TMUA: Functions
Functions: what a function does to its inputs, and what values it can output.
- Section 1 Part 1 · MM1 Algebra and functions · spec MM1.7
- Papers 1 and 2
- 7 practice questions
- No calculator
What the specification covers
- Many-to-one and one-to-one mappings
- √x always means the positive root; |x| as a function
- Domain and range questions without calculus
Key ideas
- A function can be many-to-one (like \(x^2\)) or one-to-one (like \(x^3\)).
- \(\sqrt{x}\) always means the positive square root, so \(\sqrt{x}\ge0\).
- To find a range, complete the square inside: \(x^2-6x+13=(x-3)^2+4\ge4\).
- \(|x|\) is the distance from 0: \(|x|=x\) for \(x\ge0\) and \(-x\) for \(x<0\).
Common mistakes
- \(\sqrt{x^2}=|x|\), not \(x\).
- The range of a composite depends on the range of the inner function, not its domain.
Exam tip
Sketch the function quickly: the range is what the graph covers vertically.
Worked example
Worked example
The function \(f\) is defined for all real \(x\) by \(f(x)=2-\sqrt{x^2-6x+13}\). What is the range of \(f\)?
- \(f(x) \le 0\)
- all real numbers
- \(f(x) \le 2\)
- \(f(x) \ge 0\)
- \(-2 \le f(x) \le 2\)
Answer: A: \(f(x) \le 0\)
\(x^2-6x+13=(x-3)^2+4\ge 4\), with equality at \(x=3\), and it takes every value \(\ge4\). Its (positive) square root takes every value \(\ge2\), so \(f(x)=2-\sqrt{\cdots}\) takes every value \(\le 0\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
What is the greatest value taken by \(f(x)=6-\sqrt{x^2-2x+5}\) as \(x\) varies over the real numbers?
- \(-4\)
- \(8\)
- \(4\)
- \(2\)
- \(6\)
Show the answer and solution
Answer: C: \(4\)
\(x^2-2x+5=(x-1)^2+4\ge4\), with equality at \(x=1\). The positive square root is at least \(2\), so \(f(x)\le6-2=4\), attained at \(x=1\).
Question 2
What is the greatest value taken by \(f(x)=8-\sqrt{x^2+4x+8}\) as \(x\) varies over the real numbers?
- \(-6\)
- \(6\)
- \(8\)
- \(10\)
- \(4\)
Show the answer and solution
Answer: B: \(6\)
\(x^2+4x+8=(x+2)^2+4\ge4\), with equality at \(x=-2\). The positive square root is at least \(2\), so \(f(x)\le8-2=6\), attained at \(x=-2\).
Question 3
What is the greatest value taken by \(f(x)=7-\sqrt{x^2-6x+25}\) as \(x\) varies over the real numbers?
- \(3\)
- \(11\)
- \(-3\)
- \(7\)
- \(-9\)
Show the answer and solution
Answer: A: \(3\)
\(x^2-6x+25=(x-3)^2+16\ge16\), with equality at \(x=3\). The positive square root is at least \(4\), so \(f(x)\le7-4=3\), attained at \(x=3\).
Question 4
What is the greatest value taken by \(f(x)=6-\sqrt{x^2-2x+17}\) as \(x\) varies over the real numbers?
- \(10\)
- \(-10\)
- \(-2\)
- \(2\)
- \(6\)
Show the answer and solution
Answer: D: \(2\)
\(x^2-2x+17=(x-1)^2+16\ge16\), with equality at \(x=1\). The positive square root is at least \(4\), so \(f(x)\le6-4=2\), attained at \(x=1\).
Question 5
What is the greatest value taken by \(f(x)=8-\sqrt{x^2-2x+10}\) as \(x\) varies over the real numbers?
- \(5\)
- \(8\)
- \(-5\)
- \(-1\)
- \(11\)
Show the answer and solution
Answer: A: \(5\)
\(x^2-2x+10=(x-1)^2+9\ge9\), with equality at \(x=1\). The positive square root is at least \(3\), so \(f(x)\le8-3=5\), attained at \(x=1\).
Question 6
What is the greatest value taken by \(f(x)=7-\sqrt{x^2-2x+5}\) as \(x\) varies over the real numbers?
- \(5\)
- \(7\)
- \(9\)
- \(-5\)
- \(3\)
Show the answer and solution
Answer: A: \(5\)
\(x^2-2x+5=(x-1)^2+4\ge4\), with equality at \(x=1\). The positive square root is at least \(2\), so \(f(x)\le7-2=5\), attained at \(x=1\).
Keep going
- Previous topic: Polynomials, Factor and Remainder Theorems
- Next topic: Sequences and recurrence relations
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