked solution

f(x)=3x+8. Find x when f(x)=29.

  1. Solve 3x+8=29.
  2. Then 3x=21, so x=7.

Question 72 marks

g(x)=x^2-4. Find all values of x for which g(x)=21.

Give every value, separated by commas

Hint

A square can come from either a positive input or a negative input.

Worked solution

g(x)=x^2-4. Find all values of x for which g(x)=21.

  1. Solve x^2-4=21, giving x^2=25.
  2. Therefore x=-5 or x=5.

Question 82 marks

h(x)=\sqrt{2x+9}. Find h(8).

Hint

Evaluate the expression inside the root first.

Worked solution

h(x)=\sqrt{2x+9}. Find h(8).

  1. h(8)=\sqrt{2\times8+9}.
  2. The number inside is 25, so h(8)=5.

Question 92 marks

f(x)=5-2x. Solve f(t)=t+11.

Hint

Use the rule with the input letter in the equation.

Worked solution

f(x)=5-2x. Solve f(t)=t+11.

  1. Write 5-2t=t+11.
  2. This gives -3t=6, so t=-2.

Question 103 marks

f(x)=x^2+3x-2. Find an expanded and simplified expression for f(x+1).

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Replace every input by the whole bracket, not just its first term.

Worked solution

f(x)=x^2+3x-2. Find an expanded and simplified expression for f(x+1).

  1. Substitution gives (x+1)^2+3(x+1)-2.
  2. Expand to get x^2+2x+1+3x+3-2.
  3. Collect terms: x^2+5x+2.

Question 112 marks

A workshop charges C(n)=5n+12 pounds for a group of n people. Find the charge for a group of 8 people, in pounds.

Hint

The number of people is the input to the cost function.

Worked solution

A workshop charges C(n)=5n+12 pounds for a group of n people. Find the charge for a group of 8 people, in pounds.

  1. The required charge is C(8)=5\times8+12.
  2. The charge is £52.

Question 122 marks

f(x)=ax+5. Given that f(4)=17, find a.

Hint

Use the known input and output to form an equation for the unknown coefficient.

Worked solution

f(x)=ax+5. Given that f(4)=17, find a.

  1. The information gives 4a+5=17.
  2. So 4a=12 and a=3.

Question 132 marks

f(x)=2x^2+k. Given that f(3)=25, find k.

Hint

Substitute the stated input before finding the unknown constant.

Worked solution

f(x)=2x^2+k. Given that f(3)=25, find k.

  1. 2\times3^2+k=25.
  2. This gives 18+k=25, so k=7.

Question 143 marks

g(x)=\frac{12}{x-1}, where x\ne1. Find x when g(x)=3.

Hint

Multiply the output equation by its denominator.

Worked solution

g(x)=\frac{12}{x-1}, where x\ne1. Find x when g(x)=3.

  1. Solve \frac{12}{x-1}=3.
  2. Then 12=3(x-1), so x-1=4.
  3. Therefore x=5.

Question 153 marks

f(x)=x^2-2x+4. Simplify f(a+2)-f(a).

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Substitute into both occurrences of the function, then subtract the second whole expression.

Worked solution

f(x)=x^2-2x+4. Simplify f(a+2)-f(a).

  1. Replace every x in f(x) by (a+2): f(a+2)=(a+2)^2-2(a+2)+4.
  2. Expand: a^2+4a+4-2a-4+4=a^2+2a+4.
  3. Replace every x by a: f(a)=a^2-2a+4.
  4. Subtract, keeping f(a) in a bracket: (a^2+2a+4)-(a^2-2a+4)=4a.

Question 16Challenge3 marks

f(x)=x^2-4x+6. Find all values of x for which f(x)=x.

Give every value, separated by commas

Hint

Set the function rule equal to the input, then factorise.

Worked solution

f(x)=x^2-4x+6. Find all values of x for which f(x)=x.

  1. f(x)=x means x^2-4x+6=x.
  2. Subtract x from both sides: x^2-5x+6=0.
  3. Factorise: (x-2)(x-3)=0.
  4. So x=2 or x=3. Check: f(2)=4-8+6=2 and f(3)=9-12+6=3.

Question 17Challenge3 marks

f(x)=ax+b, where a and b are constants. f(-2)=-9 and f(3)=11.

(a)

Find the value of a.

2 marks

(b)

Find the value of b.

1 mark

Hint

Substitute both input-output pairs and solve the resulting simultaneous equations.

Worked solution

f(x)=ax+b, where a and b are constants. f(-2)=-9 and f(3)=11.

Part (a)

Find the value of a.

  1. Substitute x=-2 and x=3 into ax+b: -2a+b=-9 and 3a+b=11.
  2. Subtract the first equation from the second to eliminate b: 5a=20.
  3. So a=4.

Part (b)

Find the value of b.

  1. Substitute a=4 into 3a+b=11: 12+b=11.
  2. So b=-1.

Question 18Challenge3 marks

f(x)=x^2+kx+7. Given that f(2)=f(5), find k.

Hint

Write an expression for each output and set them equal.

Worked solution

f(x)=x^2+kx+7. Given that f(2)=f(5), find k.

  1. Substitute x=2: f(2)=2^2+2k+7=11+2k.
  2. Substitute x=5: f(5)=5^2+5k+7=32+5k.
  3. Set them equal: 11+2k=32+5k, so -21=3k.
  4. Therefore k=-7.

Question 19Challenge3 marks

f(x)=\frac{2x+1}{x-3}, where x\ne3. Solve f(x)=5. Give your answer exactly.

Hint

Clear the denominator and collect the input terms.

Worked solution

f(x)=\frac{2x+1}{x-3}, where x\ne3. Solve f(x)=5. Give your answer exactly.

  1. The equation is 2x+1=5(x-3).
  2. Expand and collect: 16=3x.
  3. Therefore x=\frac{16}{3}.

Question 20Challenge4 marks

f(x)=3x^2-2x+1. Solve f(x+1)=f(x)+25.

Hint

Substitute the two inputs first; the squared terms will cancel.

Worked solution

f(x)=3x^2-2x+1. Solve f(x+1)=f(x)+25.

  1. Replace every x by (x+1): f(x+1)=3(x+1)^2-2(x+1)+1.
  2. Expand: 3(x^2+2x+1)-2x-2+1=3x^2+4x+2.
  3. The equation becomes 3x^2+4x+2=3x^2-2x+1+25.
  4. The 3x^2 terms cancel: 4x+2=-2x+26, so 6x=24.
  5. Therefore x=4.

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