Domain and Range of Functions Worked ExamplesIGCSE Maths 0580

(a) Nothing restricts the input — there is no division and no root, so any real number works.

(b) As $x$ runs over all reals, $2x + 3$ also takes every real value.

(c) Substitute.

(a) A square root is only defined when what is inside it is not negative.

Solve the inequality.

(b) A square root never returns a negative value, and it reaches $0$ when $x = 2$.

(a) Division by zero is undefined, so exclude the value making the denominator zero.

(b) A fraction is zero only when its numerator is zero, and here the numerator is the constant $1$.

So the range excludes zero, giving a horizontal asymptote at $y = 0$.

A quadratic has no domain restriction, but its range is limited by the turning point.

Complete the square to locate the vertex.

A square is never negative, so the least possible value is $-1$, reached at $x = 2$.

The $x^2$ coefficient is positive, so the curve opens upwards and rises without limit.

Evaluate at both endpoints of the domain.

Check whether the turning point lies inside the domain. It is at $x = 0$, which does.

The smallest output is the minimum at $x = 0$; the largest is the greater endpoint value.

Two separate conditions apply. First, the expression under the root must not be negative.

Second, the denominator must not be zero.

Both must hold simultaneously, so take the values satisfying the first and remove the excluded one.

(a) The range is the set of outputs that are actually reached. Arrows land on $0$, $1$ and $4$ only — nothing reaches $2$ or $3$.

(b) Use the arrow from $1$ to $0$.

Check against two more arrows: $f(3) = (3 - 1)^2 = 4$ ✓ and $f(-1) = (-2)^2 = 4$ ✓.

(c) Two different inputs, $-1$ and $3$, both map to $4$ (and $0$ and $2$ both map to $1$). An inverse would have to send $4$ back to two different values, and a function cannot have two outputs for one input.

(d) Keep only one input from each matching pair, so that every output comes from exactly one input.

(a) The curve falls from left to right, so the largest output is at the left end of the domain and the smallest at the right end. Read both from the graph.

Write the range in terms of $g(x)$.

(b) Draw the line $y = 3$. It meets the curve at $x = 2$. Check algebraically.

(c)(i) As $x$ increases the curve gets closer and closer to the $x$-axis but never reaches it: $g(x) = 0$ would need the numerator $12$ to be $0$. So $0$ is not in the range, and Priya's $\leqslant$ at $0$ is wrong.

(ii) The largest value, $6$, is still reached at $x = 0$, so that end keeps $\leqslant$.

(a) On $-1 \leqslant x \leqslant 1$ the curve is falling the whole way — the turning point at $x = 2$ is outside this domain — so the range runs between the two end values.

Write the range.

(b)(i) The lowest point of the curve is $(2, -3)$, and $f$ only takes the value $-3$ at $x = 2$. For $-3$ to be in the range, the domain must reach $x = 2$, so $k \geqslant 2$.

(ii) To the right of $x = 2$ the curve rises, and the range must not go above $6$. So $k$ is largest when $f(k) = 6$.

Factorise. $k = -1$ is the left end of the domain, so reject it. The graph confirms the curve is back at height $6$ when $x = 5$.

(iii) Any $k$ from $2$ to $5$ works: the minimum $-3$ is included, and the greatest value stays at $6$, reached at $x = -1$.

(a) The range is the set of outputs, so substitute each value in the domain into $f(x)$.

Write the outputs as a set, in the same brackets the domain uses.

(b) Substitute each of the five values. Square first, then multiply by $2$: $(-2)^2 = 4$, not $-4$.

Five inputs give only three different outputs. In a set, each value is listed once.