# Francesco Giannino » 4.Domain of a given function

### Domain of Basic Functions

A table of domain of basic and useful functions is shown.

### Domain of Basic Functions

A table of domain of basic and useful functions is shown.

### Domain of Basic Functions

A table of domain of basic and useful functions is shown.

### Domain of functions resulting from algebraic operation

Let us consider two real functions f and g with domains D1 and D2 respectively. Clearly, these domains are real number set R or subsets of R : $D_{1},D_{2}\in R$

The addition and subtraction of two real functions are defined as follows $f\pm g: D_{1}\cap D_{2}\rightarrow R$

Multiplication
The product of two real functions is defined as follows: $f g: D_{1}\cap D_{2}\rightarrow R$

### Domain of functions resulting from algebraic operation

Quotient
The quotient of two real functions involves rational form as f(x)/g(x), which is defined for g(x) ≠ 0. We need to exclude values of x for which g(x) is zero. Hence, the quotient of two real functions is defined as follows: $\frac{f}{g}: D_{1}\cap D_{2}-\{x|g(x)\neq 0\}\rightarrow R$

### Exercises

Exercise 1. Let two functions be defined as follows: $f(x)=\sqrt{x}, \, g(x)=x^{2}-5x+6$

Find domains of fg and f/g.

Solution. The function f(x) is defined for all non negative real numbers. Hence, its domain is: x ≥0
The function, g(x), being a real quadratic polynomial, is defined for all real values of x. Hence, its domain is R.

The domain of fg is the intersection of two domains, which is a non-negative interval.

### Exercises

The domain of quotient function f/g excludes values of x for which g(x) is zero. In other words, we have to exclude roots of g(x) from domain. Then: $g(x)=x^{2}-5x+6=0\Rightarrow g(x)=(x-2)(x-3)=0$ $\Rightarrow x_{1,2}=2,3$

Hence domain of f/g is intersection of two domains, without two roots of g: $[0,+\infty[-\{2,3\}$

### Exercises

Exercise 2. Find the domain of the function $f(x)= \sqrt{x-1}+\sqrt{x^{2}+x+1}=f_{1}+f_{2}$

Solution. Given function can be considered to be the addition of two separate functions. Domain of the first is: $x-1\geq 0\Rightarrow x\geq 1$
Now, for the second function, we use the sign rule: $x^{2}+x+1\geq 0$

In this case, coefficient of x2 is positive and discreminant Δ is negative. Hence, the second function is positive for all real x.
Domain of given function is intersection of two domains: [1, +∞[

### Exercises

Exercise 3. Find the domain of the function given by $f(x)=\log_{10}\frac{x^{2-5x+6}}{x^{2}+5x+9}$

Solution. The argument of logarithmic function is a rational function. We need to find values of x such that the argument of the function evaluates to a positive number. Hence, we put: $\frac{x^{2}-5x+6}{x^{2}+5x+9}>0$

The quadratic expression in the denominator has no real roots and as such can not be factorized in linear factors. Thus, the quadratic expression in the denominator is positive for all values of x as the coefficient of the squared term is positive.

### Exercises

Clearly, sign of rational function is the same as that of the quadratic expression in the numerator: $x^{2}+5x+9>0$
The roots of the quadratic expression in the numerator (see exercise 1) are: $x_{1,2}=2,3$

And the quadratic expression in the numerator evaluates to positive for intervals beyond root values: $x<2 \, \text{or} \,x>3$

Then domain of the given function is: $]-\infty,2[\cup ]3,+\infty[$ and it is sketched below.

### Exercises

Exercise 4. Find the domain of the function given by $f(x)=\sqrt{\log_{10}\frac{6x-x^{2}}{8}}$

Solution. The function is a square root of a logarithmic function. On the other hand, argument of logarithmic function is a rational function. In order to find the domain of the given function, we first determine what values of x are valid for logarithmic function. Then, we apply the condition that expression within the square root should be a non-negative number. The domain of the given function is the intersection of intervals of x obtained for each of these conditions.

