Range

Section 1.6 Range

The set of all possible output of \(f(x)\) as \(x\) runs over the domain is called a range of the function. The same \(x\) always leads to the same \(f(x).\) In other words, the set of all possible output values \(y\) for which the \(f(x)\) is defined.

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Range of \(f={f(x):x \in \text{ domain of } \, f} \)

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Steps to Find the Range:

Identify the domain: Determine the set of all valid inputs.
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Analyze the function: Determine what output values are possible by applying the function to all inputs in the domain.
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Express the range: Describe the set of all possible outputs, often using interval notation or set notation.
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Example 1.6.1.

Example 1: Linear Function

\begin{equation*} f(x) = 2x + 3 \end{equation*}

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Solution.

Domain: All real numbers, \(x \in \mathbb{R}\) for any polynomials.

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Finding the Range: let

\begin{equation*} f(x) = y= 2x+3 \end{equation*}

Now, obtain \(x\) interms of \(y\) as

\begin{equation*} y-3 =2x \, \Rightarrow \, x= \frac{y-3}{2} \end{equation*}

Since, this is also a polynomials, \(x \) ranges over all real numbers.

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Range: All real numbers, \((-\infty, \infty).\)

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Example 1.6.2.

Quadratic Function:

\begin{equation*} f(x) = x^2 \end{equation*}

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Solution.

Domain: All real numbers, \(x \in \mathbb{R}\text{.}\)

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Finding the Range:

\begin{equation*} y=x^2 \, \Rightarrow \, x = \sqrt{y} \end{equation*}

Since

\begin{equation*} y\geq 0 \end{equation*}

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Range: All non-negative real numbers, \([0, \infty)\text{.}\)

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Example 1.6.3.

Square Root Function:

\begin{equation*} f(x) = \sqrt{x} \end{equation*}

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Solution.

Domain: \(x \geq 0\) (since the square root is undefined for negative numbers in the real numbers), \([0, \infty)\text{.}\)

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Finding the Range:

\begin{equation*} y=\sqrt{x} \, \Rightarrow \, x = y^2 \end{equation*}

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For \(x = 0\text{,}\) \(f(0) = \sqrt{0} = 0\text{.}\)

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For \(x = 1\text{,}\) \(f(1) = \sqrt{1} = 1\text{.}\)

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For \(x = 4\text{,}\) \(f(4) = \sqrt{4} = 2\text{.}\)

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As \(x\) increases, \(\sqrt{x}\) produces larger non-negative numbers. The output is always non-negative since the square root of a non-negative number is non-negative.

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Range: All non-negative real numbers, \([0, \infty)\text{.}\)

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Example 1.6.4.

Constant Function:

\begin{equation*} f(x) = 5 \end{equation*}

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Solution.

Domain: All real numbers, \(x \in \mathbb{R}\text{.}\)

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Finding the Range: The function always outputs 5, regardless of the input ( \(x\) ).

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Range: \(\{5\}\) (a single value, as the output is constant).

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Example 1.6.5.

Rational Function:

\begin{equation*} f(x) = \frac{1}{x} \end{equation*}

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Solution.

Domain: All real numbers except \(x = 0\text{,}\) i.e., \(x \in \mathbb{R} \setminus \{0\}\text{.}\)

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Finding the Range: We need to determine what values \(y = \frac{1}{x}\) can take. Solve for ( \(x\) ) in terms of ( \(y \)): \(y = \frac{1}{x} \implies x = \frac{1}{y}\text{.}\) For ( \(x\) ) to be defined, \(y \neq 0\) (since \(x = \frac{1}{y}\) is undefined when \(y = 0\)). For any \(y \neq 0\text{,}\) there exists an \(x = \frac{1}{y}\) (which is not zero) such that \(f(x) = y\text{.}\) Thus, the range includes all real numbers except \(0\text{.}\)

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Range: \(\mathbb{R} \setminus \{0\}\) or \((-\infty, 0) \cup (0, \infty)\text{.}\)

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Example 1.6.6.

Sine Function:

\begin{equation*} f(x) = \sin(x) \end{equation*}

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Solution.

Domain: All real numbers, \(x \in \mathbb{R}\text{.}\)

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Finding the Range: The sine function oscillates between \(-1\) and \(1\) for all ( \(x\) ). For example, \(\sin(0) = 0\text{,}\) \(\sin(\frac{\pi}{2}) = 1\text{,}\) \(\sin(\frac{3\pi}{2}) = -1\text{.}\)

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Range: \([-1, 1]\text{.}\)

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Example 1.6.7.

