Question Statement

Find the domain and range of the following functions:

(i) $f(x)=x+7$ (ii) $f(x)=2x^2+1$ (iii) $f(x)=2\sqrt{x-5}$ (iv) $f(x)=|x-2|-3$ (v) $f(x)=1+\sin x$ (vi) $f(x)=3+\sqrt{x-2}$ (vii) $f(x)=\frac{3e^x}{7}$ (viii) $f(x)=\frac{x^2-16}{x+4}$ (ix) $f(x)=(x-1{)}^2+1$ (x) $f(x)=\frac{1}{x-1}$ (xi) $f(x)=\frac{x-2}{x+3}$ (xii) $f(x)=\frac{x^2-x-6}{x-3}$

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Background and Explanation

The domain of a function is the set of all possible input values ($x$) for which the function is defined and produces a real number. The range is the set of all possible output values ($f(x)$) that result from the domain. Common restrictions include ensuring denominators are not zero and expressions under square roots are non-negative.

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Solution

Part (i)

Function: $f(x)=x+7$

Since $f(x)$ is a linear polynomial, it is defined for all real values of $x$.

• Domain: $\mathbb{R}$

After substituting any real value of $x$, the resulting values of $f(x)$ are also real numbers covering the entire number line.

• Range: $\mathbb{R}$

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Part (ii)

Function: $f(x)=2x^2+1$

Since $f(x)$ is a polynomial, it is defined for all real values of $x$.

• Domain: $\mathbb{R}$

To find the range, note that $x^2\ge 0$ for all real $x$. Thus, the minimum value occurs at $x=0$, where $f(0)=2(0{)}^2+1=1$. As $x$ increases or decreases, $f(x)$ increases toward infinity.

• Range: $[1,+\infty )$

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Part (iii)

Function: $f(x)=2\sqrt{x-5}$

The square root function is only defined when the expression inside is non-negative: $x-5\ge 0\Rightarrow x\ge 5$

• Domain: $[5,+\infty )$

Since $\sqrt{x-5}\ge 0$ for all $x\ge 5$, multiplying by 2 maintains this inequality: $2\sqrt{x-5}\ge 0$

• Range: $[0,\infty )$

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Part (iv)

Function: $f(x)=|x-2|-3$

The absolute value function is defined for all real values of $x$.

• Domain: $\mathbb{R}$

The minimum value of the absolute value part $|x-2|$ is zero (at $x=2$). Otherwise, it is always positive. Therefore, the minimum value of $f(x)$ is $0-3=-3$, and it extends to $+\infty$.

• Range: $[-3,\infty )$

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Part (v)

Function: $f(x)=1+\sin x$

The sine function is defined for all real numbers.

• Domain: $\mathbb{R}$

We know the bounds of the sine function: $-1\le \sin x\le 1$ Adding 1 to all parts of the inequality: $1-1\le 1+\sin x\le 1+1$ $0\le 1+\sin x\le 2$ The minimum value is 0 and the maximum value is 2.

• Range: $[0,2]$

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Part (vi)

Function: $f(x)=3+\sqrt{x-2}$

The expression $\sqrt{x-2}$ is defined for $x-2\ge 0$, which means $x\ge 2$.

• Domain: $[2,+\infty )$

The minimum value of $\sqrt{x-2}$ is 0 (at $x=2$). Thus, the minimum value of $f(x)$ is $3+0=3$, and it increases to $+\infty$.

• Range: $[3,+\infty )$

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Part (vii)

Function: $f(x)=\frac{3e^x}{7}$

The exponential function $e^x$ is defined for all real numbers.

• Domain: $\mathbb{R}$

Since $e^x>0$ for all real numbers $x$, multiplying by the positive constant $\frac{3}{7}$ keeps the result strictly greater than 0.

• Range: $(0,+\infty )$

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Part (viii)

Function: $f(x)=\frac{x^2-16}{x+4}$

The function is undefined when the denominator is zero: $x+4=0\Rightarrow x=-4$. At this point, the function results in $\frac{0}{0}$, which is indeterminate.