### Exercises

Now, we know that argument of logarithmic function is a positive number and the whole logarithmic expression within the square root should be a non-negative number. This implies that we need to find the interval of x for which: $\begin{cases} \frac{6x-x^{2}}{8} > 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (1) \\ \log_{10}\frac{6x-x^{2}}{8}\geq 0 \,\,\,\,\,\,\, (2)\\ \end{cases}$

Now, we solve the first inequality (1) in the system: $\frac{6x-x^{2}}{8} > 0\Rightarrow 6x-x^{2} > 0$

Now, multiply the inequality by -1; therefore, inequality sign is reversed: $x^{2}-6x < 0$

### Exercises

Roots of corresponding quadratic equation x2 – 6x = 0 are
x = 0, 6. It means that the interval satisfying the inequality (1) of the system is : 0 < x < 6.
Now, we solve the second inequality (2) in the system: $\log_{10}\frac{6x-x^{2}}{8}\geq 0$ $\Rightarrow \frac{6x-x^{2}}{8}\geq 10^{0}\Rightarrow \frac{6x-x^{2}}{8}\geq 1$ $\Rightarrow 6x-x^{2}\geq 8 \Rightarrow 6x-x^{2}-8 \geq 0$ $\Rightarrow x^{2}-6x+8 \leq 0$

Roots of corresponding quadratic equation x2 – 6x +8 = 0 are
x = 2, 4. It means that the interval satisfying the inequality (1) of the system is : 2≤ x ≤ 4

### Exercises

Now, the interval of x valid for real values of f(x) is the one which satisfies both conditions simultaneously i.e. the interval common to two intervals determined: $\begin{cases} 0

So the domain of the given function is: $[2,4]$ and it is sketched below.

### Exercises

Exercise 5. Find the domain of the function given by $f(x)=\log_{10}\{1-\log_{10}(x^{2}-3x+12)\}$

Solution. There are two logarithmic functions composing the given function and we know that the argument of logarithmic function is a positive number. Then, the domain of the given function is the intersection of intervals of x obtained for each of the following conditions: $\begin{cases} 1-\log_{10}(x^{2}-3x+12) > 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \\ x^{2}-3x+12 > 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,(2)\\ \end{cases}$

Now, we solve the first inequality (1) in the system:

### Exercises $1-\log_{10}(x^{2}-3x+12) > 0 \Rightarrow$ $\Rightarrow \log_{10}(x^{2}-3x+12) < 1$ $\Rightarrow \log_{10}(x^{2}-3x+12) < \log_{10}10$ $\Rightarrow x^{2}-3x+12 < 10 \Rightarrow$ $\Rightarrow x^{2}-3x+2 < 0$

Roots of corresponding quadratic equation x2 – 3x + 2 = 0 are x = 1, 2. It means that interval satisfying the inequality (1) of the system is : 1 < x < 2.
Now, we solve the second inequality (2) in the system: in this case, coefficient of x2 is positive and discriminant Δ is negative. Hence, the second inequality is true for all real x.

### Exercises

Then, the domain of the given function is the intersection of the following two intervals: $]1,2[ \, \cap\, ]-\infty, +\infty[$

So the domain of the given function is: $]1,2[$ and it is  sketched below.

### Exercises

Exercise 6. Find the domain of the function given by $f(x)=\log_{2}\log_{3}\log_{4}x$

Solution. The function is formed by nesting three logarithmic functions. The base of logarithmic functions is also different. For determining domain we first find the values of x for which log4x is real, then find the values of x for which log3(log4x) is real and finally find the values of x for which f(x) is real.
The function log4x is real, if x is a positive number: x > 0
The function log3(log4x) is real, if log4x is positive: log4x > 0 $\Leftrightarrow \log_{4}x > \log_{4}1$ $\Rightarrow x > 1$

### Exercises

The function log2log3(log4x) is real, if log3log4x is positive: log3log4x > 0 $\Leftrightarrow \log_{3}\log_{4}x > \log_{3}1$ $\Rightarrow \log_{4}x > 1$ $\Leftrightarrow \log_{4}x > \log_{4}4$ $\Rightarrow x > 4$

Then, the domain of the given function is the intersection of the following three intervals: $]0,+\infty[ \, \cap\, ]1,+\infty[\, \cap\,]4, +\infty[$

### Exercises

Then domain of the given function is: $]4,+\infty[$ and it is sketched below.

Progetto "Campus Virtuale" dell'Università degli Studi di Napoli Federico II, realizzato con il cofinanziamento dell'Unione europea. Asse V - Società dell'informazione - Obiettivo Operativo 5.1 e-Government ed e-Inclusion

Fatal error: Call to undefined function federicaDebug() in /usr/local/apache/htdocs/html/footer.php on line 93