Find the domain and range of the following functions.

\begin{equation*} y=x^2+2 \end{equation*}

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\begin{equation*} y=\sqrt{x-3} \end{equation*}

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\begin{equation*} y=\frac{1}{x^2+1} \end{equation*}

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\begin{equation*} y=2^x-3 \end{equation*}

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\begin{equation*} y=\sin x+1 \end{equation*}

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\begin{equation*} y=|x-4|+1 \end{equation*}

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\begin{equation*} y=\ln (x+5) \end{equation*}

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\begin{equation*} y=\begin{cases} x+2, & \, x\lt 0 \\ 4, & \, 0\leq x\lt 3 \\ x^2, & \, x\gt 3 \end{cases} \end{equation*}

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Answer.

\begin{equation*} x\in (-\infty, \infty) \end{equation*}

\begin{equation*} y\in [2, \infty) \end{equation*}

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\begin{equation*} x\in [3, \infty) \end{equation*}

\begin{equation*} y\in [0, \infty) \end{equation*}

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\begin{equation*} x\in (-\infty, \infty) \end{equation*}

\begin{equation*} y\in (0, 1] \end{equation*}

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\begin{equation*} x\in (-\infty, \infty) \end{equation*}

\begin{equation*} y\in [-3,\infty) \end{equation*}

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\begin{equation*} x\in (-\infty, \infty) \end{equation*}

\begin{equation*} y\in [0,2] \end{equation*}

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\begin{equation*} x\in (-\infty, \infty) \end{equation*}

\begin{equation*} y\in [1,\infty) \end{equation*}

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\begin{equation*} x\in (-5, \infty) \end{equation*}

\begin{equation*} y\in (-\infty,\infty) \end{equation*}

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\begin{equation*} x\in (-\infty, \infty) \end{equation*}

\begin{equation*} y\in (-\infty,2)\cup {4} \cup (9,\infty) \end{equation*}

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Example 1.6.8.

Range of

\begin{equation*} y=\frac{x-1}{x+1} \end{equation*}

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Solution.

Convert this equation to get \(x\) in terms of \(y\)

\begin{equation*} xy+y = x-1 \end{equation*}

\begin{equation*} x = \frac{y+1}{1-y} \, \Rightarrow \, y \neq 1 \end{equation*}

Since \(x\in\mathbb{R}\text{.}\) Hence \(y\in \mathbb{R}\setminus \{1\}\)

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Example 1.6.9.

Range of

\begin{equation*} y=\frac{1}{x-1} \end{equation*}

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Solution.

Convert this equation to get \(x\) in terms of \(y\)

\begin{equation*} xy-y = 1 \end{equation*}

\begin{equation*} x = \frac{1+y}{y} \, \Rightarrow \, y \neq 0 \end{equation*}

Since \(x\in\mathbb{R}\text{.}\) Hence \(y\in \mathbb{R}\setminus \{0\}\)

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Example 1.6.10.

Find the range of

\begin{equation*} y = 2\sin x +1 \end{equation*}

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\begin{equation*} y = \frac{2}{3-2\sin x} \end{equation*}

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\begin{equation*} y = \frac{2}{3-2\{x\}} \, \text{ where } \, \{\cdot \} \, \text {is fractional part function } \end{equation*}

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Solution.

\begin{align*} -1 \amp \leq \sin x \leq 1 \\ -2 \amp \leq 2\sin x \leq 2 \\ -1 \amp \leq 2\sin x +1 \leq 3 \end{align*}

Hence, the Range: \([-1,3]\)

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\begin{align*} -1 \amp \leq \sin x \leq 1 \\ 2 \amp \geq -2\sin x \geq -2 \\ 5 \amp \geq 3-2\sin x \geq 1 \\ \frac{1}{5} \amp \leq \frac{1}{3-2\sin x } \leq 1 \\ \frac{2}{5} \amp \leq \frac{2}{3-2\sin x } \leq 2 \end{align*}

Hence, the Range: \([\frac{2}{5},2]\)

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\begin{align*} 0 \amp \leq \{x\} \lt 1 \\ 0 \amp \geq -2 \{x\} \gt -2 \\ 3 \amp \geq 3-2 \{x\} \gt 1 \\ \frac{1}{3} \amp \leq \frac{1}{3-2 \{x\}} \lt 1 \\ \frac{2}{3} \amp \leq \frac{2}{3-2 \{x\} } \lt 2 \end{align*}

Hence, the Range: \([\frac{2}{3},2]\)

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Example 1.6.11.

Find the range of following:

\begin{equation*} y=\sqrt{x^2+4} \end{equation*}

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\begin{equation*} y=\sqrt{4-x^2} \end{equation*}

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\begin{equation*} y=\sqrt{x} \end{equation*}

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Solution.

\begin{equation*} y=\sqrt{x^2+4} \end{equation*}

\begin{align*} x^2+4 \amp \geq 4 \, \text{ so domain: } \, (-\infty,\infty)\\ or, \, 2 \amp \leq \sqrt{x^2 +4} \lt \infty \end{align*}

Hence the range: \([2, \infty)\)

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\begin{equation*} y=\sqrt{4-x^2} \end{equation*}

\begin{align*} -2 \amp \leq \sqrt{4-x^2} \leq 2 \, \text{ so domain: } \, (-2, 2] \end{align*}

Hence the range: \([0, 2]\)

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\begin{equation*} y=\sqrt{x} \end{equation*}

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Example 1.6.12.