• Domain: $\mathbb{R}-\{-4\}$

To find the range, simplify the function for $x\ne -4$: $f(x)=\frac{(x-4)(x+4)}{(x+4)}=x-4$ If $x$ could be $-4$, $f(-4)$ would be $-4-4=-8$. Since $-4$ is excluded from the domain, $-8$ must be excluded from the range.

• Range: $\mathbb{R}-\{-8\}$

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Part (ix)

Function: $f(x)=(x-1{)}^2+1$

The function is defined for all real values of $x$.

• Domain: $\mathbb{R}$

The squared term $(x-1{)}^2$ is always $\ge 0$. Therefore, the minimum value of $f(x)$ is $0+1=1$, and the maximum value is $+\infty$.

• Range: $[1,+\infty )$

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Part (x)

Function: $f(x)=\frac{1}{x-1}$

The function is undefined when $x-1=0$, which means $x=1$.

• Domain: $\mathbb{R}-\{1\}$

To find the range, let $y=f(x)$ and solve for $x$: $y=\frac{1}{x-1}$ $xy-y=1$ $xy=y+1$ $x=\frac{y+1}{y}$ The expression for $x$ is undefined when $y=0$.

• Range: $\mathbb{R}-\{0\}$

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Part (xi)

Function: $f(x)=\frac{x-2}{x+3}$

The function is undefined when $x+3=0$, which means $x=-3$.

• Domain: $\mathbb{R}-\{-3\}$

To find the range, let $y=f(x)$ and solve for $x$: $y=\frac{x-2}{x+3}$ $y(x+3)=x-2$ $xy+3y=x-2$ $3y+2=x-xy$ $x(1-y)=3y+2$ $x=\frac{3y+2}{1-y}$ The expression for $x$ is undefined when $1-y=0$, which means $y=1$.

• Range: $\mathbb{R}-\{1\}$

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Part (xii)

Function: $f(x)=\frac{x^2-x-6}{x-3}$

The function is undefined when $x-3=0$, which means $x=3$.

• Domain: $\mathbb{R}-\{3\}$

Simplify the function by factoring the numerator: $f(x)=\frac{x^2-3x+2x-6}{x-3}$ $f(x)=\frac{x(x-3)+2(x-3)}{(x-3)}$ $f(x)=\frac{(x+2)(x-3)}{(x-3)}=x+2(\text{for }x\ne 3)$ If $x$ could be $3$, $f(3)=3+2=5$. Since $x=3$ is excluded from the domain, $y=5$ is excluded from the range.

• Range: $\mathbb{R}-\{5\}$

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Key Formulas or Methods Used

• Rational Functions: Domain is $\mathbb{R}$ excluding values where the denominator is zero.
• Square Root Functions: Domain requires the radicand $\ge 0$.
• Absolute Value: $|x|\ge 0$ for all $x$.
• Trigonometric Bounds: $-1\le \sin x\le 1$.
• Exponential Functions: $e^x>0$ for all real $x$.
• Range via Inverse: Solving $y=f(x)$ for $x$ to find restrictions on $y$.
• Holes in Graphs: For functions like $\frac{g(x)}{h(x)}$ where a factor cancels, the excluded domain value creates an excluded range value.

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Summary of Steps

1. Identify Domain Restrictions: Look for denominators, square roots, or logs to find where the function is undefined.
2. Simplify the Function: For rational functions, factor and cancel terms to identify "holes" (removable discontinuities).
3. Analyze Function Behavior: Use known bounds of functions (like $\sin x$, $x^2$, or $|x|$) to determine the minimum and maximum output values.
4. Solve for $x$ (Algebraic Method): Set $y=f(x)$, rearrange to express $x$ in terms of $y$, and identify values of $y$ that make the expression undefined.
5. State Results: Write the domain and range clearly using interval notation or set-builder notation.