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Functions - Domain and range

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Domain of a function is the complete set of possible values for the independent variable (xx), which make the function 'work' and will output real yy-values. Graphically, it is the 'width' of the function along the xx-axis.

Graph of y = sqrt(x+2) showing the domain starts from x = -2 and extends to the right.
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The Range of a function is the complete set of all possible resulting values of the dependent variable (yy), after we have substituted the domain. Graphically, it is the 'height' or vertical spread of the function.

Graph of y = x^2 + 1 showing the range starts from the vertex at y = 1 and goes upwards.
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For rational functions f(x)=P(x)Q(x)f(x) = \frac{P(x)}{Q(x)}, the domain excludes any values where the denominator Q(x)=0Q(x) = 0. These exclusions often appear as vertical asymptotes on a graph.

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For square root functions f(x)=g(x)f(x) = \sqrt{g(x)}, the domain is restricted to values where the radicand is non-negative: g(x)≥0g(x) \geq 0. This ensures the output remains within the set of real numbers.

📐Formulae

f(x)=yf(x) = y

For f(x)=P(x)Q(x), Domain: {x∈R∣Q(x)≠0}\text{For } f(x) = \frac{P(x)}{Q(x)}, \text{ Domain: } \{x \in \mathbb{R} \mid Q(x) \neq 0\}

For f(x)=g(x), Domain: {x∈R∣g(x)≥0}\text{For } f(x) = \sqrt{g(x)}, \text{ Domain: } \{x \in \mathbb{R} \mid g(x) \geq 0\}

Range of f(x)=x2+k⇒y∈[k,∞) (if coefficient of x2>0)\text{Range of } f(x) = x^2 + k \Rightarrow y \in [k, \infty) \text{ (if coefficient of } x^2 > 0)

Range of f(x)=ax (where a>0)⇒y∈(0,∞)\text{Range of } f(x) = a^x \text{ (where } a > 0) \Rightarrow y \in (0, \infty)

💡Examples

Problem 1:

Determine the domain of the function f(x)=5x−7f(x) = \frac{5}{x - 7}.

Solution:

x−7≠0  ⟹  x≠7x - 7 \neq 0 \implies x \neq 7 Domain: {x∈R∣x≠7}\{x \in \mathbb{R} \mid x \neq 7\}

Explanation:

Since division by zero is undefined, the denominator x−7x - 7 cannot be equal to 00. Solving for xx gives the restriction.

Problem 2:

Find the range of the function f(x)=x2+4f(x) = x^2 + 4 for the domain x∈Rx \in \mathbb{R}.

Solution:

x2≥0x^2 \geq 0 x2+4≥4x^2 + 4 \geq 4 Range: [4,∞)[4, \infty) or y≥4y \geq 4

Explanation:

The smallest value of x2x^2 is 00 (when x=0x=0). Therefore, the smallest value the function can take is 0+4=40 + 4 = 4. The graph opens upwards, so all values above 44 are included.

Problem 3:

A ball is thrown in the air and its height hh in meters after tt seconds is given by h(t)=−5t2+20t+2h(t) = -5t^2 + 20t + 2. If the ball hits the ground at t≈4.1t \approx 4.1 seconds, find the practical domain and range.

Solution:

Domain: [0,4.1][0, 4.1] To find the maximum height (Range), find the vertex: t=−b2a=−202(−5)=2t = -\frac{b}{2a} = -\frac{20}{2(-5)} = 2 h(2)=−5(2)2+20(2)+2=−20+40+2=22h(2) = -5(2)^2 + 20(2) + 2 = -20 + 40 + 2 = 22 Range: [0,22][0, 22]

Explanation:

In a real-world context, time tt cannot be negative and stops when the ball hits the ground (h=0h=0). The range starts from the ground (00) up to the maximum height (the yy-coordinate of the vertex).

Problem 4:

Identify the domain and range of the function f(x)=1x−2+3f(x) = \frac{1}{x - 2} + 3 based on its graph and algebraic properties.

Graph of y = 1/(x-2) + 3 showing asymptotes at x=2 and y=3.

Solution:

  1. Domain: The denominator cannot be zero. x−2≠0⇒x≠2x - 2 \neq 0 \Rightarrow x \neq 2. Thus, Domain is {x∈R∣x≠2}\{x \in \mathbb{R} \mid x \neq 2\}.
  2. Range: As x→∞x \to \infty or x→−∞x \to -\infty, the fraction 1x−2→0\frac{1}{x-2} \to 0, so f(x)→3f(x) \to 3. The function never actually reaches the value y=3y = 3. Thus, Range is {y∈R∣y≠3}\{y \in \mathbb{R} \mid y \neq 3\}.

Explanation:

The graph of a reciprocal function has vertical and horizontal asymptotes. The vertical asymptote at x=2x=2 defines the exclusion in the domain, and the horizontal asymptote at y=3y=3 defines the exclusion in the range.

Problem 5:

A restricted function is defined as f(x)=4−x2f(x) = 4 - x^2 for the domain −1≤x≤2-1 \leq x \leq 2. Find the range of this function.

Graph of y = 4 - x^2 restricted to the domain [-1, 2].

Solution:

  1. Evaluate the function at the boundaries and vertex:
  • At x=−1x = -1, f(−1)=4−(−1)2=3f(-1) = 4 - (-1)^2 = 3.
  • At x=0x = 0 (the vertex), f(0)=4−02=4f(0) = 4 - 0^2 = 4.
  • At x=2x = 2, f(2)=4−22=0f(2) = 4 - 2^2 = 0.
  1. Identify the minimum and maximum yy-values: The highest point is at the vertex (y=4y=4) and the lowest point within the interval is at x=2x=2 (y=0y=0).
  2. Range is [0,4][0, 4].

Explanation:

Since the vertex x=0x=0 lies within the domain [−1,2][-1, 2], the maximum value of the range is the yy-coordinate of the vertex. The minimum value is the lower of the two endpoint values.