Find the range of following quadratic equations:

\begin{equation*} y= x^2+x+1 \end{equation*}

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\begin{equation*} y= -x^2-2x+3 \end{equation*}

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\begin{equation*} y= x^2+2x+4; \, 1\leq x\leq 3 \end{equation*}

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\begin{equation*} y= x^2+2x+4; \, -3\leq x\leq -2 \end{equation*}

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\begin{equation*} y= 2x^2+x+1; \, -1\leq x\leq 1 \end{equation*}

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Solution.

For a quadratic equation

\begin{equation*} y=ax^2+bx+c; \, \quad \text{Range:}\, \left[ \left. -\frac{D}{4a}, \infty\right)\right. \, \text{ for } \, a \gt 0. \end{equation*}

here \(D=b^2-4ac; \, \) is a discriminant.

\begin{align*} y \amp = x^2+x+1 \\ D \amp = 1^2-4\times 1\times 1 = -3 \end{align*}

Hence, the range: \(\left[ \left. \frac{3}{4}, \infty\right)\right.\)

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For a quadratic equation

\begin{equation*} y=-ax^2-bx+c; \, \quad \text{Range:}\, \left. \left( -\infty, -\frac{D}{4a} \right. \right]\, \text{ for } \, a \lt 0. \end{equation*}

here \(D=b^2-4ac; \, \) is a discriminant.

\begin{align*} y \amp = -x^2-2x+3 \\ D \amp = 4+4\times 1\times 3 = 16 \end{align*}

Hence, the range: \(\left. \left( -\infty, 4 \right. \right]\)

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Find the

\begin{equation*} Vertex = \left(-\frac{b}{2a}, -\frac{D}{4a}\right) \end{equation*}

Now find the value of function corresponding to the vertex and at \(x=0.\) These value give the range of these functions.

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Example 1.6.13.

Find the range of \(2^x+2^{-x}\) and \(2^x+2^{-x}+3^x+3^{-x}\text{.}\)

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Solution.

Arithmatic Mean (AM) \(\geq \) Geometric Mean (GM). Therefore

\begin{equation*} \frac{2^x+2^{-x}}{2}\geq (2^x\times 2^{-x})^{1/2} \end{equation*}

\begin{equation*} \Rightarrow \, (2^x+2^{-x}) \geq 2 \times (1)^{1/2} \end{equation*}

\begin{equation*} \therefore \, (2^x+2^{-x}) \geq 2 \end{equation*}

so the range is \([2,\infty)\)

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find another by yourself.

sum of the number and its reciprocal is always \(\geq 2\text{.}\)

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Example 1.6.14.

Find the range of

\begin{equation*} y=2^{\sin x} \end{equation*}

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\begin{equation*} y=5^{(x^2+1)} \end{equation*}

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\begin{equation*} y=\left(\frac{1}{2}\right)^{(2\sin x+2)} \end{equation*}

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\begin{equation*} y=\log_{10}{(2\sin x+10)} \end{equation*}

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\begin{equation*} y=\log_{10}{(x^2+1)} \end{equation*}

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\begin{equation*} y=\log_{10}{(x^2-5x+6)} \end{equation*}

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Solution.

For

\begin{equation*} y=2^{\sin x} \end{equation*}

let, \(\sin x =t\) and we know that

\begin{align*} -1 \amp \leq \sin x\leq 1 \\ -1 \amp \leq t \leq 1 \\ 2^{-1} \amp \leq 2^{t} \leq 2^{1} \\ \frac{1}{2} \amp \leq 2^{t} \leq 2 \end{align*}

Range: \([\frac{1}{2},2]\)

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For

\begin{equation*} y=5^{(x^2+1)} \end{equation*}

let, \(x^2+1 =t\) and we know that

\begin{align*} 1 \amp \leq t \lt \infty \\ 5^{1} \amp \leq 5^{t} \lt \infty \\ 5 \amp \leq 5^{t} \lt \infty \end{align*}

Range: \([5,\infty)\)

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For

\begin{equation*} y=\left(\frac{1}{2}\right)^{(2\sin x+2)} \end{equation*}

let, \(2\sin x+2 =t\) and we know that

\begin{align*} -1 \amp \leq \sin x\leq 1 \\ -2 \amp \leq 2\sin x \leq 2 \\ 0 \amp \leq 2\sin x+2 \leq 4 \\ 0 \amp \leq t \leq 4 \\ \left(\frac{1}{2}\right)^{0} \amp \geq \left(\frac{1}{2}\right)^{t} \geq \left(\frac{1}{2}\right)^{4} \\ 1 \amp \geq \left(\frac{1}{2}\right)^{t} \geq \frac{1}{16} \end{align*}

Range: \([\frac{1}{16},1]\)

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